Re: [PHP-DB] elseif

2003-01-27 Thread Paul Chvostek
On Mon, Jan 27, 2003 at 06:26:06PM -0600, Addison Ellis wrote:
> 
> hello and thank you for your time.
>   will the below work to check if one field or the other is 
> filled in upon submission? either contact_phone or contact_email is 
> required but not both. best, addison

I would use something like:

$error = "";
if (strlen(trim($name)) < 1) {
$error .= "You must enter a company name.\n";
}
...
if (! ($contact_phone || $contact_email) ) {
$error .= "You must enter a contact phone number or email address.\n";
}
if (strlen(trim($contact_email)) > 0 && 
eregi('[a-z0-9_.-]+@([a-z0-9]+\.)+[a-z][a-z]+',$contact_email)) {
$error .= "Email address has an invalid format.\n";
}
if (!$error) {
$q = "INSERT INTO stores blah blah";
...
}

> $error = "";
> if ($name == "")
>   {
>   $error = "$errorYou must enter a company name.";
>   }
> if ($address == "")
>  {
>  $error = "$errorYou must enter an address.";
>  }
> if ($city == "")
>  {
>  $error = "$errorYou must enter a city.";
>  }
> if ($state == "")
>  {
>  $error = "$errorYou must enter a state.";
>  }
> if ($zip == "")
>  {
>  $error = "$errorYou must enter a zip.";
>  }
> if ($contact_phone == "")
>  {
>  $error = "$errorYou must enter a contact phone number or 
> email address.";
>  }
> elseif ($contact_email == "")
>  {
>  $error = "$errorYou must enter a contact phone number or 
> email address.";
>  }
> if ($error == "")
>   {
>   $query = "insert into stores  set
> -- 
> Addison Ellis
> small independent publishing co.
> 114 B 29th Avenue North
> Nashville, TN 37203
> (615) 321-1791
> [EMAIL PROTECTED]
> [EMAIL PROTECTED]
> subsidiaries of small independent publishing co.
> [EMAIL PROTECTED]
> [EMAIL PROTECTED]
-- 
  Paul Chvostek <[EMAIL PROTECTED]>
  Operations / Abuse / Whatever
  it.canada, hosting and development   http://www.it.ca/


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RE: [PHP-DB] elseif

2003-01-27 Thread Ryan Jameson (USA)
You should probably replace $name == "" with strlen($name) < 1.
Also, you should probably used $error .= instead of $error = so that you can list all 
the errors instead of just the last one.

<>< Ryan


-Original Message-
From: Addison Ellis [mailto:[EMAIL PROTECTED]]
Sent: Monday, January 27, 2003 5:26 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] elseif


hello and thank you for your time.
will the below work to check if one field or the other is 
filled in upon submission? either contact_phone or contact_email is 
required but not both. best, addison

$error = "";
if ($name == "")
{
$error = "$errorYou must enter a company name.";
}
if ($address == "")
 {
 $error = "$errorYou must enter an address.";
 }
if ($city == "")
 {
 $error = "$errorYou must enter a city.";
 }
if ($state == "")
 {
 $error = "$errorYou must enter a state.";
 }
if ($zip == "")
 {
 $error = "$errorYou must enter a zip.";
 }
if ($contact_phone == "")
 {
 $error = "$errorYou must enter a contact phone number or 
email address.";
 }
elseif ($contact_email == "")
 {
 $error = "$errorYou must enter a contact phone number or 
email address.";
 }
if ($error == "")
{
$query = "insert into stores  set
-- 
Addison Ellis
small independent publishing co.
114 B 29th Avenue North
Nashville, TN 37203
(615) 321-1791
[EMAIL PROTECTED]
[EMAIL PROTECTED]
subsidiaries of small independent publishing co.
[EMAIL PROTECTED]
[EMAIL PROTECTED]

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[PHP-DB] elseif

2003-01-27 Thread Addison Ellis
hello and thank you for your time.
	will the below work to check if one field or the other is 
filled in upon submission? either contact_phone or contact_email is 
required but not both. best, addison

$error = "";
if ($name == "")
	{
	$error = "$errorYou must enter a company name.";
	}
if ($address == "")
{
$error = "$errorYou must enter an address.";
}
if ($city == "")
{
$error = "$errorYou must enter a city.";
}
if ($state == "")
{
$error = "$errorYou must enter a state.";
}
if ($zip == "")
{
$error = "$errorYou must enter a zip.";
}
if ($contact_phone == "")
{
$error = "$errorYou must enter a contact phone number or 
email address.";
}
elseif ($contact_email == "")
{
$error = "$errorYou must enter a contact phone number or 
email address.";
}
if ($error == "")
	{
	$query = "insert into stores  set
--
Addison Ellis
small independent publishing co.
114 B 29th Avenue North
Nashville, TN 37203
(615) 321-1791
[EMAIL PROTECTED]
[EMAIL PROTECTED]
subsidiaries of small independent publishing co.
[EMAIL PROTECTED]
[EMAIL PROTECTED]

RE: [PHP-DB] elseif with header

2003-01-21 Thread Bruno Pereira
I Larry is correct you havo to use (==) not (=).
The first one is comparing two variables and the second is that var is equal
to another var.

