[PHP] Code check please

2003-09-17 Thread James Johnson 
Hi,

Can anyone see why this code isn't working

$SID = $_SESSION['svUserID'];
$AdID = $_GET['AdID'];

// get the campuses assigned
$qCampusID = SELECT inst_id FROM ads_campuses WHERE ad_id = $AdID;
$CampusIDList = mysql_query($qCampusID, $CCB) or die(mysql_error());
$row_CampusIDList = mysql_fetch_assoc($CampusIDList);
//var_dump($row_CampusIDList);

$tmp = array();
while( $rowData = mysql_fetch_array( $CampusIDList )) {
//print(ROW Data $rowData);
if(!in_array($rowData['inst_id'],$tmp)){
  $tmp[] = $rowData['inst_id'];
 }
}
//var_dump($tmp);
$cidList = implode( ',', $tmp );
//var_dump($cidList);

If I do a var_dump($row_CampusIDList); I get a value, so I know there's data
to work with. But if I do a var_dump($tmp) and var_dump($cidList), they're
both empty.

This particular value query only has 1 inst_id of 100. Other queries with
one value (99,45, etc) will work.

Very strange...
Thanks,
James

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RE: [PHP] Code check please

2003-09-17 Thread James Johnson 
Hmm, ok I took out the 
$row_CampusIDList = mysql_fetch_assoc($CampusIDList); 

and it works But how come?

J

-Original Message-
From: James Johnson [mailto:[EMAIL PROTECTED] 
Sent: Wednesday, September 17, 2003 9:17 AM
To: [EMAIL PROTECTED]
Subject: [PHP] Code check please


Hi,

Can anyone see why this code isn't working

$SID = $_SESSION['svUserID'];
$AdID = $_GET['AdID'];

// get the campuses assigned
$qCampusID = SELECT inst_id FROM ads_campuses WHERE ad_id = $AdID;
$CampusIDList = mysql_query($qCampusID, $CCB) or die(mysql_error());
$row_CampusIDList = mysql_fetch_assoc($CampusIDList);
//var_dump($row_CampusIDList);

$tmp = array();
while( $rowData = mysql_fetch_array( $CampusIDList )) {
//print(ROW Data $rowData);
if(!in_array($rowData['inst_id'],$tmp)){
  $tmp[] = $rowData['inst_id'];
 }
}
//var_dump($tmp);
$cidList = implode( ',', $tmp );
//var_dump($cidList);

If I do a var_dump($row_CampusIDList); I get a value, so I know there's data
to work with. But if I do a var_dump($tmp) and var_dump($cidList), they're
both empty.

This particular value query only has 1 inst_id of 100. Other queries with
one value (99,45, etc) will work.

Very strange...
Thanks,
James

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Re: [PHP] Code check please

2003-09-17 Thread CPT John W. Holmes 
From: James Johnson [EMAIL PROTECTED]

 Can anyone see why this code isn't working

 $SID = $_SESSION['svUserID'];
 $AdID = $_GET['AdID'];

 // get the campuses assigned
 $qCampusID = SELECT inst_id FROM ads_campuses WHERE ad_id = $AdID;

You say down below that this query only selects one row...

 $CampusIDList = mysql_query($qCampusID, $CCB) or die(mysql_error());
 $row_CampusIDList = mysql_fetch_assoc($CampusIDList);

and this mysql_fetch_assoc() call will return that one row into
$row_CampusIDList.

 //var_dump($row_CampusIDList);

 $tmp = array();
 while( $rowData = mysql_fetch_array( $CampusIDList )) {

This while will never execute as there are no more rows to select.

 //print(ROW Data $rowData);
 if(!in_array($rowData['inst_id'],$tmp)){
  $tmp[] = $rowData['inst_id'];
 }
 }
 //var_dump($tmp);
 $cidList = implode( ',', $tmp );
 //var_dump($cidList);

 If I do a var_dump($row_CampusIDList); I get a value, so I know there's
data
 to work with. But if I do a var_dump($tmp) and var_dump($cidList), they're
 both empty.

 This particular value query only has 1 inst_id of 100. Other queries with
 one value (99,45, etc) will work.

---John Holmes...

