Re: [PHP-DB] Newbie Question $2

[Michael Stowe]

Can we kill this thread?  Or can you guys continue the conversation between
yourselves.  We now have 8 emails pertaining to the technical question, and
8 emails ranting about him asking it.

I understand that some people do not believe this is the appropriate forum
for that question - but personal attacks and rantings accomplish nothing
other than to provide bully tactics and express your outrage at "list
spamming" while spamming the list.

Thank you for your consideration.

- Mike


On Wed, Jun 18, 2014 at 11:49 AM, Jim Giner 
wrote:

> On 6/18/2014 2:16 PM, Aziz Saleh wrote:
>
>> On Wed, Jun 18, 2014 at 2:13 PM, Karl DeSaulniers 
>> wrote:
>>
>>
>>>
>>> Sent from losPhone
>>>
>>>  On Jun 18, 2014, at 7:56 AM, Jim Giner 

>>> wrote:
>>>

  On 6/18/2014 12:31 AM, Ethan Rosenberg, PhD wrote:
>
>> On 06/17/2014 12:02 PM, onatawah...@yahoo.ca wrote:
>> Hi Ethan,
>>
>> Here are some things to clean up your code:
>>
>> Your line:
>>
>> $phn = $_POST[phone];
>>
>> should use quotations as follows:
>>
>> $phn = $_POST['phone'];
>>
>> Your line:
>>
>> $sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
>>
>> Should use double quotes if you need the variable to be interpreted:
>>
>> $sql1 ="select Lname, Fname from Customers where Phone = $Phn ";
>>
>> Lastly, as people have mentioned PDO is probably the best way to go.
>> Try connecting to your database with PDO. Look on Google for "PDO
>> prepared statements" and use those instead of the mysql escape string
>> method.
>>
>> Hope this helps,
>>
>> -Kevin
>>
>> Sent from Yahoo Mail on Android
>>
> IT WORKS!!!
>
> Here is the code -
>
>  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
> http://www.w3.org/1999/xhtml";>
>
> 
> $bla = 1;
> ?>
>  
>  
>  
>  
>  
>  
>  
>  
>  
>  
>   error_reporting(-1);
>  require '/home/ethan/PHP/ethan.inc';
>  $db = "Store";
>  $cxn = mysqli_connect($host,$user,$password,$db);
>
>  $phn = $_POST[phone];
>  $phn = (string)$phn;
>  $dsh = '-';
>  $Phn =
>
>  $phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$
>>> phn[6].$phn[7].$phn[8].$phn[9];
>>>

>  $sql1 ="select Lname, Fname from Customers where Phone =
> '$Phn' ";
>  $result1 = mysqli_query($cxn, $sql1);
>  if(!$result)
>  {
> ?>
>  
>
>  No Match Found
>  
>  
>   }
>
> ?>
>  
>   rules="all" frame="box">
>  
>  Last Name
>  First Name
> 
>  while($row1 = mysqli_fetch_row($result1))
>  {
>
>  $Lname = $row1[0];
>  $Fname = $row1[1];
>
>
>
> ?>  
>
>
>  
> }
> ?>
>
>  >
>  
> 
>
> As you [those that replied] accurately noted, the problem was with the
> quoting.
>
> I appreciate all your comments, take them seriously and will use the
> information contained in them for future programming.
>
> No matter how much skill in programming I have, I will remain a NEWBIE;
> ie, someone who wishes to grrow in knowledge and acknowledges that
> there
> are many programmers much more skilled than I.
>
> Thanks again.
>
> Ethan
>
 happy to hear you got it working.  Sad to see that you didn't heed the

>>> tips provided to you and alter your code, and that you still have errors
>>> in
>>> it.  oh, well
>>>


>>> Wow. Just wow. I though when I signed up on this list that if I did what
>>> Ethan did I would be shunned from the list. But I guess I was wrong. You
>>> can be an ask hole on here and people will still try and help. Kudos to
>>> the
>>> good souls who try.
>>>
>>> Karl
>>> --
>>> PHP Database Mailing List (http://www.php.net/)
>>> To unsubscribe, visit: http://www.php.net/unsub.php
>>>
>>>
>>>  There are lots of people who have free time on their hands to teach the
>> basics, which I think is a good thing. Personally, if someone doesn't care
>> enough to read the manual or attempt to understand the basics, I wouldn't
>> spend too much time on their problems.
>>
>>  And despite Ethan's continual ignorance of the manual and the basic
> principles espoused by those taking

Re: [PHP-DB] Newbie Question $2

[Jim Giner]

On 6/18/2014 2:16 PM, Aziz Saleh wrote:

On Wed, Jun 18, 2014 at 2:13 PM, Karl DeSaulniers 
wrote:




Sent from losPhone


On Jun 18, 2014, at 7:56 AM, Jim Giner 

wrote:



On 6/18/2014 12:31 AM, Ethan Rosenberg, PhD wrote:

On 06/17/2014 12:02 PM, onatawah...@yahoo.ca wrote:
Hi Ethan,

Here are some things to clean up your code:

Your line:

$phn = $_POST[phone];

should use quotations as follows:

$phn = $_POST['phone'];

Your line:

$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';

Should use double quotes if you need the variable to be interpreted:

$sql1 ="select Lname, Fname from Customers where Phone = $Phn ";

Lastly, as people have mentioned PDO is probably the best way to go.
Try connecting to your database with PDO. Look on Google for "PDO
prepared statements" and use those instead of the mysql escape string
method.

Hope this helps,

-Kevin

Sent from Yahoo Mail on Android

IT WORKS!!!

Here is the code -

http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
http://www.w3.org/1999/xhtml";>



 
 
 
 
 
 
 
 
 
 

$phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];


 $sql1 ="select Lname, Fname from Customers where Phone =
'$Phn' ";
 $result1 = mysqli_query($cxn, $sql1);
 if(!$result)
 {
?>
 

 No Match Found
 
 

 
 
 
 Last Name
 First Name
  
   
   
 

   
 >
 


As you [those that replied] accurately noted, the problem was with the
quoting.

I appreciate all your comments, take them seriously and will use the
information contained in them for future programming.

No matter how much skill in programming I have, I will remain a NEWBIE;
ie, someone who wishes to grrow in knowledge and acknowledges that there
are many programmers much more skilled than I.

Thanks again.

Ethan

happy to hear you got it working.  Sad to see that you didn't heed the

tips provided to you and alter your code, and that you still have errors in
it.  oh, well




Wow. Just wow. I though when I signed up on this list that if I did what
Ethan did I would be shunned from the list. But I guess I was wrong. You
can be an ask hole on here and people will still try and help. Kudos to the
good souls who try.

Karl
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php



There are lots of people who have free time on their hands to teach the
basics, which I think is a good thing. Personally, if someone doesn't care
enough to read the manual or attempt to understand the basics, I wouldn't
spend too much time on their problems.

And despite Ethan's continual ignorance of the manual and the basic 
principles espoused by those taking the time to respond to him we still 
do it.  Aren't we all amazing?


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Re: [PHP-DB] Newbie Question $2

[Aziz Saleh]

On Wed, Jun 18, 2014 at 2:13 PM, Karl DeSaulniers 
wrote:

