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inquisitive

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Here is the error; Warning: mysql_fetch_array(): 7 is not a valid MySQL...and the line is that errors is the while statement at the end...

 

The following is the code:

 

<?php

if(isset($_POST['mysubmit']))

switch($_POST['type_account']) {

case "pr_specialist":

$type = "PRSpecialist";

break;

case "contractors":

$type = "Contractor";

break;

case "students":

$type = "Student";

break;

case "admin":

$type = "admin";

break;

/* more */

}

$query = "SELECT * FROM users WHERE pending = 0 AND type = '$type'";

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

while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {

/* parse rows */

?>

     

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The code inside of the while() loop is overwriting $result.

 

How do I know this? Your mysql_query() has an or die() statement that would prevent the while loop from being executed at all if there was no result resource (and this has nothing to do with how many rows there are in the result set because a result set with zero rows returns a result resource.)

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Try this

 

<?php
if(isset($_POST['mysubmit'])){
switch($_POST['type_account']) {
	case "pr_specialist":
		$type = "PRSpecialist";
		break;
	case "contractors":
		$type = "Contractor";
		break;
	case "students":
		$type = "Student";
		break;
	case "admin":
		$type = "admin";
	break;
}
$query = "SELECT * FROM users WHERE pending = 0 AND type = '$type'";
$result = mysql_query($query) or die(mysql_error());
while ($line = mysql_fetch_assoc($result)) {
	print_r($line);
}
}
?>

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