Some Error ...

[techiefreak05]

I have a script that updates an image"headline" where the image id is .... what it is .. well heres the script:

[code]<h4>Your Uploaded Images</h4>
Here are the images you currently have. You can add a comment or headline to each image. A headline is a sentence that will display when a user views the image.<br>
<br>
<?php
// LINE 60 (see below) $sql = "SELECT * FROM `images` WHERE `userid` = '$_SESSION[id]'";
$query=mysql_query($sql);
while($row=mysql_fetch_array($query)){
$userID=$row['userid'];
$imgID=$row['id'];
$imgPATH=$row['imgPath'];
$headline=$row['headline'];
?>
<center>
<?php
if($_POST['update']){
echo "<b><font color=red>Comment Changed!</font></b><br><br>";
$sql = "UPDATE `images` SET `headline` = '$_POST[picComment]' WHERE `id` = '$_POST[imgID]' LIMIT 1";
$query = mysql_query($sql);
}
?>
<a href="imageView.php?imgID=<?php echo $imgID; ?>&userID=<?php echo $userID; ?>"><img src="<?php echo $imgPATH; ?>" width="250" height="250"></a><br>
<br>
<form action="" method="post">
      <table align="center" border="0" cellspacing="0" cellpadding="3">
      <input type="hidden" value="<?php echo $imgID; ?>" name="imgID">
      <br>
<tr><td ><font color=black>Add A Headline:</font></td><td><input type="text" name="picComment" maxlength="100" value="<?php echo $headline; ?>"></td></tr>
<tr><td colspan="2" align="right"><input type="submit" name="update" value="-Update-"></td></tr>
</table>
</form>
<hr width="55%">
</center>
<?php
}[/code]

everything is fine and dandy .. until i click "Update"  .. i mean it still updates the headline but it gives me this error alseo:
[code]Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\imageUpload.php on line 60[/code]

[akitchin]

first of all, you've identified line 60 incorrectly.  when looking for a line number, always kill wrapping, as this will artificially inflate the current line counter.  line 60 is actually the line that uses mysql_fetch_array(), as you might guess from the error.

when you get an invalid resource error, it's often because your query is failing and has not returned a legitimate resource ID.  easiest way to avoid these in the future is to add an or die() clause to all of your queries to make sure you're notified IMMEDIATELY if a MySQL query is failing:

[code]$query=mysql_query($sql) or die(mysql_error());[/code]

my guess is that you don't actually have anything in $_SESSION[id], or it's being echoed improperly.  add the or die() clause and MySQL should spit out a clear enough error.