>
>
> Sent from losPhone
>
> > On Jun 18, 2014, at 7:56 AM, Jim Giner 
> wrote:
> >
> >> On 6/18/2014 12:31 AM, Ethan Rosenberg, PhD wrote:
> >>> On 06/17/2014 12:02 PM, onatawah...@yahoo.ca wrote:
> >>> Hi Ethan,
> >>>
> >>> Here are some things to clean up your code:
> >>>
> >>> Your line:
> >>>
> >>> $phn = $_POST[phone];
> >>>
> >>> should use quotations as follows:
> >>>
> >>> $phn = $_POST['phone'];
> >>>
> >>> Your line:
> >>>
> >>> $sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
> >>>
> >>> Should use double quotes if you need the variable to be interpreted:
> >>>
> >>> $sql1 ="select Lname, Fname from Customers where Phone = $Phn ";
> >>>
> >>> Lastly, as people have mentioned PDO is probably the best way to go.
> >>> Try connecting to your database with PDO. Look on Google for "PDO
> >>> prepared statements" and use those instead of the mysql escape string
> >>> method.
> >>>
> >>> Hope this helps,
> >>>
> >>> -Kevin
> >>>
> >>> Sent from Yahoo Mail on Android
> >> IT WORKS!!!
> >>
> >> Here is the code -
> >>
> >>  >> "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
> >> http://www.w3.org/1999/xhtml";>
> >>
> >> 
> >>  >>   $bla = 1;
> >> ?>
> >> 
> >> 
> >> 
> >> 
> >> 
> >> 
> >> 
> >> 
> >> 
> >> 
> >>  >> error_reporting(-1);
> >> require '/home/ethan/PHP/ethan.inc';
> >> $db = "Store";
> >> $cxn = mysqli_connect($host,$user,$password,$db);
> >>
> >> $phn = $_POST[phone];
> >> $phn = (string)$phn;
> >> $dsh = '-';
> >> $Phn =
> >>
> $phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];
> >>
> >> $sql1 ="select Lname, Fname from Customers where Phone =
> >> '$Phn' ";
> >> $result1 = mysqli_query($cxn, $sql1);
> >> if(!$result)
> >> {
> >> ?>
> >> 
> >>
> >> No Match Found
> >> 
> >> 
> >>  >> }
> >>
> >> ?>
> >> 
> >>  >> rules="all" frame="box">
> >> 
> >> Last Name
> >> First Name
> >>  >>
> >> while($row1 = mysqli_fetch_row($result1))
> >> {
> >>
> >> $Lname = $row1[0];
> >> $Fname = $row1[1];
> >>
> >>
> >>
> >> ?>  
> >>   
> >>   
> >> 
> >>  >>   }
> >> ?>
> >>   
> >> >
> >> 
> >> 
> >>
> >> As you [those that replied] accurately noted, the problem was with the
> >> quoting.
> >>
> >> I appreciate all your comments, take them seriously and will use the
> >> information contained in them for future programming.
> >>
> >> No matter how much skill in programming I have, I will remain a NEWBIE;
> >> ie, someone who wishes to grrow in knowledge and acknowledges that there
> >> are many programmers much more skilled than I.
> >>
> >> Thanks again.
> >>
> >> Ethan
> > happy to hear you got it working.  Sad to see that you didn't heed the
> tips provided to you and alter your code, and that you still have errors in
> it.  oh, well
> >
>
> Wow. Just wow. I though when I signed up on this list that if I did what
> Ethan did I would be shunned from the list. But I guess I was wrong. You
> can be an ask hole on here and people will still try and help. Kudos to the
> good souls who try.
>
> Karl
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
There are lots of people who have free time on their hands to teach the
basics, which I think is a good thing. Personally, if someone doesn't care
enough to read the manual or attempt to understand the basics, I wouldn't
spend too much time on their problems.

Re: [PHP-DB] Newbie Question $2

[Karl DeSaulniers]

Sent from losPhone

> On Jun 18, 2014, at 7:56 AM, Jim Giner  wrote:
> 
>> On 6/18/2014 12:31 AM, Ethan Rosenberg, PhD wrote:
>>> On 06/17/2014 12:02 PM, onatawah...@yahoo.ca wrote:
>>> Hi Ethan,
>>> 
>>> Here are some things to clean up your code:
>>> 
>>> Your line:
>>> 
>>> $phn = $_POST[phone];
>>> 
>>> should use quotations as follows:
>>> 
>>> $phn = $_POST['phone'];
>>> 
>>> Your line:
>>> 
>>> $sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
>>> 
>>> Should use double quotes if you need the variable to be interpreted:
>>> 
>>> $sql1 ="select Lname, Fname from Customers where Phone = $Phn ";
>>> 
>>> Lastly, as people have mentioned PDO is probably the best way to go.
>>> Try connecting to your database with PDO. Look on Google for "PDO
>>> prepared statements" and use those instead of the mysql escape string
>>> method.
>>> 
>>> Hope this helps,
>>> 
>>> -Kevin
>>> 
>>> Sent from Yahoo Mail on Android
>> IT WORKS!!!
>> 
>> Here is the code -
>> 
>> > "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
>> http://www.w3.org/1999/xhtml";>
>> 
>> 
>> >   $bla = 1;
>> ?>
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> 
>> > error_reporting(-1);
>> require '/home/ethan/PHP/ethan.inc';
>> $db = "Store";
>> $cxn = mysqli_connect($host,$user,$password,$db);
>> 
>> $phn = $_POST[phone];
>> $phn = (string)$phn;
>> $dsh = '-';
>> $Phn =
>> $phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];
>> 
>> $sql1 ="select Lname, Fname from Customers where Phone =
>> '$Phn' ";
>> $result1 = mysqli_query($cxn, $sql1);
>> if(!$result)
>> {
>> ?>
>> 
>> 
>> No Match Found
>> 
>> 
>> > }
>> 
>> ?>
>> 
>> > rules="all" frame="box">
>> 
>> Last Name
>> First Name
>> > 
>> while($row1 = mysqli_fetch_row($result1))
>> {
>> 
>> $Lname = $row1[0];
>> $Fname = $row1[1];
>> 
>> 
>> 
>> ?>  
>>   
>>   
>> 
>> >   }
>> ?>
>>   
>> >
>> 
>> 
>> 
>> As you [those that replied] accurately noted, the problem was with the
>> quoting.
>> 
>> I appreciate all your comments, take them seriously and will use the
>> information contained in them for future programming.
>> 
>> No matter how much skill in programming I have, I will remain a NEWBIE;
>> ie, someone who wishes to grrow in knowledge and acknowledges that there
>> are many programmers much more skilled than I.
>> 
>> Thanks again.
>> 
>> Ethan
> happy to hear you got it working.  Sad to see that you didn't heed the tips 
> provided to you and alter your code, and that you still have errors in it.  
> oh, well
> 

Wow. Just wow. I though when I signed up on this list that if I did what Ethan 
did I would be shunned from the list. But I guess I was wrong. You can be an 
ask hole on here and people will still try and help. Kudos to the good souls 
who try. 

Karl
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php

Re: [PHP-DB] Newbie Question $2

[Jim Giner]

On 6/18/2014 12:31 AM, Ethan Rosenberg, PhD wrote:

On 06/17/2014 12:02 PM, onatawah...@yahoo.ca wrote:

Hi Ethan,

Here are some things to clean up your code:

Your line:

$phn = $_POST[phone];

should use quotations as follows:

$phn = $_POST['phone'];

Your line:

$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';

Should use double quotes if you need the variable to be interpreted:

$sql1 ="select Lname, Fname from Customers where Phone = $Phn ";

Lastly, as people have mentioned PDO is probably the best way to go.
Try connecting to your database with PDO. Look on Google for "PDO
prepared statements" and use those instead of the mysql escape string
method.

Hope this helps,

-Kevin

Sent from Yahoo Mail on Android



IT WORKS!!!

Here is the code -

http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>
http://www.w3.org/1999/xhtml";>



 
 
 
 
 
 
 
 
 
 

 

 No Match Found
 
 

 
 
 
 Last Name
 First Name
  
   
   
 

   
 >
 


As you [those that replied] accurately noted, the problem was with the
quoting.

I appreciate all your comments, take them seriously and will use the
information contained in them for future programming.

No matter how much skill in programming I have, I will remain a NEWBIE;
ie, someone who wishes to grrow in knowledge and acknowledges that there
are many programmers much more skilled than I.

Thanks again.

Ethan

happy to hear you got it working.  Sad to see that you didn't heed the 
tips provided to you and alter your code, and that you still have errors 
in it.  oh, well


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To unsubscribe, visit: http://www.php.net/unsub.php

Re: [PHP-DB] Newbie Question $2

[Ethan Rosenberg, PhD]

On 06/17/2014 12:02 PM, onatawah...@yahoo.ca wrote:

Hi Ethan,

Here are some things to clean up your code:

Your line:

$phn = $_POST[phone];

should use quotations as follows:

$phn = $_POST['phone'];

Your line:

$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';

Should use double quotes if you need the variable to be interpreted:

$sql1 ="select Lname, Fname from Customers where Phone = $Phn ";

Lastly, as people have mentioned PDO is probably the best way to go. Try connecting to 
your database with PDO. Look on Google for "PDO prepared statements" and use 
those instead of the mysql escape string method.

Hope this helps,

-Kevin

Sent from Yahoo Mail on Android



IT WORKS!!!

Here is the code -

"http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd";>

http://www.w3.org/1999/xhtml";>













			$Phn = 
$phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];

$sql1 ="select Lname, Fname from Customers where Phone = 
'$Phn' ";
$result1 = mysqli_query($cxn, $sql1);
if(!$result)
{
?>   


No Match Found




			frame="box">


Last Name
First Name
  
  
  


  
>



As you [those that replied] accurately noted, the problem was with the 
quoting.


I appreciate all your comments, take them seriously and will use the 
information contained in them for future programming.


No matter how much skill in programming I have, I will remain a NEWBIE; 
ie, someone who wishes to grrow in knowledge and acknowledges that there 
are many programmers much more skilled than I.


Thanks again.

