mysql_fetch_array

[dazzclub]

Hi guys and girls..

 

I was originally having a problem with something else which made me want to view my site in note pad from the view source in the browser...I was reading the code and i had noticed this..

 

<br />

<b>Warning</b>:  mysql_fetch_array(): supplied argument is not a valid MySQL result resource in <b>C:\wamp\www\darrensPAINT\density-graphs.php</b> on line <b>94</b><br />

</select>

 

I was surpised i had not seen it in the browser first..anyway..

 

I tracked it down to line 94 and looked at the lines previous to help me spot the problem..I then read a tutorial by Ben Smithers  on,  "Debugging: A Beginner's guide" on phpfreaks and i tried to echo out the problem, as this was suggested to be a good method but i think i messed up somewhere.

 

here is the code thats causing the error, along with my attempt at echoing out the problem

//third dropdown list

echo "<select name='time'><option value=''>Select time</option>";

while($selectiontime= mysql_fetch_array($querytime) or trigger_error(mysql_error().'<br />Query was:'.$selectiontime);

echo '<br />'.Query was:'.$selectiontime.'<br />';

 

If anyone can help, that would be great

 

cheers

Darren

 

 

 

[xtopolis]

$selectiontime is an array that holds the results of the resource $querytime has from mysql.

 

You need to echo the same $variable that is in the "mysql_query($variable)" area in order to debug to see that what you're asking mysql for.

Also, it would help us to have more code, specifically what sql you're sending.