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Re: [PHP] Code check please

2001-06-21 Thread Philip Olson 

You want mysql_error(), call it somewhere after your query.

  echo mysql_error();

Regarding the query, the following should work just fine :

  $sql = INSERT INTO tablefoo (a,b,c) VALUES ('$a','$b','$c');

Be sure to add slashes sometime to your values beforehand. Anyway,
mysql_error() is what you're wanting.

  http://www.php.net/manual/function.mysql-error.php

regards,
philip


On Thu, 21 Jun 2001, Andreas Skarin wrote:

 I'm sorry guys, neither of the snippets work. I must have screwed something else up
 too. Is there any way I can provoke an error message from your code examples below?
 If anyone manages to find out what's wrong, please tell me. I'm not giving up until
 I smash this bug :-)
 
 // Andreas
 
  Rich Cavanaugh wrote:
 
   try:
  
   $sql = INSERT INTO tabell (fornamn, efternamn, email) values ('{$fornamn}',
   '{$efternamn}', '{$email}');
 
  Sebastian Wenleder wrote:
 
   I'd use this SQL-query:
  
   $sql = INSERT INTO tabell \
   (fornamn,efternamn,email) \
   VALUES(.$fornamn., \
   .$efternamn., .$email.);
 
 
 
 
 --
 Andreas Skarin
 Svenska Dream Theater-Sällskapet
 http://www.sdts.nu - mailto:[EMAIL PROTECTED]
 
 
 
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RE: [PHP] Code check please

2001-06-21 Thread Warren Vail 

Andreas,

I'm not sure about code that appears to be missing like an execute query
function or checking the number of affected rows to make sure that the
insert worked, but clearly your insert code is not properly formed.  The
query could also be impacted by whether the columns are defined with special
attributes like DATE, etc.  Assuming that all columns are defined as
character strings the query should look like;

$sql = INSERT INTO tabell (fornamn, efternamn, email)  .
VALUES(\$fornamn\, \$efternamn\, \$email\);

I've found that I often have problems trying to use single quotes in sql
queries so I avoid them.  I notice that you also included a semi-colon in
the query text, while this works when piping a file of queries into the
mysql command, it does not work from a program.  The semi-colon in a command
stream into the mysql command actually signifies a point to stop and execute
the preceding text as a query and after executing that query, continue on
from the next character.

hope you got it working,

Warren Vail

-Original Message-
From: Andreas Skarin [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 20, 2001 9:00 AM
To: PHP General
Subject: [PHP] Code check please


I've tried to get this working for over an hour
now, and it still won't. I don't even get an error
message to help me find the problem so I was
hoping that someone could check my code for me.

I'm fooling around with a basic form that is
supposed to send one's name, surname and e-mail
address to receive.php. receive.php is then
supposed to take the information and add it to a
table called tabell in a database called
databas, but it doesn't work. I think there
might be something wrong with my MySQL query.

- - - - - - - - - - - FORM - - - - - - - - - - - -
- -

form action=receive.php method=post
PFouml;rnamn:br
input type=text name=fornamn
size=25/p
pEfternamn:br
input type=text name=efternamn
size=25/p
pE-mailadress:br
input type=text name=email
size=25/p
input type=submit name=submit
value=Log in
/form

- - - - - - - - - - - - - - - - - - - - - - - - -
- - -

- - - - - - - - - - - RECEIVE.PHP - - - - - - - -
- - -

?php

// connection to MySQL
$connection = mysql_connect(localhost,
username, password);
if (!$connection) {
echo (PUnable to connect to the database
server at this time./P );
exit();
}

//select database
if (! @mysql_select_db(databas) ) {
echo (PUnable to locate the database at
this time./P);
exit();
}

// MySQL query
$sql = INSERT INTO tabell SET .
fornamn ='$fornamn', .
efternamn='$efternamn', .
email='$email';;
?

- - - - - - - - - - - - - - - - - - - - - - - - -
- - -

Thanks in advance!