Ethan


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To unsubscribe, visit: http://www.php.net/unsub.php

Re: [PHP-DB] Newbie Question $2

[onatawah...@yahoo.ca]

Hi Ethan,

Here are some things to clean up your code:

Your line: 

$phn = $_POST[phone]; 

should use quotations as follows:

$phn = $_POST['phone'];

Your line:

$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';

Should use double quotes if you need the variable to be interpreted:

$sql1 ="select Lname, Fname from Customers where Phone = $Phn ";

Lastly, as people have mentioned PDO is probably the best way to go. Try 
connecting to your database with PDO. Look on Google for "PDO prepared 
statements" and use those instead of the mysql escape string method.

Hope this helps,

-Kevin

Sent from Yahoo Mail on Android

Re: [PHP-DB] Newbie Question $2

[Mike Stowe]

Oh a few quick things. 

First, you can use substr to break up the phone instead of grabbing characters- 
might be a little easier to read long term. 

Secondly, mysql_real_escape_string will return the cleaned string, but doesn't 
change the original variable. So you'll need $phn = 
mysql_real_escape_string($phn);

Thirdly anytime you use a single quote the strong is interpreted literally. 
You'll want to switch out the single quotes with double quotes, and then wrap 
$phn in single quotes in order to not break your query. 

"Select ... Where phn = '$phn'"

I'd also really suggest looking at using PDO or even the mysqli extension tho 
instead of just plain mysql (believe this has been deprecated). 

Sorry for the quick reply, on mobile. But feel free to email me directly and 
I'll be happy to help out more. 

- Mike

Sent from my iPhone

> On Jun 16, 2014, at 7:58 PM, Ethan Rosenberg 
>  wrote:
> 
> Dear List -
> 
> I have the following code:
> 
> The input from the form is a 10 digit string [1234567890] which is converted 
> to phone number format [123-456-7890]
> 
> $phn = $_POST[phone];
> $phn = (string)$phn;
> $dsh = '-';
> $Phn = 
> $phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];
>  
>echo $Phn; // this is folded by Thunderbird.  In the script, it is //all 
> on one line
> 
>mysql_real_escape_string($Phn);
>$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
>echo $sql1; //this always shows $phn as Phn and not as a numerical 
> //string.
>$result1 = mysqli_query($cxn, $sql1);
> 
> TIA
> 
> Ethan
> 
> -- 
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 

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Re: [PHP-DB] Newbie Question $2

[Karl DeSaulniers]

On Jun 16, 2014, at 10:05 PM, Karl DeSaulniers  wrote:

> On Jun 16, 2014, at 9:58 PM, Ethan Rosenberg 
>  wrote:
> 
>> Dear List -
>> 
>> I have the following code:
>> 
>> The input from the form is a 10 digit string [1234567890] which is converted 
>> to phone number format [123-456-7890]
>> 
>> $phn = $_POST[phone];
>> $phn = (string)$phn;
>> $dsh = '-';
>> $Phn = 
>> $phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];
>>  
>>   echo $Phn; // this is folded by Thunderbird.  In the script, it is //all 
>> on one line
>> 
>>   mysql_real_escape_string($Phn);
>>   $sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
>>   echo $sql1; //this always shows $phn as Phn and not as a numerical 
>> //string.
>>   $result1 = mysqli_query($cxn, $sql1);
>> 
>> TIA
>> 
>> Ethan
>> 
> 
> Well, from first glance you're combining mysql and mysqli. 
> Don't know if that is wise or permissible since I think mysql has been 
> depreciated. 
> Go with mysqli. Next you may want to try...
> 
> $sql1 = 'SELECT Lname, Fname FROM Customers WHERE Phone = '.$Phn;
> 
> Best,
> 
> Karl DeSaulniers
> Design Drumm
> http://designdrumm.com
> 

Also, you may want to store the number in your database without the dash and 
just apply the dash when displaying the number in HTML.
Not that this is entirely necessary, more of a personal choice. 
If you have a large number of phone numbers stored lets say, 
numbers with no dash take up less space in the grand scheme of things I guess.

Best,

Karl DeSaulniers
Design Drumm
http://designdrumm.com
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Re: [PHP-DB] Newbie Question $2

[Aziz Saleh]

On Mon, Jun 16, 2014 at 10:58 PM, Ethan Rosenberg <
erosenb...@hygeiabiomedical.com> wrote:

> Dear List -
>
> I have the following code:
>
> The input from the form is a 10 digit string [1234567890] which is
> converted to phone number format [123-456-7890]
>
> $phn = $_POST[phone];
>  $phn = (string)$phn;
>  $dsh = '-';
>  $Phn = $phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$
> phn[6].$phn[7].$phn[8].$phn[9];
> echo $Phn; // this is folded by Thunderbird.  In the script, it is
> //all on one line
>
> mysql_real_escape_string($Phn);
> $sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
> echo $sql1; //this always shows $phn as Phn and not as a numerical
> //string.
> $result1 = mysqli_query($cxn, $sql1);
>
> TIA
>
> Ethan
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>
>
This page should help you:
http://www.php.net//manual/en/language.types.string.php understand the
difference between single and double quotes.

Re: [PHP-DB] Newbie Question $2

[Karl DeSaulniers]

On Jun 16, 2014, at 9:58 PM, Ethan Rosenberg  
wrote:

> Dear List -
> 
> I have the following code:
> 
> The input from the form is a 10 digit string [1234567890] which is converted 
> to phone number format [123-456-7890]
> 
> $phn = $_POST[phone];
> $phn = (string)$phn;
> $dsh = '-';
> $Phn = 
> $phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9];
>  
>echo $Phn; // this is folded by Thunderbird.  In the script, it is //all 
> on one line
> 
>mysql_real_escape_string($Phn);
>$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
>echo $sql1; //this always shows $phn as Phn and not as a numerical 
> //string.
>$result1 = mysqli_query($cxn, $sql1);
> 
> TIA
> 
> Ethan
> 

Well, from first glance you're combining mysql and mysqli. 
Don't know if that is wise or permissible since I think mysql has been 
depreciated. 
Go with mysqli. Next you may want to try...

$sql1 = 'SELECT Lname, Fname FROM Customers WHERE Phone = '.$Phn;

Best,

Karl DeSaulniers
Design Drumm
http://designdrumm.com
> 


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[PHP-DB] Newbie Question $2

[Ethan Rosenberg]

Dear List -

I have the following code:

The input from the form is a 10 digit string [1234567890] which is 
converted to phone number format [123-456-7890]


$phn = $_POST[phone];
 $phn = (string)$phn;
 $dsh = '-';
 $Phn = 
$phn[0].$phn[1].$phn[2].$dsh.$phn[3].$phn[4].$phn[5].$dsh.$phn[6].$phn[7].$phn[8].$phn[9]; 

echo $Phn; // this is folded by Thunderbird.  In the script, it is 
//all on one line


mysql_real_escape_string($Phn);
$sql1 ='select Lname, Fname from Customers where Phone = $Phn ';
echo $sql1; //this always shows $phn as Phn and not as a numerical 
//string.

$result1 = mysqli_query($cxn, $sql1);

TIA

Ethan

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[PHP-DB] Newbie question on MDB2 error handling

[P.A.]

Hello all,

I am quite new to php and especially MDB2 package. Despite some thorough  
research I could not find an answer yet.
When I try to connect to a database using MDB2::connect method, and the  
database server is not running, I only get an
error code of -1 and a message saying "MDB2 Error: unknown error". Looking  
at MDB2 source it rather should say "connect failed" with a code of -24  
(MDB2_ERROR_CONNECT_FAILED). Is my expectation wrong, or how can I get  
some meaningful message and error code in this situation?


The code I use is:

$dsn = "mysqli://mysql:my...@127.0.0.1/MYDB"; // This works when mysql  
server is running

$con = & MDB2::connect($dsn);

if(MDB2::isError($con))
{
print_r($con->getCode());
print_r($con->getMessage());
}

Any hint is warmly appreciated!

Thanks,

Pascal.

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[PHP-DB] newbie question on PHP & Mysql...

[Evert Meulie]

Hi all!

I'm taking my first steps with PHP & MySQL.

Can anyone give me a hint on why this would not work?

*

$result = mysql_query('SELECT SUM(AcctInputOctets), SUM(AcctOutputOctets)  FROM 
radacct WHERE username = $argv[1] ');
echo mysql_result($result,0), "\n";
echo mysql_result($result,0,1);

*


I get: Warning: mysql_result(): supplied argument is not a valid MySQL result 
resource



Regards,
Evert

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Re: [PHP-DB] Newbie Question - mysql_select_db fails with access denied.

[Christian Schlaefcke]

When I switch back to user root with empty password and use
mysql_connect("localhost", "root") it works.

I just found out when I set the permission for "myuser" to "localhost"
instead of "any" it works also for "myuser".

Now I have to check out why I canÂt use passwords. Maybe the PHP Version
IÂm running is not ready to work with MySQL 4.1.4? Any Ideas?