Cumprimentos

Bruno Pereira
[EMAIL PROTECTED]

-Original Message-
From: Larry E. Ullman [mailto:[EMAIL PROTECTED]]
Sent: terca-feira, 21 de Janeiro de 2003 0:29
To: Addison Ellis
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] elseif with header


> if ($category = 2 and $subcategory = 52)
> {
>header("Location:
> http://www.vanderbilt.edu/register/ca/vacation_rentals.php";);
> }
> elseif ($category = 2 and $subcategory = 50)
> {
>header("Location:
> http://www.vanderbilt.edu/register/ca/rentals.php";);
> }  , etc.
>
>  for some reason all the elseifs are going to the if header.
> any input as to how i can correct this is greatly appreciated.

You are using the assignment operator (=) instead of the equals
comparison operator (==) which makes your if condition always true.

Larry


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RE: [PHP-DB] elseif with header

2003-01-20 Thread zmn
You can just as simply include the files you need, there is no need for
using the headers.

Did the script return any errors?

-Oorspronkelijk bericht-
Van: Addison Ellis [mailto:[EMAIL PROTECTED]] 
Verzonden: dinsdag 21 januari 2003 1:19
Aan: [EMAIL PROTECTED]
Onderwerp: [PHP-DB] elseif with header

hello and thank you...
i have two pages where on 1-category is selected. on 2 
subcategory is selected. on submit from page two it goes to 
direct.php. there i have
if ($category = 2 and $subcategory = 52)
{
header("Location: 
http://www.vanderbilt.edu/register/ca/vacation_rentals.php";);
}
elseif ($category = 2 and $subcategory = 50)
{
header("Location:
http://www.vanderbilt.edu/register/ca/rentals.php";);
}  , etc.

  for some reason all the elseifs are going to the if header.
any input as to how i can correct this is greatly appreciated.
thank you again. addison
-- 
Addison Ellis
small independent publishing co.
114 B 29th Avenue North
Nashville, TN 37203
(615) 321-1791
[EMAIL PROTECTED]
[EMAIL PROTECTED]
subsidiaries of small independent publishing co.
[EMAIL PROTECTED]
[EMAIL PROTECTED]


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Re: [PHP-DB] elseif with header

2003-01-20 Thread Larry E. Ullman
if ($category = 2 and $subcategory = 52)
{
   header("Location: 
http://www.vanderbilt.edu/register/ca/vacation_rentals.php";);
}
elseif ($category = 2 and $subcategory = 50)
{
   header("Location: 
http://www.vanderbilt.edu/register/ca/rentals.php";);
}  , etc.

 for some reason all the elseifs are going to the if header.
any input as to how i can correct this is greatly appreciated.

You are using the assignment operator (=) instead of the equals 
comparison operator (==) which makes your if condition always true.

Larry


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[PHP-DB] elseif with header

2003-01-20 Thread Addison Ellis
hello and thank you...
	i have two pages where on 1-category is selected. on 2 
subcategory is selected. on submit from page two it goes to 
direct.php. there i have
if ($category = 2 and $subcategory = 52)
{
   header("Location: 
http://www.vanderbilt.edu/register/ca/vacation_rentals.php";);
}
elseif ($category = 2 and $subcategory = 50)
{
   header("Location: http://www.vanderbilt.edu/register/ca/rentals.php";);
}  , etc.

 for some reason all the elseifs are going to the if header.
any input as to how i can correct this is greatly appreciated.
thank you again. addison
--
Addison Ellis
small independent publishing co.
114 B 29th Avenue North
Nashville, TN 37203
(615) 321-1791
[EMAIL PROTECTED]
[EMAIL PROTECTED]
subsidiaries of small independent publishing co.
[EMAIL PROTECTED]
[EMAIL PROTECTED]

RE: [PHP-DB] elseif statement syntax

2002-03-20 Thread Andrea Caldwell

Thanks for responding Steve,

My error reporting is turned on, because I always get errors.  I found the
problem, I had two ;; at the end of one of the statements.

