[PHP] Fatal error: Function name must be a string

[Adam Williams]

I'm getting the following error and I don't see whats wrong with my 
line.  Any ideas?


*Fatal error*: Function name must be a string in 
*/var/www/sites/intra-test/contract/perform.php* on line *57*


and my snippet of code is:

if ( $_POST[perform] == View Contracts )
   {

   $mysqli_get_userid = SELECT user_id from user where email =
'.$_SESSION[username].';

   $mysqli_get_userid_result = $mysqli_query($mysqli,  // line 57
$mysqli_get_userid) or die(mysqli_error($mysqli));

   while ($userid_result = 
mysqli_fetch_array($mysqli_get_userid_result))

   {
   $user_id = $userid_result[user_id];
   }
   }

Re: [PHP] Fatal error: Function name must be a string

[Daniel Brown]

On Jan 2, 2008 4:58 PM, Adam Williams [EMAIL PROTECTED] wrote:
 I'm getting the following error and I don't see whats wrong with my
 line.  Any ideas?

 *Fatal error*: Function name must be a string in
 */var/www/sites/intra-test/contract/perform.php* on line *57*

 and my snippet of code is:

 if ( $_POST[perform] == View Contracts )
 {

 $mysqli_get_userid = SELECT user_id from user where email =
 '.$_SESSION[username].';

 $mysqli_get_userid_result = $mysqli_query($mysqli,  // line 57
 $mysqli_get_userid) or die(mysqli_error($mysqli));

 while ($userid_result =
 mysqli_fetch_array($mysqli_get_userid_result))
 {
 $user_id = $userid_result[user_id];
 }
 }


Change:

$mysqli_query($mysqli,

 to:

mysqli_query($mysqli,

Otherwise you're declaring mysqli_query() as a variable.

-- 
Daniel P. Brown
[Phone Numbers Go Here!]
[They're Hidden From View!]

If at first you don't succeed, stick to what you know best so that you
can make enough money to pay someone else to do it for you.

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