Regards,

Chris

Am Fr, den 24.09.2004 um 12:04 Uhr -0300 schrieb Pablo M. Rivas:
> You're right:
> if you use mysql_connect("localhost", "root","");
> you''ll get "acces denied for user [EMAIL PROTECTED]
> 
> if you use mysql_connect("localhost");
> you'll get "Access denied for user @localhost"
> I'll google for a while... but.. to solve YOUR problem, just use:
> mysql_connect("localhost","user","");
> 
> Good luck!
> 
> 
> On Fri, 24 Sep 2004 14:56:35 +0200, Christian Schlaefcke
> <[EMAIL PROTECTED]> wrote:
> > I have chosen an empty password. I know that this is a security problem.
> > IÂm using it for testing at the moment only. I use it because there also
> > seems to be an issue with php connecting to mysql servers with verions >
> > 4.1.
> > 
> > When trying to connect with password I get this:
> > Connect failed : Client does not support authentication protocol
> > requested by server; consider upgrading MySQL client.
> > 
> > Because I want to track down issue after issue and not anything at once
> > I decided to find out why mysql_select_db fails first.
> > 
> > What really makes me wonder in my tiny example is that the error message
> > does not say something like:
> > 
> > SELECT_DB failed: Access denied for user 'myuser'@'localhost' to
> > database 'mydb'
> > 
> > instead it says:
> > SELECT_DB failed: Access denied for user ''@'localhost' to database
> > 'mydb'
> > 
> > So I think that the username is not interpreted at all. Why not?
> > 
> > Regards,
> > 
> > Chris
> > 
> > Am Fr, den 24.09.2004 um 13:43 Uhr +0100 schrieb Peter Lovatt:
> > 
> > 
> > > Hi
> > >
> > > try $db = mysql_connect("localhost", "myuser","mypassword") - without it  you 
> > > are not  submitting a password
> > >
> > > Peter
> > >
> > > > -Original Message-
> > > > From: Christian Schlaefcke [mailto:[EMAIL PROTECTED]
> > > > Sent: 24 September 2004 13:31
> > > > To: [EMAIL PROTECTED]
> > > > Subject: [PHP-DB] Newbie Question - mysql_select_db fails with access
> > > > denied.
> > > >
> > > >
> > > > Hi Folks,
> > > >
> > > > I want to do something like the following:
> > > > $db = mysql_connect("localhost", "myuser") or die("Connect failed : " .
> > > > mysql_error());
> > > >
> > > > mysql_select_db("mydb",$db) or die("SELECT_DB failed : " . mysql_error
> > > > ());
> > > >
> > > > $result = mysql_query("SELECT * FROM mytable", $db) or die("Query
> > > > failed : " . mysql_error());
> > > >
> > > > The script does not proceed beyond the "mysql_select_db" line. I get
> > > > this error:
> > > > SELECT_DB failed: Access denied for user ''@'localhost' to database
> > > > 'mydb'
> > > >
> > > > What happened to the user information. I have granted all neccessary
> > > > rights to "myuser". But it seems that itÂs even not caring about the
> > > > user.
> > > >
> > > > I have MySQL 4.1.4 and PHP 4.3.8-2.1 on a Fedore Core 2 System  running.
> > > >
> > > > Thanks & Regards,
> > > >
> > > > Chris
> > > >
> > > > --
> > > > PHP Database Mailing List (http://www.php.net/)
> > > > To unsubscribe, visit: http://www.php.net/unsub.php
> > > >
> > >
> > >
> > 
> > --
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> > 
> > 
> 
> 
> 
> -- 
> Pablo M. Rivas. http://pmrivas.ipupdater.com http://www.r3soft.com.ar
> ---
> 

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Re: [PHP-DB] Newbie Question - mysql_select_db fails with access denied.

[Pablo M. Rivas]

You're right:
if you use mysql_connect("localhost", "root","");
you''ll get "acces denied for user [EMAIL PROTECTED]

if you use mysql_connect("localhost");
you'll get "Access denied for user @localhost"
I'll google for a while... but.. to solve YOUR problem, just use:
mysql_connect("localhost","user","");

Good luck!


On Fri, 24 Sep 2004 14:56:35 +0200, Christian Schlaefcke
<[EMAIL PROTECTED]> wrote:
> I have chosen an empty password. I know that this is a security problem.
> I´m using it for testing at the moment only. I use it because there also
> seems to be an issue with php connecting to mysql servers with verions >
> 4.1.
> 
> When trying to connect with password I get this:
> Connect failed : Client does not support authentication protocol
> requested by server; consider upgrading MySQL client.
> 
> Because I want to track down issue after issue and not anything at once
> I decided to find out why mysql_select_db fails first.
> 
> What really makes me wonder in my tiny example is that the error message
> does not say something like:
> 
> SELECT_DB failed: Access denied for user 'myuser'@'localhost' to
> database 'mydb'
> 
> instead it says:
> SELECT_DB failed: Access denied for user ''@'localhost' to database
> 'mydb'
> 
> So I think that the username is not interpreted at all. Why not?
> 
> Regards,
> 
> Chris
> 
> Am Fr, den 24.09.2004 um 13:43 Uhr +0100 schrieb Peter Lovatt:
> 
> 
> > Hi
> >
> > try $db = mysql_connect("localhost", "myuser","mypassword") - without it  you are 
> > not  submitting a password
> >
> > Peter
> >
> > > -Original Message-
> > > From: Christian Schlaefcke [mailto:[EMAIL PROTECTED]
> > > Sent: 24 September 2004 13:31
> > > To: [EMAIL PROTECTED]
> > > Subject: [PHP-DB] Newbie Question - mysql_select_db fails with access
> > > denied.
> > >
> > >
> > > Hi Folks,
> > >
> > > I want to do something like the following:
> > > $db = mysql_connect("localhost", "myuser") or die("Connect failed : " .
> > > mysql_error());
> > >
> > > mysql_select_db("mydb",$db) or die("SELECT_DB failed : " . mysql_error
> > > ());
> > >
> > > $result = mysql_query("SELECT * FROM mytable", $db) or die("Query
> > > failed : " . mysql_error());
> > >
> > > The script does not proceed beyond the "mysql_select_db" line. I get
> > > this error:
> > > SELECT_DB failed: Access denied for user ''@'localhost' to database
> > > 'mydb'
> > >
> > > What happened to the user information. I have granted all neccessary
> > > rights to "myuser". But it seems that it´s even not caring about the
> > > user.
> > >
> > > I have MySQL 4.1.4 and PHP 4.3.8-2.1 on a Fedore Core 2 System  running.
> > >
> > > Thanks & Regards,
> > >
> > > Chris
> > >
> > > --
> > > PHP Database Mailing List (http://www.php.net/)
> > > To unsubscribe, visit: http://www.php.net/unsub.php
> > >
> >
> >
> 
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 
> 



-- 
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RE: [PHP-DB] Newbie Question - mysql_select_db fails with access denied.

[Christian Schlaefcke]

I have chosen an empty password. I know that this is a security problem.
IÂm using it for testing at the moment only. I use it because there also
seems to be an issue with php connecting to mysql servers with verions >
4.1. 

When trying to connect with password I get this:
Connect failed : Client does not support authentication protocol
requested by server; consider upgrading MySQL client.

Because I want to track down issue after issue and not anything at once
I decided to find out why mysql_select_db fails first.

What really makes me wonder in my tiny example is that the error message
does not say something like:

SELECT_DB failed: Access denied for user 'myuser'@'localhost' to
database 'mydb'

instead it says:
SELECT_DB failed: Access denied for user ''@'localhost' to database
'mydb'

So I think that the username is not interpreted at all. Why not?

Regards,

Chris

Am Fr, den 24.09.2004 um 13:43 Uhr +0100 schrieb Peter Lovatt:
> Hi
> 
> try $db = mysql_connect("localhost", "myuser","mypassword") - without it  you are 
> not  submitting a password 
> 
> Peter
> 
> > -Original Message-
> > From: Christian Schlaefcke [mailto:[EMAIL PROTECTED]
> > Sent: 24 September 2004 13:31
> > To: [EMAIL PROTECTED]
> > Subject: [PHP-DB] Newbie Question - mysql_select_db fails with access
> > denied.
> > 
> > 
> > Hi Folks,
> > 
> > I want to do something like the following:
> > $db = mysql_connect("localhost", "myuser") or die("Connect failed : " .
> > mysql_error());
> > 
> > mysql_select_db("mydb",$db) or die("SELECT_DB failed : " . mysql_error
> > ());
> > 
> > $result = mysql_query("SELECT * FROM mytable", $db) or die("Query
> > failed : " . mysql_error());
> > 
> > The script does not proceed beyond the "mysql_select_db" line. I get
> > this error:
> > SELECT_DB failed: Access denied for user ''@'localhost' to database
> > 'mydb'
> > 
> > What happened to the user information. I have granted all neccessary
> > rights to "myuser". But it seems that itÂs even not caring about the
> > user.
> > 
> > I have MySQL 4.1.4 and PHP 4.3.8-2.1 on a Fedore Core 2 System  running.
> > 
> > Thanks & Regards,
> > 
> > Chris
> > 
> > -- 
> > PHP Database Mailing List (http://www.php.net/)
> > To unsubscribe, visit: http://www.php.net/unsub.php
> > 
> 
> 

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RE: [PHP-DB] Newbie Question - mysql_select_db fails with access denied.

