Code Not Working!

[Dysan]

Why do I get the following error message, when output items from the database?

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Documents and Settings\User\Desktop\Xampp\htdocs\1.php on line 17

 

foreach($array as $item)
{
  $result = mysql_query("SELECT * FROM names WHERE id = '$id'");
  while($row = mysql_fetch_array($result))     //This is line 17
  {
    $row['Firstname']." - ". $row['Surname'] ." ".$row['Sex'];
  }
}

[GingerRobot]

This means there is either something wrong with your query, or there are no results. Try this:

 

<?php
foreach($array as $item){
$result = mysql_query("SELECT * FROM names WHERE id = '$id'") or die(mysql_error());
$num = mysql_num_rows($result);
if($num > 0){
	while($row = mysql_fetch_array($result)){
		$row['Firstname']." - ". $row['Surname'] ." ".$row['Sex'];
	}
}
}
?>

[helraizer]

Ginger's code might work but normally for the mysql_query you should probably do it like:

 

<?php 

$sql = "SELECT * FROM `names` WHERE `id`='$id'"

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

?>

 

I think you'll find that that should work (haven't tried it but it should)

[revraz]

Tick marks shouldn't make any difference in this case.