Matthew,
It seems, from looking at your supplied code, that you have a variable-naming
typographic error.
The variables have to match:
$projects_results = mysql_query($projects, $conn) or die(mysql_error());
$project_data = mysql_fetch_array($project_results))
--
Jeff White is:
[EMAIL PROTECTED]
-------------- Original message --------------
From: Matthew Gonzales <[EMAIL PROTECTED]>
> 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
> $display_block = '
>
> Project Name
> Project Type
> View Project
> Project Created
>
';
>
> //Extract data to table
> while ($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 .='
>
> ' . $project_name .
> '
> ' . $project_type .
> '
> View
> class="heading">' . $project_date .
> '
>
';
> }
>
> //Close display block
> $display_block .='';
>
> 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: [EMAIL PROTECTED]
> Phone: (706)542-9538
>
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