[Peter Lovatt]

Hi

try $db = mysql_connect("localhost", "myuser","mypassword") - without it  you are not  
submitting a password 

Peter

> -Original Message-
> From: Christian Schlaefcke [mailto:[EMAIL PROTECTED]
> Sent: 24 September 2004 13:31
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Newbie Question - mysql_select_db fails with access
> denied.
> 
> 
> Hi Folks,
> 
> I want to do something like the following:
> $db = mysql_connect("localhost", "myuser") or die("Connect failed : " .
> mysql_error());
> 
> mysql_select_db("mydb",$db) or die("SELECT_DB failed : " . mysql_error
> ());
> 
> $result = mysql_query("SELECT * FROM mytable", $db) or die("Query
> failed : " . mysql_error());
> 
> The script does not proceed beyond the "mysql_select_db" line. I get
> this error:
> SELECT_DB failed: Access denied for user ''@'localhost' to database
> 'mydb'
> 
> What happened to the user information. I have granted all neccessary
> rights to "myuser". But it seems that itÂs even not caring about the
> user.
> 
> I have MySQL 4.1.4 and PHP 4.3.8-2.1 on a Fedore Core 2 System  running.
> 
> Thanks & Regards,
> 
> Chris
> 
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> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 

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Re: [PHP-DB] Newbie Question - mysql_select_db fails with access denied.

[John]

This might be a stupid idea, but did you remember to issue the Flush Privelages; 
command to the mysql database after granting your user all the rights?
 
--John


Christian Schlaefcke <[EMAIL PROTECTED]> wrote:
Hi Folks,

I want to do something like the following:
$db = mysql_connect("localhost", "myuser") or die("Connect failed : " .
mysql_error());

mysql_select_db("mydb",$db) or die("SELECT_DB failed : " . mysql_error
());

$result = mysql_query("SELECT * FROM mytable", $db) or die("Query
failed : " . mysql_error());

The script does not proceed beyond the "mysql_select_db" line. I get
this error:
SELECT_DB failed: Access denied for user ''@'localhost' to database
'mydb'

What happened to the user information. I have granted all neccessary
rights to "myuser". But it seems that it´s even not caring about the
user.

I have MySQL 4.1.4 and PHP 4.3.8-2.1 on a Fedore Core 2 System running.

Thanks & Regards,

Chris

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[PHP-DB] Newbie Question - mysql_select_db fails with access denied.

[Christian Schlaefcke]

Hi Folks,

I want to do something like the following:
$db = mysql_connect("localhost", "myuser") or die("Connect failed : " .
mysql_error());

mysql_select_db("mydb",$db) or die("SELECT_DB failed : " . mysql_error
());

$result = mysql_query("SELECT * FROM mytable", $db) or die("Query
failed : " . mysql_error());

The script does not proceed beyond the "mysql_select_db" line. I get
this error:
SELECT_DB failed: Access denied for user ''@'localhost' to database
'mydb'

What happened to the user information. I have granted all neccessary
rights to "myuser". But it seems that itÂs even not caring about the
user.

I have MySQL 4.1.4 and PHP 4.3.8-2.1 on a Fedore Core 2 System  running.

Thanks & Regards,

Chris

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RE: [PHP-DB] Newbie question

[Hutchins, Richard]

Try:





> -Original Message-
> From: Gajo Csaba [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, March 24, 2004 4:10 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Newbie question
> 
> 
> Hi!
> I'm new to PHP, so this may be an easy question. I tried to make a 
> simple form-page, where the user enters his name, and on the 
> following 
> page it says "Hello $name". What happens is that on the 
> second page it 
> says soemthing like "unknown identifier $name".
> 
> I'm using Apache 2 so that may be causing the problem, cause 
> this is a 
> clear textbook example:
> 
> -- first.php --
> 
> 
> First
> 
>  vlink="#808040"
> alink="#808040">
> 
> Whats your name?
>  value="">
> 
> 
> 
> 
> --- second.php 
> 
> 
> Second
> 
>  vlink="#808040"
> alink="#808040">
>print "Hello $name!";
> ?>
> 
> 
> 
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> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
> 
> 

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[PHP-DB] Newbie question

[Gajo Csaba]

Hi!
I'm new to PHP, so this may be an easy question. I tried to make a 
simple form-page, where the user enters his name, and on the following 
page it says "Hello $name". What happens is that on the second page it 
says soemthing like "unknown identifier $name".

I'm using Apache 2 so that may be causing the problem, cause this is a 
clear textbook example:

-- first.php --


First



Whats your name?





--- second.php 


Second






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Re: [PHP-DB] newbie question

[Peter Beckman]

On Fri, 13 Feb 2004, t wrote:

> hi,
>
> i am relatively new to php and new to this email list.
>
> i have what i think are fairly simple questions about using mysql and
> php. i have done some research and can't seem to find the answer i
> need.
>
> # 1. i want to set the date format to display dates in a format other
> than the standard mysql -mm-dd format. i have tried using the mysql
> DATE_FORMAT but i can's seem to get it to work...
>
> ideally i'd like to display dates as 2 digit date followed by three
> letter month abbreviation  and leave the year off completely...
>
> example:   13 feb

 Use date()
 Documentation:
 http://php.net/date

> # 2. i want to hide entries that are newer than the current date AND
> hide entries older than 365 days.

 Limit your SQL to

 ... where date>now() and dateunix_timestamp() and date<(unix_timestamp()-(365*86400)) ...

 Documentation:
 http://mysql.com/date_sub (should redirect to
 http://www.mysql.com/doc/en/Date_and_time_functions.html)


---
Peter Beckman  Internet Guy
[EMAIL PROTECTED] http://www.purplecow.com/
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[PHP-DB] newbie question

[t]

hi,

i am relatively new to php and new to this email list.

i have what i think are fairly simple questions about using mysql and 
php. i have done some research and can't seem to find the answer i 
need.

# 1. i want to set the date format to display dates in a format other 
than the standard mysql -mm-dd format. i have tried using the mysql 
DATE_FORMAT but i can's seem to get it to work...

ideally i'd like to display dates as 2 digit date followed by three 
letter month abbreviation  and leave the year off completely...

example:   13 feb



# 2. i want to hide entries that are newer than the current date AND 
hide entries older than 365 days.

i apologize if this is the wrong place to ask these questions. if 
someone could point me in the right direction i would appreciate it.



here is the url and code (minus the login info)

http://www.broadcastatic.com



$db = mysql_connect("localhost", "[loginnamehere]", "[passwdhere]");

mysql_select_db("[databasenamehere]",$db);

// display individual record

if ($date) {

   $result = mysql_query("SELECT * FROM [tablenamehere] WHERE date= 
'$date' ",$db);

   $myrow = mysql_fetch_array($result);

// do not use now   printf("\n", $myrow["date"]);

   printf("hosted by %s\n", $myrow["dj"]);

   printf("broadcast on %s\n", $myrow["date"]);

   printf("%s\n", $myrow["location"]);

   printf("--> %s\n", $myrow["entry"]);

   printf("mp3 link\n", 
$myrow["date"]);

   include("archive/index/$date.php");

} else {

// display list of shows by date

   $result = mysql_query("SELECT * FROM [tablenamehere] WHERE 1 ORDER 
BY 'date' DESC",$db);

if ($myrow = mysql_fetch_array($result)) {

  // display list if there are records to display

  do {

printf("%s 

   %s%s\n",
   $PHP_SELF, $myrow["date"], $myrow["date"], 
$myrow["date"], $myrow["dj"], $myrow["entry"]);

  } while ($myrow = mysql_fetch_array($result));

} else {

  // no records to display

  echo "Sorry, no records were found!";

}

}



?>



THANKS!

tommy birchett
http://www.broadcastatic.com
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Re: [PHP-DB] newbie question on data accumulation

[CPT John W. Holmes]

From: "Joffrey Leevy" <[EMAIL PROTECTED]>

> Curious as to what happens after data is repeatedly
> selected from a mysql table overtime.  Does it
> accumulate as junk data, stored at some location and
> eventually slow down the database/program/server?
> Does any purging have to take place?

I would imagine it would be thrown away when ever the connection was closed
or the result set was retrieved entirely. Why not ask this on a MySQL list?
You'd probably get much better answers.