// Andreas


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RE: [PHP] Code check please

2001-06-21 Thread Warren Vail 

It just occurred to me that one thing that will kill a query is special
query characters imbedded in your strings, like quotes (single or double).
This can be resolved by the

$resultstring = addslashes($sourcestring);

this should escape special characters normally used to signal key components
to the mysql query processor.

of course when you select the column you need to run it thru;

$resultstring = stripslashes($dbcolumnvalue);

to get back your original value.

good luck,

Warren Vail

-Original Message-
From: Andreas Skarin [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 20, 2001 3:46 PM
To: PHP General
Subject: Re: [PHP] Code check please


I'm sorry guys, neither of the snippets work. I must have screwed something
else up
too. Is there any way I can provoke an error message from your code examples
below?
If anyone manages to find out what's wrong, please tell me. I'm not giving
up until
I smash this bug :-)

// Andreas

 Rich Cavanaugh wrote:

  try:
 
  $sql = INSERT INTO tabell (fornamn, efternamn, email) values
('{$fornamn}',
  '{$efternamn}', '{$email}');

 Sebastian Wenleder wrote:

  I'd use this SQL-query:
 
  $sql = INSERT INTO tabell \
  (fornamn,efternamn,email) \
  VALUES(.$fornamn., \
  .$efternamn., .$email.);




--
Andreas Skarin
Svenska Dream Theater-Sällskapet
http://www.sdts.nu - mailto:[EMAIL PROTECTED]



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Re: [PHP] Code check please

2001-06-21 Thread Robert Vetter 



Andreas Skarin wrote:
 
 I've tried to get this working for over an hour
 now, and it still won't. I don't even get an error
 message to help me find the problem so I was
 hoping that someone could check my code for me.
 
 I'm fooling around with a basic form that is
 supposed to send one's name, surname and e-mail
 address to receive.php. receive.php is then
 supposed to take the information and add it to a
 table called tabell in a database called
 databas, but it doesn't work. I think there
 might be something wrong with my MySQL query.
 
 - - - - - - - - - - - FORM - - - - - - - - - - - -
 - -
 
 form action=receive.php method=post
 PFouml;rnamn:br
 input type=text name=fornamn
 size=25/p
 pEfternamn:br
 input type=text name=efternamn
 size=25/p
 pE-mailadress:br
 input type=text name=email
 size=25/p
 input type=submit name=submit
 value=Log in
 /form
 
 - - - - - - - - - - - - - - - - - - - - - - - - -
 - - -
 
 - - - - - - - - - - - RECEIVE.PHP - - - - - - - -
 - - -
 
 ?php
 
 // connection to MySQL
 $connection = mysql_connect(localhost,
 username, password);
 if (!$connection) {
 echo (PUnable to connect to the database
 server at this time./P );
 exit();
 }
 
 //select database
 if (! @mysql_select_db(databas) ) {
 echo (PUnable to locate the database at
 this time./P);
 exit();
 }
 
 // MySQL query
 $sql = INSERT INTO tabell SET .
 fornamn ='$fornamn', .
 efternamn='$efternamn', .
 email='$email';;
 ?

Hello,

and where is the mysql_query($sql) function call?

Robert

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[PHP] Code check please

2001-06-20 Thread Andreas Skarin 

I've tried to get this working for over an hour
now, and it still won't. I don't even get an error
message to help me find the problem so I was
hoping that someone could check my code for me.

I'm fooling around with a basic form that is
supposed to send one's name, surname and e-mail
address to receive.php. receive.php is then
supposed to take the information and add it to a
table called tabell in a database called
databas, but it doesn't work. I think there
might be something wrong with my MySQL query.

- - - - - - - - - - - FORM - - - - - - - - - - - -
- -

form action=receive.php method=post
PFouml;rnamn:br
input type=text name=fornamn
size=25/p
pEfternamn:br
input type=text name=efternamn
size=25/p
pE-mailadress:br
input type=text name=email
size=25/p
input type=submit name=submit
value=Log in
/form

- - - - - - - - - - - - - - - - - - - - - - - - -
- - -

- - - - - - - - - - - RECEIVE.PHP - - - - - - - -
- - -

?php

// connection to MySQL
$connection = mysql_connect(localhost,
username, password);
if (!$connection) {
echo (PUnable to connect to the database
server at this time./P );
exit();
}

//select database
if (! @mysql_select_db(databas) ) {
echo (PUnable to locate the database at
this time./P);
exit();
}

// MySQL query
$sql = INSERT INTO tabell SET .
fornamn ='$fornamn', .
efternamn='$efternamn', .
email='$email';;
?

- - - - - - - - - - - - - - - - - - - - - - - - -
- - -

Thanks in advance!

