mysql_fetch_assoc very strange

Question

Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result

okay so here is my code.

<?php

include 'connect.php';


$id = addslashes($_REQUEST['id']);


$image = "SELECT * FROM images WHERE id=$id";
$result = mysql_query($image);
$imagearray = mysql_fetch_assoc($result);
$realimage = $imagearray['image'];

header("Content-type: image/jpeg");

echo $realimage;

?>

the page uses GET to get the id but then it gives me this error:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/a6289454/public_html/get.php on line 9

@MuhammadAbrar Even an empty table returns a valid resource from `mysql_query` (it will cause `mysql_fetch_assoc` to return false). The error is likely on the query. — Sony Santos, Feb 12 '12 at 00:11

score 3 · Accepted Answer · answered Feb 11 '12 at 23:47

3

var_dump(mysql_error()); will help you find the error message. But most likely thing is you need quotes:

$image = "SELECT * FROM `images` WHERE `id`='".$id."'";

answered Feb 11 '12 at 23:47

Niet the Dark Absol

320,036
81
464
592

I also suggest to use [`mysql_real_escape_string`](http://www.php.net/manual/en/function.mysql-real-escape-string.php) instead of `addslashes`. – Sony Santos Feb 12 '12 at 00:15

score 0 · Answer 2 · answered Feb 11 '12 at 23:59

0

I think the problem might be in addslashes(), try mysql_real_escape_string() instead.

answered Feb 11 '12 at 23:59

iDifferent

2,190
1
14
19

nope im still getting mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/a6289454/public_html/get.php on line 9 – Matt Williams Feb 12 '12 at 00:11
No reason to escape an integer.. – Richard Feb 12 '12 at 00:14
Sure, no reason to escape it if it's an integer but you never know what it might be since it comes from the GET. – iDifferent Feb 12 '12 at 00:21

score 0 · Answer 3 · answered Feb 12 '12 at 00:10

Why are you adding slashes to $_REQUEST['id']? It should be an integer

Try this code:

if(is_numeric($_REQUEST['id'])){

    $q = "SELECT * FROM images WHERE id = " . $id; // Don't put variables in strings!
    $rs= mysql_query($q) or die("There is an error in the query:<br>" . $q . "<br><br>The error is: ". mysql_error()); // Throw and error if there is one.
    $result = mysql_fetch_assoc($rs);

} else { die('Not a valid ID'); }

Also for good practice use require_once('connect.php'); rather than include 'connect.php';

score 0 · Answer 4 · answered Feb 12 '12 at 00:11

0

 $image = "SELECT * FROM images WHERE id=$id";
 $result = mysql_query($image) or die(mysql_error());

Try that

Also for your $id use type casting it's a simple solution to a simple problem

 $id = (int) $_REQUEST['id'];

answered Feb 12 '12 at 00:11

dead

364
2
10

mysql_fetch_assoc very strange

4 Answers4