---John Holmes...

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[PHP-DB] newbie question on data accumulation

[Joffrey Leevy]

Hi all:

Curious as to what happens after data is repeatedly
selected from a mysql table overtime.  Does it
accumulate as junk data, stored at some location and
eventually slow down the database/program/server? 
Does any purging have to take place?

Thanks
J

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RE: [PHP-DB] Newbie Question - PHP and MSACCESS

[Gary . Every]

//Place your request into a variable:
$result = odbc_do($connection,"select * from events where id=$id");
$var = odbc_fetch_row($result);

// Now you have an array to work on
foreach($var as $line){
//Do your formatting thusly:
echo '
' . $line[0] . '
' . $line[1] . '
...

}


Hi,
 
Am using php to open a customer's msaccess file.
 
First time for everything.  It's a very small database of less then 25
rows, and will always be this small.
 
Am needing to return all the rows to a web page using php at all times.
Very simplistic.
 
Am apparently set up correctly on the DSN, and am able to do an
odbc_connect:
 
$connection = odbc_connect("meetings","","");
 
Am able to fetch rows, or so it seems:
 
$result = odbc_do($connection,"select * from events where id=$id");
odbc_fetch_row($result);
 
Am not understanding how to "print()" the fields in the rows, one after
the next, on the same line, and then do the same with the next record,
until all records have been presented.
 
Sorry to be lame.  Any help would be gratefully received.
 
 
-Joe

RE: [PHP-DB] Newbie Question - PHP and MSACCESS

[Dennis Cole]

$cnx = odbc_connect( 'DSN' , 'USER', 'PASS' );
if (!$cnx) {
?>Could not connect to ODBCmailto:[EMAIL PROTECTED]]
Sent: Sunday, February 16, 2003 4:01 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Newbie Question - PHP and MSACCESS


Hi,

Am using php to open a customer's msaccess file.

First time for everything.  It's a very small database of less then 25
rows, and will always be this small.

Am needing to return all the rows to a web page using php at all times.
Very simplistic.

Am apparently set up correctly on the DSN, and am able to do an
odbc_connect:

$connection = odbc_connect("meetings","","");

Am able to fetch rows, or so it seems:

$result = odbc_do($connection,"select * from events where id=$id");
odbc_fetch_row($result);

Am not understanding how to "print()" the fields in the rows, one after
the next, on the same line, and then do the same with the next record,
until all records have been presented.

Sorry to be lame.  Any help would be gratefully received.


-Joe


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[PHP-DB] Newbie Question - PHP and MSACCESS

[J.M. Cocchini]

Hi,
 
Am using php to open a customer's msaccess file.
 
First time for everything.  It's a very small database of less then 25
rows, and will always be this small.
 
Am needing to return all the rows to a web page using php at all times.
Very simplistic.
 
Am apparently set up correctly on the DSN, and am able to do an
odbc_connect:
 
$connection = odbc_connect("meetings","","");
 
Am able to fetch rows, or so it seems:
 
$result = odbc_do($connection,"select * from events where id=$id");
odbc_fetch_row($result);
 
Am not understanding how to "print()" the fields in the rows, one after
the next, on the same line, and then do the same with the next record,
until all records have been presented.
 
Sorry to be lame.  Any help would be gratefully received.
 
 
-Joe

RE: [PHP-DB] Newbie-Question: What's that @?

[Marcus Fleige]

great!

@ryan & josh: thank you for the quick answers! :-)

marcus


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RE: [PHP-DB] Newbie-Question: What's that @?

[Josh Johnson]

It suppresses error messages and warnings when you call a function,
regardless of your level of error reporting. 

So yeah, it is like batch programming :)

-- Josh

-Original Message-
From: Marcus Fleige [mailto:marcus.fleige@;gmx.de] 
Sent: Wednesday, October 30, 2002 7:05 PM
To: PHP-Mailinglist
Subject: [PHP-DB] Newbie-Question: What's that @?

hi there,

i've got a general question about php-scripts:

whats the difference between

@mysql_query($query)

and

@mysql_query($query)?

is it the same as in batch programming?
like, ignore all return messages?

thanks and greetings from germany,

marcus

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RE: [PHP-DB] Newbie-Question: What's that @?

[Ryan Jameson (USA)]

Well the only difference I see is the ? ... hehe... but if you are refering to the @ 
symbol then you are correct ... ignore errors... :-)

<>< Ryan

-Original Message-
From: Marcus Fleige [mailto:marcus.fleige@;gmx.de]
Sent: Wednesday, October 30, 2002 5:05 PM
To: PHP-Mailinglist
Subject: [PHP-DB] Newbie-Question: What's that @?


hi there,

i've got a general question about php-scripts:

whats the difference between

@mysql_query($query)

and

@mysql_query($query)?

is it the same as in batch programming?
like, ignore all return messages?

thanks and greetings from germany,

marcus

--
^v^
[EMAIL PROTECTED]



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[PHP-DB] Newbie-Question: What's that @?

[Marcus Fleige]

hi there,

i've got a general question about php-scripts:

whats the difference between

@mysql_query($query)

and

@mysql_query($query)?

is it the same as in batch programming?
like, ignore all return messages?

thanks and greetings from germany,

marcus

--
^v^
[EMAIL PROTECTED]



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[PHP-DB] ****Newbie Question****

[Ron]

I am use to running MySql & PHP but now I am running into a 
situation, where the website I am admin at is changing the 
server over to MSSql is the and easy way to convert all the 
existing data and webpages from MySql to MSSql. Any assistance 
with this would be greatly appreciated..TIA

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[PHP-DB] ***Newbie Question***

[Ron]

I have an issue I need to see if I can resolve. I am
and admin of a company website that uses MySql for the
database, They are going to change the server to MSSql
is there and easy way to convert on the existing Info
over from MySql to MSSql any help would greatly be
appreciated . TIA


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[PHP-DB] newbie question: fetchrow or smth else ?

[Alecs]

 Could some one please check the following code and guesstimate where am I
 wrong ?
 (the only thing I got from browser is: Fatal error: Call to undefined
 function: fetchrow() pc_list.php on line 14, but on that line I have
 fetchRow() not fetchrow(). To make it worse even when I am replacing
 fetchRow() with FetchRow the same error is reported)

 PS. I used PEAR that comes with php-4.1.1 and php-4.0.4pl1. nothing changed
 :(

 ---
 require_once "DB.php" ;

 $db = DB::connect("$db_type://$db_user:$db_pass@$db_host/$db_name",
 $db_pipe) ;
 $req = "SELECT pc.hostname, m.logo, pc.mac, pc.ip FROM computer pc, mac m
 WHERE m.id = LEFT(pc.mac) " ;

 $p_req = $db->prepare($req) ;
 $res = $db->execute($p_req) ;

 echo "  \n" ;
 echo "   Hostname   Logo   MAC   IP

 \n" ;

 $counter = 0 ;
 while ( $arr=$res->fetchRow() ) {
 echo "   $arr[0]   
  " ;
 echo "  $arr[2]   $arr[3]   \n" ;
 $counter++ ;
 }




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[PHP-DB] newbie question: request & response

[Luke Crouch]

Does PHP have built-in support for using request and response objects?
Thanks,

-L



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Re: [PHP-DB] Newbie question - other way than insert statement?

[Daniel Schwab]

ok, thnx

> No, but SQL is your friend
*lol*

greez
daniel




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Re: [PHP-DB] Newbie question - other way than insert statement?

[Miles Thompson]

No, but SQL is your friend, and is actually easier than messing around with 
recordsets, once you get the hang of it.

The exact form of your INSERT statement will depend a bit on the database 
are you using, but the following is typical. This is MySQL syntax, I 
believe PostgresSQL is similar.

INSERT INTO tablename ( firstname, lastname, phone_number) VALUES ( 
'$strFirst', '$strLast', '$strPhoneNo' )

and if you are truly confident and believe your table structure will NEVER 
change, you could shorten it to

INSERT tablename VALUES ( '$strFirst', '$strLast', '$strPhoneNo' )

Sometimes you just UPDATE

UPDATE tablename SET firstname = '$strFirst', lastname = '$strLast', 
phone_number = '$strPhoneNo' WHERE primary_key_field = '$UniqueID'

If any rows are returned by a SELECT statement determines whether you use 
an INSERT or an UPDATE, but then you have to do that to determine whether 
you .AddNew or .Edit.

Hope this helps - Miles Thompson

PS Of course _nothing_ beats the ease of use and clarity of FoxPro and its 
descendants. 