Thanks- Andrea

-Original Message-
From: Steve Cayford [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, March 20, 2002 10:27 AM
To: Andrea Caldwell
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP-DB] elseif statement syntax


You probably have error reporting turned off in your php.ini file or you
may be directing errors into your log file, try checking there.

Why are you tacking "or die(mysql_error())" on the end of each of your
query assignments? You're just assigning a string to a variable so
there's nothing to fail, or at least no mysql errors.

-Steve


On Wednesday, March 20, 2002, at 10:41  AM, Andrea Caldwell wrote:

> Hi All,   I'm pretty new at this, so go easy on me please ;-)
>
> What is wrong w/ this syntax?  If the search results are 0, it just
> displays
> a blank screen instead of echoing the error message if numresults ==0
> or the
> mysql_error message.  If data is found, everything is fine.  Thanks in
> advance for your help!
>
> if($searchterm){
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory
> where
> realname like '%".$searchterm."%'" or die (mysql_error());
> }
> elseif($location){
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory
> where
> location like '%".$location."%'" or die (mysql_error());
> }
> else{
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory
> where
> location like '%".$searchloc."%'" or die (mysql_error());
> }
>
> $result = mysql_query($query) or die (mysql_error());
> $num_results = mysql_num_rows($result)or die (mysql_error());;
>
>  if($num_results==0){
>  echo "Sorry, nothing matched your search request.  Please go back and
> try
> again.";
>  }
>
>  else {
>  echo "Number of Entries Found:
> ".$num_results."";
>  }
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>



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Re: [PHP-DB] elseif statement syntax

2002-03-20 Thread Steve Cayford

You probably have error reporting turned off in your php.ini file or you 
may be directing errors into your log file, try checking there.

Why are you tacking "or die(mysql_error())" on the end of each of your 
query assignments? You're just assigning a string to a variable so 
there's nothing to fail, or at least no mysql errors.

-Steve


On Wednesday, March 20, 2002, at 10:41  AM, Andrea Caldwell wrote:

> Hi All,   I'm pretty new at this, so go easy on me please ;-)
>
> What is wrong w/ this syntax?  If the search results are 0, it just 
> displays
> a blank screen instead of echoing the error message if numresults ==0 
> or the
> mysql_error message.  If data is found, everything is fine.  Thanks in
> advance for your help!
>
> if($searchterm){
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory 
> where
> realname like '%".$searchterm."%'" or die (mysql_error());
> }
> elseif($location){
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory 
> where
> location like '%".$location."%'" or die (mysql_error());
> }
> else{
> $query = "select directory.realname, directory.phone, directory.ext,
> directory.phone2, directory.email, directory.location from directory 
> where
> location like '%".$searchloc."%'" or die (mysql_error());
> }
>
> $result = mysql_query($query) or die (mysql_error());
> $num_results = mysql_num_rows($result)or die (mysql_error());;
>
>  if($num_results==0){
>  echo "Sorry, nothing matched your search request.  Please go back and 
> try
> again.";
>  }
>
>  else {
>  echo "Number of Entries Found:
> ".$num_results."";
>  }
>
>
>
> --
> PHP Database Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
>


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[PHP-DB] elseif statement syntax

2002-03-20 Thread Andrea Caldwell

Hi All,   I'm pretty new at this, so go easy on me please ;-)

What is wrong w/ this syntax?  If the search results are 0, it just displays
a blank screen instead of echoing the error message if numresults ==0 or the
mysql_error message.  If data is found, everything is fine.  Thanks in
advance for your help!

if($searchterm){
$query = "select directory.realname, directory.phone, directory.ext,
directory.phone2, directory.email, directory.location from directory where
realname like '%".$searchterm."%'" or die (mysql_error());
}
elseif($location){
$query = "select directory.realname, directory.phone, directory.ext,
directory.phone2, directory.email, directory.location from directory where
location like '%".$location."%'" or die (mysql_error());
}
else{
$query = "select directory.realname, directory.phone, directory.ext,
directory.phone2, directory.email, directory.location from directory where
location like '%".$searchloc."%'" or die (mysql_error());
}

$result = mysql_query($query) or die (mysql_error());
$num_results = mysql_num_rows($result)or die (mysql_error());;

 if($num_results==0){
 echo "Sorry, nothing matched your search request.  Please go back and try
again.";
 }

 else {
 echo "Number of Entries Found:
".$num_results."";
 }



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