Hooch Posted July 17, 2009 Share Posted July 17, 2009 Hey all.The following script worked in PHP4, but I cannot see the problem for it not to work in PHP5Here is the error... Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/public_html/admin_cats.php on line 74 if(isset($_GET['cat'])) { $catTbl = $_GET['cat']; $query = mysql_query("SELECT * FROM `cat" . $cat . "`"); //Show the new button 1st echo ' <span class="txtbox-small"> <a href="admin_cats.php?new_cat=' . $catTbl . '" class="light-10-link">New Category</a> </span>'; echo ' for ' . $catLevel; echo '<br><br>'; while($r = mysql_fetch_array($query)) { echo '<div class="left_box_100">'; echo '<a href="admin_cats.php?edit_cat=' . $r['id'] . '&catTbl=' . $cat . '" class="light-14-link">Edit</a>'; echo ' <a href="admin_cats.php?delete_cat=' . $r['id'] . '&catTbl=' . $cat . '" class="light-14-link">Delete</a>'; echo '</div>'; echo '<div class="right_box"> ' . stripslashes($r['cat']) . '</div>'; echo '<hr>'; } } Line 74 is while($r = mysql_fetch_array($query)) Thank you for your time and help Link to comment Share on other sites More sharing options...
Hooch Posted July 17, 2009 Author Share Posted July 17, 2009 I found the problem $query = mysql_query("SELECT * FROM `cat" . $cat . "`"); " . $cat . "seems to be the culprit.This has to be a variable for the code to work.I'll post back if I fix the error.In the meantime I will keep checking for suggestions.ty**EDIT**Found the error.$cat should be $catTblI am so worried about the transfer to 5 I am expecting big problems when it'sthe normal typos.Thanks for looking anyway. Link to comment Share on other sites More sharing options...
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