supplied argument is not a valid MySQL result resource

[jakerow]

Thanks for reading! I'm working on a site and it won't display my records from the database. I have a little filter system that did work, but now it doesn't. I get the following error message:

 

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/vhosts/60608.glrdmd.eu/httpdocs/jakesite/select3.php on line 26

 

 

Relevant code to the error is:

 

<?php 
include("connection.php");

if(isset($_POST['Graphic']) &&
   $_POST['Graphic'] == 'yes')
{
$result1 = "graphic";
}
if(isset($_POST['Interactive']) &&
   $_POST['Interactive'] == 'yes')
{
$result2 = "interactive";
}
if(isset($_POST['Websites']) &&
   $_POST['Websites'] == 'yes')
{
$result3 = "websites";
}
$order = $_POST['group1'];
$sql = ("SELECT * FROM 'jake_portfolio' WHERE CAT = '$result1' OR CAT = '$result2' OR CAT = '$result3' ORDER BY '$order' LIMIT 0, 4");
$result = mysql_query($sql);
$checked = "checked";
$row = mysql_fetch_array($result);
?>

 

Does anyone see what I am doing wrong here?

[Pikachu2000]

The query is failing because table names do not get enclosed in quotes. Beyond that if there are other errors, echo mysql_error and see what they are.

[jakerow]

I have removed those quotes, I didn't have them there in the first place but I read somewhere that it might fix the problem?

 

Anyway I put this in:

 

$row = if(!mysql_fetch_array($result)){

echo mysql_error()};

 

And now I get the following error:

Parse error: syntax error, unexpected T_IF in /var/www/vhosts/60608.glrdmd.eu/httpdocs/jakesite/select3.php on line 26

[Pikachu2000]

This instead:

$result = mysql_query($sql) or die( "<br>Query string: $sql<br>Produced error: " . mysql_error() );

 

This would be for debugging only, BTW. On a live site, you would never want to echo the actual query string or MySQL error to the user.

[jakerow]

Your suggested debugging solution provided me with this:

 

Query string: SELECT * FROM jake_portfolio WHERE CAT = OR CAT = OR CAT = ORDER BY LIMIT 0, 4

Produced error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'OR CAT = OR CAT = ORDER BY LIMIT 0, 4' at line 1

 

So what is wrong with with my coding in that part? I have removed the ''s

[Pikachu2000]

All of your variables are empty. You'd need to post the form markup and any other relevant code so we can see what's up with it and offer suggestions . . .

[jakerow]

The form is as follows:

 

<div id="filtermenu">
    <form action="select3.php" method="post">
      Search:
      <p>
        <input name="Graphic" type="checkbox" value="yes" <?PHP if($result1 == 'graphic'){print $checked ; }?>/>
        Graphic<br />
        <input name="Interactive" type="checkbox" value="yes"<?PHP if($result2 == 'interactive'){print $checked ; }?>/>
        Interactive<br />
        <input name="Websites" type="checkbox" value="yes" <?PHP if($result3 == 'websites'){print $checked ; }?>/>
        Websites</p>
      Sort by:
      <p>
        <input name="group1" type="radio" value="DATE DESC" <?PHP if($order == 'DATE DESC'){print $checked ; } elseif($order != 'DATE ASC' or 'VIEWS'){print $checked ;}?>/>
        Newest<br />
        <input name="group1" type="radio" value="DATE ASC" <?PHP if($order== 'DATE ASC'){print $checked ; }?>/>
        Oldest<br />
        <input name="group1" type="radio" value="VIEWS" <?PHP if($order == 'VIEWS'){print $checked ; }?>/>
        Most viewed</p>
      <input name="Search!" type="submit" value="Filter" id="button"/>
    </form>

 

thanks for taking your time for me by the way.

[Pikachu2000]

One other thought. Is this problem occurring only before the form has been submitted, or after the form has been submitted also?

[jakerow]

I use the form to remember which boxes are checked etc. but it doesn't give me anything from the database. It happens before and after though.

[Pikachu2000]

To stop the issue before submission, you need to make sure the form has been submitted before allowing the query to run. To do that, enclose everything after the include() in a conditional that checks whether the form has been sent. After that, see what errors are returned when the form is submitted, and post them along with the revised code.

 

<?php
include("connection.php");

if( strtolower($_SERVER['REQUEST_METHOD']) === 'post' ) {

// rest of code

}

[jakerow]

Hey Pikachu, allthough it didn't complete the problem it lead to some insights and it did finally work. Thanks a lot for your help!