// Andreas


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Re: [PHP] Code check please

2001-06-20 Thread J Smith 


Your code is [mostly] fine, except for two things:

1. Although you connected to your database with mysql_connect() and
   mysql_select_db(), you didn't do anything with connection.

   Try using mysql_query() to send a query to the db, i.e.
mysql_query($sql).


2. Your SQL is a bit out of whack. The SET clause is used for
   UPDATEs, not INSERTs.

   Instead, try:

   $sql = INSERT INTO tabell (fornamn, efternamn, email) VALUES ( .
  '$fornamn', '$efternamn', '$email');;


Hope that works.

J



Andreas Skarin [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 I've tried to get this working for over an hour
 now, and it still won't. I don't even get an error
 message to help me find the problem so I was
 hoping that someone could check my code for me.

 I'm fooling around with a basic form that is
 supposed to send one's name, surname and e-mail
 address to receive.php. receive.php is then
 supposed to take the information and add it to a
 table called tabell in a database called
 databas, but it doesn't work. I think there
 might be something wrong with my MySQL query.

 - - - - - - - - - - - FORM - - - - - - - - - - - -
 - -

 form action=receive.php method=post
 PFouml;rnamn:br
 input type=text name=fornamn
 size=25/p
 pEfternamn:br
 input type=text name=efternamn
 size=25/p
 pE-mailadress:br
 input type=text name=email
 size=25/p
 input type=submit name=submit
 value=Log in
 /form

 - - - - - - - - - - - - - - - - - - - - - - - - -
 - - -

 - - - - - - - - - - - RECEIVE.PHP - - - - - - - -
 - - -

 ?php

 // connection to MySQL
 $connection = mysql_connect(localhost,
 username, password);
 if (!$connection) {
 echo (PUnable to connect to the database
 server at this time./P );
 exit();
 }

 //select database
 if (! @mysql_select_db(databas) ) {
 echo (PUnable to locate the database at
 this time./P);
 exit();
 }

 // MySQL query
 $sql = INSERT INTO tabell SET .
 fornamn ='$fornamn', .
 efternamn='$efternamn', .
 email='$email';;
 ?

 - - - - - - - - - - - - - - - - - - - - - - - - -
 - - -

 Thanks in advance!

 // Andreas


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 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] Code check please

2001-06-20 Thread Rich Cavanaugh 

You're using UPDATE syntax for your INSERT

try:

$sql = INSERT INTO tabell (fornamn, efternamn, email) values ('{$fornamn}',
'{$efternamn}', '{$email}');

Rich Cavanaugh

-Original Message-
From: Andreas Skarin [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, June 20, 2001 12:00 PM
To: PHP General
Subject: [PHP] Code check please


I've tried to get this working for over an hour
now, and it still won't. I don't even get an error
message to help me find the problem so I was
hoping that someone could check my code for me.

I'm fooling around with a basic form that is
supposed to send one's name, surname and e-mail
address to receive.php. receive.php is then
supposed to take the information and add it to a
table called tabell in a database called
databas, but it doesn't work. I think there
might be something wrong with my MySQL query.

- - - - - - - - - - - FORM - - - - - - - - - - - -
- -

form action=receive.php method=post
PFouml;rnamn:br
input type=text name=fornamn
size=25/p
pEfternamn:br
input type=text name=efternamn
size=25/p
pE-mailadress:br
input type=text name=email
size=25/p
input type=submit name=submit
value=Log in
/form

- - - - - - - - - - - - - - - - - - - - - - - - -
- - -

- - - - - - - - - - - RECEIVE.PHP - - - - - - - -
- - -

?php

// connection to MySQL
$connection = mysql_connect(localhost,
username, password);
if (!$connection) {
echo (PUnable to connect to the database
server at this time./P );
exit();
}

//select database
if (! @mysql_select_db(databas) ) {
echo (PUnable to locate the database at
this time./P);
exit();
}

// MySQL query
$sql = INSERT INTO tabell SET .
fornamn ='$fornamn', .
efternamn='$efternamn', .
email='$email';;
?

- - - - - - - - - - - - - - - - - - - - - - - - -
- - -

Thanks in advance!