At 11:00 PM 11/20/2001 +0100, Daniel Schwab wrote:
>Hi
>
>I came from ASP and I'm sort of a newbie. I'm sure you'll see
>that after my question :-)
>
>I'm looking for a way to insert data in an mysql db but not with an
>insert into statement. Is there any way i can do some sort of:
>
>mydb.myfieldname = $myvalue
>
>??
>
>(same as in asp with recordsets: rs!fieldname = value)
>any way to do that in a similar manner?
>
>thanks for your help
>daniel
>
>
>
>
>
>
>
>
>
>
>
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>To contact the list administrators, e-mail: [EMAIL PROTECTED]


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[PHP-DB] Newbie question - other way than insert statement?

[Daniel Schwab]

Hi

I came from ASP and I'm sort of a newbie. I'm sure you'll see
that after my question :-)

I'm looking for a way to insert data in an mysql db but not with an
insert into statement. Is there any way i can do some sort of:

mydb.myfieldname = $myvalue

??

(same as in asp with recordsets: rs!fieldname = value)
any way to do that in a similar manner?

thanks for your help
daniel











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Re: [PHP-DB] Newbie Question

[Kodrik]

>>Hi Kodrik:
I just looked at your function and would like to use it myself, but shouldn't 
the ampersand line:
$copy=eregi_replace ("&", "&", $copy);
look like this instead? So it matches the other lines that have a symbol 
replaced by HTML??
$copy=eregi_replace ("&", "&", $copy);
Cheers.
Russ<<

No, because the & sign is needed to represent the ascii values.
So I'm actually doing it to put it back where it got converted by 
htmlspecialchars.

If you are going to use it for a forum, you might also be interested by  
hyperlinkor.
It converts urls into links, however they are entered (with a www. or not, 
httpd or not...)
http://zc8.com/zc8/ZC8news/shownews.php?articleid=108

You run your string through this function before displaying. and the links 
are anchored.
Do not run it before entering a string in a database, only before echoing the 
string.

My reply to you was bounced back, so I replied to your message on the list 
instead.

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Re: [PHP-DB] Newbie Question

[Russ Michell]

Yes there is:

Using a php function called: nl2br()
When the data is called from the db and printed to the page use $textdata = 
nl2br($textdata)
This will preserve the enter button being pressed.

Regards:
Russ

On Thu, 08 Nov 2001 09:54:16 -0600 Jay Fitzgerald <[EMAIL PROTECTED]> wrote:

> Ok, I am still fairly new at PHP and MySQL also, so please bear with me.
> 
> 
> TASK: I have a client that wants to have job openings listed on their site 
> and they want to be able to add, edit and delete the postings themselves. I 
> would do this in flat-file format but there is the risk of that file size 
> getting too large and slowing down the server.
> 
> 
> SOLUTION: I have created a MySQL database that will hold all the postings 
> in a table called 'jobs' and have created a PHP form that will post this 
> jobs into the db.
> 
> PROBLEM: When I go to the PHP form and enter all of the pertinent job 
> information, there is one specific field that will have to have carriage 
> returns/line breaks in it between paragraphs. Everything is working except 
> for this. Is there a way whenever the user presses , that either 
> PHP/MySQL will convert this into a  tag only when being displayed in a 
> browser and not in the db??
> 
> 
> Can anyone out there please help me with this? I am available off-list as 
> well if it will be easier to pass code back and forth. Any assistance is 
> greatly appreciated!
> 
> 
> 
> Should you have any questions, comments or concerns, feel free to call me 
> at 318-338-2034.
> 
> Thank you for your time,
> 
> Jay Fitzgerald, Design Director - CSBW-A, CPW-A, CWD-A, CEMS-A
> ==
> Bayou Internet..(888) 
> 30-BAYOUhttp://www.bayou.com
> Mississippi Internet...(800) 
> MISSISSIPPI...http://www.mississippi.net
> Vicksburg Online..(800) 
> MISSISSIPPIhttp://www.vicksburg.com
> ==
> Tel: (318) 338-2034ICQ: 38823829 Fax: 
> (318) 323-5053
> 
> 
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> PHP Database Mailing List (http://www.php.net/)
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> 

#---#

  "Believe nothing - consider everything"   
  
  Russ Michell
  Anglia Polytechnic University Webteam
  Room 1C 'The Eastings' East Road, Cambridge
  
  e: [EMAIL PROTECTED]
  w: www.apu.ac.uk/webteam

  www.theruss.com

#---#


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[PHP-DB] Newbie Question

[Jay Fitzgerald]

Ok, I am still fairly new at PHP and MySQL also, so please bear with me.


TASK: I have a client that wants to have job openings listed on their site 
and they want to be able to add, edit and delete the postings themselves. I 
would do this in flat-file format but there is the risk of that file size 
getting too large and slowing down the server.


SOLUTION: I have created a MySQL database that will hold all the postings 
in a table called 'jobs' and have created a PHP form that will post this 
jobs into the db.

PROBLEM: When I go to the PHP form and enter all of the pertinent job 
information, there is one specific field that will have to have carriage 
returns/line breaks in it between paragraphs. Everything is working except 
for this. Is there a way whenever the user presses , that either 
PHP/MySQL will convert this into a  tag only when being displayed in a 
browser and not in the db??


Can anyone out there please help me with this? I am available off-list as 
well if it will be easier to pass code back and forth. Any assistance is 
greatly appreciated!



Should you have any questions, comments or concerns, feel free to call me 
at 318-338-2034.

Thank you for your time,

Jay Fitzgerald, Design Director - CSBW-A, CPW-A, CWD-A, CEMS-A
==
Bayou Internet..(888) 
30-BAYOUhttp://www.bayou.com
Mississippi Internet...(800) 
MISSISSIPPI...http://www.mississippi.net
Vicksburg Online..(800) 
MISSISSIPPIhttp://www.vicksburg.com
==
Tel: (318) 338-2034ICQ: 38823829 Fax: 
(318) 323-5053


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RE: [PHP-DB] Newbie Question: Convert SELECT to DELETE statement in mysql

[Rick Emery]

No need to defrag/compact...MySQL will reuse the space.

Since posting my original message, I've learned that version 4.0 (alpha)  of
MySQL was released on 15 October.  In Paragraph 6.4.6 of the MySQL manual,
we read:
The multi table delete format is supported starting from MySQL 4.0.0. 

Bottom line: if you don't mind unstable, ever-changing, in-development
application for mission-critical applications, you can use vers 4.0.
Otherwise, you need to follow the algorithm you've laid out.  Or, if your
web-host has the option, investigate table of TYPE=INNODB.  These tables
support FOREIGN keys which may allow multi-table deletion.  In version 3.23,
your web-host must be running the MYSQLD-MAX binary.  In version 4.0, INNODB
support is in all binaries.

-Original Message-
From: Shaun Honsvick [mailto:[EMAIL PROTECTED]]
Sent: Thursday, October 18, 2001 6:35 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] Newbie Question: Convert SELECT to DELETE
statement in mysql


No, I am not trying to delete with a select statement. That was just there
to show my tables, fields, etc.

So if I understand what you are saying I can not do a delete statement with
a join in it. I would have to query each individual row for the ID that need
to be deleted, delete the row in the first table, then delete the row in the
second table, third table, etc. Then loop for the next intance. Theres got
to be a better way.  Also, after this is done should I run some sort compact
or defrag on the database that restore wasted space?

Thanks,
Shaun


"Rick Emery" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
> Are you trying to delete records with a SELECT statement?  That can't be
> done.
>
> To delete from two different tables, you must use two different DELETE
> statements.
>
> -Original Message-
> From: Shaun Honsvick [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, October 16, 2001 3:59 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Newbie Question: Convert SELECT to DELETE statement in
> mysql
>
>
> I am trying to run a delete on a mysql database. The delete query needs to
> delete from 2 tables. The select statement is:
>
> SELECT tinvoices.*, ttrackingnumbers.* FROM tinvoices LEFT JOIN
> ttrackingnumbers ON tinvoices.InvoiceID = ttrackingnumbers.InvoiceID WHERE
> Date < '2001-08-01';
>
> I can't seem to get a delete statement to work. (This is my first project)
>
> Thanks,
> Shaun
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
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> To contact the list administrators, e-mail: [EMAIL PROTECTED]
>



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Re: [PHP-DB] Newbie Question: Convert SELECT to DELETE statement in mysql

[Heikki Tuuri]

Hi!

In MySQL-4.0.0 you can delete from several tables in a join:

DELETE t1, t2 FROM t1, t2, t3 WHERE ...;

But 4.0.0 is still alpha, and it will take some months for it to
stabilize. No compact or defrag is normally needed with
MySQL.

