Unable to execute MYSQL query inside a PHP function

Question

PHP:

function generate_uid() {  
  $uid = mt_rand();  

  $sql = "SELECT user_id  
            FROM user_registration";  

  if (!mysql_query($sql,$con)) {  
    die('Error: ' . mysql_error());  
  }  

  $result = mysql_query($sql);  
  $availability = TRUE;  

  while($row = mysql_fetch_array($result)) {  
    if($row['user_id'] == $uid) {  
      $availability = FALSE; 
    }  
  }

  if($availability == FALSE) {
    generate_uid();
  }  
}

The error

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in E:\Web Design EC\register2.php on line 8 Error:

But when i execute this function as normal php there is no error and the $uid is generated. What maybe the problem??

Tatu Ulmanen · Accepted Answer · 2010-10-20 18:27:06Z

$con is not defined in the current variable scope. You can pass the connection to function as a parameter:

function generate_uid($con) {
    ...
    mysql_query($sql, $con);
}

$uid = generate_uid($con);

Or you can use global:

function generate_uid() {
    global $con;
    ...
    mysql_query($sql, $con);
}

$uid = generate_uid();

Or, simply leave the connection variable out, and the last opened connection will be used:

function generate_uid() {
    ...
    mysql_query($sql); 
}

$uid = generate_uid();

All of those should work.

To learn more about variable scope, check out the PHP manual on the subject at:

http://www.php.net/manual/en/language.variables.scope.php

Wizzard · Accepted Answer · 2010-10-20 18:27:18Z

You are using $con in your function, but $con is not defined in your function scope.

http://php.net/manual/en/language.variables.scope.php

You need to either

a) Pass $con to your function or

b) Set $con as a global

c) Not use $con at all eg.

    if (!mysql_query($sql))

however make sure that your app is only connecting to ONE database, otherwise you will run into issues because when you don't specify a connection it will use the last connection opened.

dnagirl · Accepted Answer · 2010-10-20 18:28:11Z

Inside your function, your MySQL connection doesn't exist. You either need to pass it in, or use the global keyword:

$con = mysql_connect('localhost', 'mysql_user', 'mysql_password');

function generate_uid()
{
global $con;
$uid = mt_rand();
$sql="SELECT user_id FROM user_registration";
if (!mysql_query($sql,$con))
{
die('Error: ' . mysql_error());
}
$result = mysql_query($sql);
$availability=TRUE;
while($row = mysql_fetch_array($result))
{
if($row['user_id']==$uid)
{
$availability=FALSE; }
} if($availability==FALSE) { generate_uid(); }
}

Claudiu · Accepted Answer · 2010-10-20 18:38:25Z

0

while($row = mysql_fetch_array($result))

You can't use that if you plan to use $row as an associative array. Use:

while($row = mysql_fetch_assoc($result))

instead.

EDIT

MYSQL_BOTH is turned on by default, forgot about that. The problem is pointed out by the other users, regarding the connection not being available in the function scope.

edited Oct 20, 2010 at 18:38

answered Oct 20, 2010 at 18:28

Claudiu

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2 Comments

Tatu Ulmanen Over a year ago

mysql_fetch_array fetches a result row by default as a numeric array AND as an associative array.

Claudiu Over a year ago

Forgot that's turned on by default, sorry

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