mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[Gingechilla]

I have read through and tried the results of questions like this but found nothing that helped.

 

// If current page number, use it 
// if not, set one! 

if(!isset($_GET['page'])){ 
   $page = 1; 
} else { 
   $page = $_GET['page']; 
} 

$search=$s;
$search=$_POST["search"];

$CATSEARCH=$CAT;
$CATSEARCH=$_POST["CAT"];
$CATPAGE=$_POST["CAT"];



// Define the number of results per page 

if($mR > 50){
$max_results = 50;
} else { 
$max_results = $mR;	
} 

// Figure out the limit for the query based 
// on the current page number. 
$from = (($page * $max_results) - $max_results); 

// Perform MySQL query on only the current page number's results 

$sql = "SELECT * FROM items WHERE title LIKE snowball ORDER BY title LIMIT 10";
$query = mysql_query($sql);
while($r = mysql_fetch_array($query)); { [color=red]<<<<<<<<<<<<<<<<< LINE 76 >>>>>>>>>>>>>>[/color]

  $title=$r["title"];
  $code=$r["code"];
  $message=$r["message"];
  $no problem=$r["no problem"];
  $cat1=$r["cat1"];
  $cat2=$r["cat2"];
  $cat3=$r["cat3"];
  $id=$r["id"];


   // Build formatted results here. 
  echo "$title<p>";
}

 

I get this error:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /results.php on line 76

 

I have checked the names of the tables and fields over and over, I have also tried the "die" result and that did not tell me anything.

[Yesideez]

Immediately after your line $sql= add this line:

echo $sql;

This will show you what is being called. If reading what you get in your browser still doesn't help you post it in here.

[Gingechilla]

That just shows me what I asked to search the database for :--/

 

SELECT * FROM items WHERE title LIKE snowball ORDER BY title LIMIT 10

[Gingechilla]

Ok I sorta got it going, you have to do this to get anything:

 

SELECT * FROM items WHERE title LIKE $VARIABLENAME ORDER BY title LIMIT 10

 

You need to make a variable somewhere, no idea why it works like this though O-o

[Yesideez]

That just shows me what I asked to search the database for :--/

Exactly! It shows you the exact query being called after the variables have been filled into the $sql string. This can be helpful debugging now and then as you may unexpectedly be changing the contents of one variable without realising it - I've done this when tired.

 

What do you mean "no idea why it works like this though"?

 

Sorry for late reply, had my head in the javascript forum.