Regards,

Heikki
http://www.innodb.com

Shaun Honsvick wrote in message <[EMAIL PROTECTED]>...
>No, I am not trying to delete with a select statement. That was just there
>to show my tables, fields, etc.
>
>So if I understand what you are saying I can not do a delete statement with
>a join in it. I would have to query each individual row for the ID that
need
>to be deleted, delete the row in the first table, then delete the row in
the
>second table, third table, etc. Then loop for the next intance. Theres got
>to be a better way.  Also, after this is done should I run some sort
compact
>or defrag on the database that restore wasted space?
>
>Thanks,
>Shaun
>
>
>"Rick Emery" <[EMAIL PROTECTED]> wrote in message
>news:[EMAIL PROTECTED]...
>> Are you trying to delete records with a SELECT statement?  That can't be
>> done.
>>
>> To delete from two different tables, you must use two different DELETE
>> statements.
>>
>> -Original Message-
>> From: Shaun Honsvick [mailto:[EMAIL PROTECTED]]
>> Sent: Tuesday, October 16, 2001 3:59 PM
>> To: [EMAIL PROTECTED]
>> Subject: [PHP-DB] Newbie Question: Convert SELECT to DELETE statement in
>> mysql
>>
>>
>> I am trying to run a delete on a mysql database. The delete query needs
to
>> delete from 2 tables. The select statement is:
>>
>> SELECT tinvoices.*, ttrackingnumbers.* FROM tinvoices LEFT JOIN
>> ttrackingnumbers ON tinvoices.InvoiceID = ttrackingnumbers.InvoiceID
WHERE
>> Date < '2001-08-01';
>>
>> I can't seem to get a delete statement to work. (This is my first
project)
>>
>> Thanks,
>> Shaun
>>
>>
>>
>> --
>> PHP Database Mailing List (http://www.php.net/)
>> To unsubscribe, e-mail: [EMAIL PROTECTED]
>> For additional commands, e-mail: [EMAIL PROTECTED]
>> To contact the list administrators, e-mail: [EMAIL PROTECTED]
>>
>
>



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Re: [PHP-DB] Newbie Question: Convert SELECT to DELETE statement in mysql

[Shaun Honsvick]

No, I am not trying to delete with a select statement. That was just there
to show my tables, fields, etc.

So if I understand what you are saying I can not do a delete statement with
a join in it. I would have to query each individual row for the ID that need
to be deleted, delete the row in the first table, then delete the row in the
second table, third table, etc. Then loop for the next intance. Theres got
to be a better way.  Also, after this is done should I run some sort compact
or defrag on the database that restore wasted space?

Thanks,
Shaun


"Rick Emery" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]...
> Are you trying to delete records with a SELECT statement?  That can't be
> done.
>
> To delete from two different tables, you must use two different DELETE
> statements.
>
> -Original Message-
> From: Shaun Honsvick [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, October 16, 2001 3:59 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] Newbie Question: Convert SELECT to DELETE statement in
> mysql
>
>
> I am trying to run a delete on a mysql database. The delete query needs to
> delete from 2 tables. The select statement is:
>
> SELECT tinvoices.*, ttrackingnumbers.* FROM tinvoices LEFT JOIN
> ttrackingnumbers ON tinvoices.InvoiceID = ttrackingnumbers.InvoiceID WHERE
> Date < '2001-08-01';
>
> I can't seem to get a delete statement to work. (This is my first project)
>
> Thanks,
> Shaun
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, e-mail: [EMAIL PROTECTED]
> For additional commands, e-mail: [EMAIL PROTECTED]
> To contact the list administrators, e-mail: [EMAIL PROTECTED]
>



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Re: [PHP-DB] Newbie question

[~BD~]

EXACTLY what I was looking for...
Thanks!

http://www.bustdustr.net
Home of Radio Free Bd


- Original Message -
From: CC Zona <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, May 03, 2001 1:42 PM
Subject: Re: [PHP-DB] Newbie question


> In article <034101c0d3f3$0dd747e0$41041dd8@winbox>,
>  [EMAIL PROTECTED] ("~BD~") wrote:
>
> > at the end of my script, I run a set of closing and clean-up
procedures..
> > what I would like to be able to do is that when a user bails out of the
> > script by closing the browser or leaving the site/page, detect that
they've
> > gone, and then go ahead and run my closing procedures.. is this
possible?
>
> http://php.net/manual/en/features.connection-handling.php
> http://php.net/register-shutdown-function
>
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Re: [PHP-DB] Newbie question

[CC Zona]

In article <034101c0d3f3$0dd747e0$41041dd8@winbox>,
 [EMAIL PROTECTED] ("~BD~") wrote:

> at the end of my script, I run a set of closing and clean-up procedures..
> what I would like to be able to do is that when a user bails out of the
> script by closing the browser or leaving the site/page, detect that they've
> gone, and then go ahead and run my closing procedures.. is this possible?

http://php.net/manual/en/features.connection-handling.php
http://php.net/register-shutdown-function

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[PHP-DB] Newbie question

[~BD~]

I've gone thru everything I can find, and I'm sure I'm just missing it (it's
happened before), but...

at the end of my script, I run a set of closing and clean-up procedures..
what I would like to be able to do is that when a user bails out of the
script by closing the browser or leaving the site/page, detect that they've
gone, and then go ahead and run my closing procedures.. is this possible?

I'd appreciate a point in the right direction, with or without an RTFM ...
:)

TIA
~BD~

http://www.bustdustr.net
Home of Radio Free Bd



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Re: [PHP-DB] newbie question on form..

[Andreas D. Landmark]

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

At 25.03.2001 22:36, you wrote:
>Hello boys that's ok??  well 
>I have this question
>I connect to a database (on interbase 6) and retrive some datas from a
>table...
>now  i want to create a form where the input field show old datas into
>the respective input text box...

echo "";

should create a textbox with $from_database as it's value
and then it's up to you how you'd like to fill $from_database, either from
a database (duh) or by any other means...

Regards,


- -- 
Andreas Landmark / noXtension
[EMAIL PROTECTED]
"Martin Luther King, Malcom X, Freedom of speech is as good as sex."
 - Madonna, Perfomer
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Version: PGPfreeware 7.0.3 for non-commercial use 

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[PHP-DB] newbie question on form..

[TONIATI ORAZIO]

Hello boys that's ok??  well 
I have this question
I connect to a database (on interbase 6) and retrive some datas from a
table...
now  i want to create a form where the input field show old datas into
the respective input text box...

anyone can help me??

thank's at all...

by  Davide Marchesin


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[PHP-DB] Newbie Question - Importing a table

[Michael J. Waters]

I'm new to MySQL and am having trouble importing a table.  The table is a 
text file.  Where I think I'm having trouble is on the options.  The table 
has no text delimiter and uses "," as a field separator.  Where I think I'm 
having problems is with the options.  What I'm using is [-c, --columns= and 
then naming all the columns then the name of the database and the table I 
want it to create.

Any help would be appreciated.

Thanks,
Mike Waters


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Re: [PHP-DB] newbie question

[Lisa Elita]

Thanks Victor, I guess I have to change the extension to .html.
How about http://www.blablabla.com/blabla ? Can all search engines index it
?

Thanks,
Lisa E
- Original Message -
From: Victor Foitzik <[EMAIL PROTECTED]>
To: Lisa Elita <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Monday, February 05, 2001 9:49 AM
Subject: Re: [PHP-DB] newbie question


> Hi Lisa,
>
> >Can search engines index any HTML pages that have .php as its extension
(any
> >.php pages without a '?' or '&' in the URL) ? For example:
> >http://www.blablabla.com/blabla.php
> >
> >Thanks,
> >Lisa E
>
> yes they can, but my experience is, that most search engine robots do
_not_
> index pages ending in .php and .php3 due to their dynamic nature (please
> correct me or give me a hint how to change this). We had to change the
> extensions to .html to avoid being not indexed by search engines.
>
> Hope this helps
> Victor
>
>


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Re: [PHP-DB] newbie question

[Victor Foitzik]

Hi Lisa,

>Can search engines index any HTML pages that have .php as its extension (any
>.php pages without a '?' or '&' in the URL) ? For example:
>http://www.blablabla.com/blabla.php
>
>Thanks,
>Lisa E

yes they can, but my experience is, that most search engine robots do _not_ 
index pages ending in .php and .php3 due to their dynamic nature (please 
correct me or give me a hint how to change this). We had to change the 
extensions to .html to avoid being not indexed by search engines.

Hope this helps
Victor


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Re: [PHP-DB] newbie question

[jaskirat singh]

Yes,
It will index the page as any html page

At 10:58 PM 1/27/01 +0700, Lisa Elita wrote:
>Can search engines index any HTML pages that have .php as its extension (any
>.php pages without a '?' or '&' in the URL) ? For example:
>http://www.blablabla.com/blabla.php
>
>Thanks,
>Lisa E
>
>
>
>
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> 

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[PHP-DB] newbie question

[Lisa Elita]

Can search engines index any HTML pages that have .php as its extension (any
.php pages without a '?' or '&' in the URL) ? For example:
http://www.blablabla.com/blabla.php

Thanks,
Lisa E




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