Re: Error Question!

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Matt,

  You may want to watch out for is sql injection attacks. Just to be on
the safe side consider something like:
$projects = "select name, type, location, date_added from uploads where
user_id = '" . mysql_real_escape_string($_SESSION['user_id'], $conn) .
"'";

The reason for the error is $project_results != $projects_results.

Thanks,
James


On Fri, 2008-05-23 at 13:42 -0400, Matthew Gonzales wrote:
> Hello Everyone,
> 
> I keep getting this error: *Warning*: mysql_fetch_array(): supplied 
> argument is not a valid MySQL result resource in 
> *C:\wamp\www\login\testroom\upload.php* on line *107, *for this 
> particular area of code.
> 
> 
> Code:
> 
> //Check to see what files they have uploaded in the past
>     $projects = "select name, type, location, date_added from uploads 
> where user_id = '" . $_SESSION['user_id'] . "' ";
>     $projects_results = mysql_query($projects, $conn) or die(mysql_error());
>    
>     //Create table for projects
>     $display_block = '<table>
>                             <tr>
>                                 <th>Project Name</th>
>                                 <th>Project Type</th>
>                                 <th>View Project</th>
>                                 <th>Project Created</th>
>                             </tr>';
>                            
>     //Extract data to table
>     while ($project_data = mysql_fetch_array($project_results))  * Error 
> occurs here!*
>     {
>         $project_name = $project_data['name'];
>         $project_type = $project_data['type'];
>         $project_link = $project_data['loction'];
>         $project_date = $project_data['fmt_date_added'];
>        
>         //insert project data   
>         $display_block .='
>                             <tr>
>                                 <td class="heading">' . $project_name . 
> '</td>
>                                 <td class="heading">' . $project_type . 
> '</td>
>                                 <td class="heading"><a href="' . 
> $project_link . '">View</a></td>
>                                 <td> class="heading">' . $project_date . 
> '</td>
>                             </tr>';
>     }
>    
>     //Close display block
>     $display_block .='</table>';
> 
> Does anyone have any idea what is going on. I have used a similar code 
> liek this before and it worked. I can output the mysql string to the 
> browser and then copy and paste it to PHPmyadmin and do an sql query and 
> it comes back with the correct results. I don't know what is up.
> 
> Thanks!
> 
> Matt G
> 
> **
> -- 
> Matthew Gonzales
> IT Professional Specialist
> Enterprise Information Technology Services
> University of Georgia
> Email: matt323@xxxxxxx <mailto:matt323@xxxxxxx>
> Phone: (706)542-9538
> 


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