// Andreas


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Re: [PHP] Code check please

2001-06-20 Thread Sebastian Wenleder 

At 18:00 Uhr +0200 20.06.2001, Andreas Skarin wrote:
I've tried to get this working for over an hour
now, and it still won't. I don't even get an error
message to help me find the problem so I was
hoping that someone could check my code for me.

[snip]

$sql = INSERT INTO tabell SET .
 fornamn ='$fornamn', .
 efternamn='$efternamn', .
 email='$email';;


Hi Andreas

I'd use this SQL-query:

$sql = INSERT INTO tabell \
(fornamn,efternamn,email) \
VALUES(.$fornamn., \
.$efternamn., .$email.);


Hope it works for you!

Best,
Sebastian

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Re: [PHP] Code check please

2001-06-20 Thread Lenar Lõhmus 

UPDATE syntax for INSERT is allowed in MySQL, so that should not be the
problem as far as you include all fields not having default value, timestamp
type or auto_increment attribute in your statement. What's the error message
if there is any?

lenar

Rich Cavanaugh [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 You're using UPDATE syntax for your INSERT

 try:

 $sql = INSERT INTO tabell (fornamn, efternamn, email) values
('{$fornamn}',
 '{$efternamn}', '{$email}');

 Rich Cavanaugh

 -Original Message-
 From: Andreas Skarin [mailto:[EMAIL PROTECTED]]
 Sent: Wednesday, June 20, 2001 12:00 PM
 To: PHP General
 Subject: [PHP] Code check please


 I've tried to get this working for over an hour
 now, and it still won't. I don't even get an error
 message to help me find the problem so I was
 hoping that someone could check my code for me.

 I'm fooling around with a basic form that is
 supposed to send one's name, surname and e-mail
 address to receive.php. receive.php is then
 supposed to take the information and add it to a
 table called tabell in a database called
 databas, but it doesn't work. I think there
 might be something wrong with my MySQL query.

 - - - - - - - - - - - FORM - - - - - - - - - - - -
 - -

 form action=receive.php method=post
 PFouml;rnamn:br
 input type=text name=fornamn
 size=25/p
 pEfternamn:br
 input type=text name=efternamn
 size=25/p
 pE-mailadress:br
 input type=text name=email
 size=25/p
 input type=submit name=submit
 value=Log in
 /form

 - - - - - - - - - - - - - - - - - - - - - - - - -
 - - -

 - - - - - - - - - - - RECEIVE.PHP - - - - - - - -
 - - -

 ?php

 // connection to MySQL
 $connection = mysql_connect(localhost,
 username, password);
 if (!$connection) {
 echo (PUnable to connect to the database
 server at this time./P );
 exit();
 }

 //select database
 if (! @mysql_select_db(databas) ) {
 echo (PUnable to locate the database at
 this time./P);
 exit();
 }

 // MySQL query
 $sql = INSERT INTO tabell SET .
 fornamn ='$fornamn', .
 efternamn='$efternamn', .
 email='$email';;
 ?

 - - - - - - - - - - - - - - - - - - - - - - - - -
 - - -

 Thanks in advance!

 // Andreas


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Re: [PHP] Code check please

2001-06-20 Thread Andreas Skarin 

I'm sorry guys, neither of the snippets work. I must have screwed something else up
too. Is there any way I can provoke an error message from your code examples below?
If anyone manages to find out what's wrong, please tell me. I'm not giving up until
I smash this bug :-)

// Andreas

 Rich Cavanaugh wrote:

  try:
 
  $sql = INSERT INTO tabell (fornamn, efternamn, email) values ('{$fornamn}',
  '{$efternamn}', '{$email}');

 Sebastian Wenleder wrote:

  I'd use this SQL-query:
 
  $sql = INSERT INTO tabell \
  (fornamn,efternamn,email) \
  VALUES(.$fornamn., \
  .$efternamn., .$email.);




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Andreas Skarin
Svenska Dream Theater-Sällskapet
http://www.sdts.nu - mailto:[EMAIL PROTECTED]



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RE: [PHP] Code check please

2001-06-20 Thread Jason Murray 

 2. Your SQL is a bit out of whack. The SET clause is used for
UPDATEs, not INSERTs.
 
Instead, try:
 
$sql = INSERT INTO tabell (fornamn, efternamn, email) VALUES ( .
   '$fornamn', '$efternamn', '$email');;

MySQL actually lets you use SET for INSERTS, though.

Jason

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