Ninjakreborn Posted May 13, 2007 Share Posted May 13, 2007 <?php $category = mysql_real_escape_string($_GET['category']); $select = "SELECT * FROM embarrassing WHERE category = '$category' AND approved = 'yes'"; while ($row = mysql_fetch_array($select)) { echo '<a href="embarrassing2.php?id=' . $row['id'] . '" title="' . stripslashes($row['title']) . '">' . stripslashes($row['title']) . '</a>'; echo '<br />'; } echo mysql_error(); ?> That is returning Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /homepages/################# on line 31 Any advice on what could be wrong with that query. The data is passing properly, very strange... Any advice? Quote Link to comment https://forums.phpfreaks.com/topic/51239-solved-error/ Share on other sites More sharing options...
AndyB Posted May 13, 2007 Share Posted May 13, 2007 Glad it's solved, but do you ever think of error trapping your queries? That would have pointed out the solution in a flash. Quote Link to comment https://forums.phpfreaks.com/topic/51239-solved-error/#findComment-252411 Share on other sites More sharing options...
Ninjakreborn Posted May 14, 2007 Author Share Posted May 14, 2007 I use to, but I sort of ended up stopping. I suppose it's better to, but I ended up just falling out of habit over time, because I don't normally get them wrong (I was in a hurry when I created that one). Most of the time they get right the first time. If they don't I normally do error handling on them. Quote Link to comment https://forums.phpfreaks.com/topic/51239-solved-error/#findComment-252545 Share on other sites More sharing options...
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