mysql_fetch_assoc() error

[toddmarx]

Hello!

I am getting this error message when loading this page:

Warning: mysql_fetch_assoc(): 2 is not a valid MySQL result resource in c:\program files\easyphp\www\computerzonehq\admin\admin3\vfeedback.php on line 32

Here is the code.

<?php require_once('../../Connections/mysqlconnect.php'); ?>
<?php
mysql_select_db($database_mysqlconnect, $mysqlconnect);
$query_ViewFeedback = "SELECT * FROM feedback ORDER BY id ASC";
$ViewFeedback = mysql_query($query_ViewFeedback, $mysqlconnect) or die(mysql_error());
$row_ViewFeedback = mysql_fetch_assoc($ViewFeedback);
$totalRows_ViewFeedback = mysql_num_rows($ViewFeedback);

mysql_free_result($ViewFeedback);
?>
<table width="752" border="0" cellpadding="0" cellspacing="0">
  <!--DWLayoutTable-->
  <tr> 
    <td width="752" height="145">&nbsp; <table border="1">
        <tr> 
          <td><div align="center">id</div></td>
          <td><div align="center">lname</div></td>
          <td><div align="center">fname</div></td>
          <td><div align="center">email</div></td>
          <td><div align="center">state</div></td>
          <td><div align="center">comment</div></td>
        </tr>
        <?php do { ?>
        <tr> 
          <td><?php echo $row_ViewFeedback['id']; ?></td>
          <td><?php echo $row_ViewFeedback['lname']; ?></td>
          <td><?php echo $row_ViewFeedback['fname']; ?></td>
          <td><?php echo $row_ViewFeedback['email']; ?></td>
          <td><?php echo $row_ViewFeedback['state']; ?></td>
          <td><?php echo $row_ViewFeedback['comment']; ?></td>
        </tr>
        <?php } while ($row_ViewFeedback = mysql_fetch_assoc($ViewFeedback)); ?>
      </table>
</td>
  </tr>
</table>

Line 32 is:

<?php } while ($row_ViewFeedback = mysql_fetch_assoc($ViewFeedback)); ?>

I am having a hard time with this one. This is straight out of DreamweaverMX. I have never seed it mess up a code like this before. I have looked at other codes that I have that have the same process and I dont see what is wrong.

BTW, I have tried to echo $ViewFeedback and this is what I get:
(Resource id #2) = ViewFeedback
I thought that this was kind of odd.....

Thanks for all your help

[barand]

Might have something to do with your

mysql_free_result($ViewFeedback);

[toddmarx]

like?

[tuf-web_co_uk]

changing your code to this should fix the problem

<?php require_once('../../Connections/mysqlconnect.php'); ?>
<?php
mysql_select_db($database_mysqlconnect, $mysqlconnect);
$query_ViewFeedback = "SELECT * FROM feedback ORDER BY id ASC";
$ViewFeedback = mysql_query($query_ViewFeedback, $mysqlconnect) or die(mysql_error());
$row_ViewFeedback = mysql_fetch_assoc($ViewFeedback);
$totalRows_ViewFeedback = mysql_num_rows($ViewFeedback);

while ($row_ViewFeedback = mysql_fetch_assoc($ViewFeedback)) {
?>
<table width="752" border="0" cellpadding="0" cellspacing="0">
  <!--DWLayoutTable-->
  <tr> 
    <td width="752" height="145">&nbsp; <table border="1">
        <tr> 
          <td><div align="center">id</div></td>
          <td><div align="center">lname</div></td>
          <td><div align="center">fname</div></td>
          <td><div align="center">email</div></td>
          <td><div align="center">state</div></td>
          <td><div align="center">comment</div></td>
        </tr>
                <tr> 
          <td><?php echo $row_ViewFeedback['id']; ?></td>
          <td><?php echo $row_ViewFeedback['lname']; ?></td>
          <td><?php echo $row_ViewFeedback['fname']; ?></td>
          <td><?php echo $row_ViewFeedback['email']; ?></td>
          <td><?php echo $row_ViewFeedback['state']; ?></td>
          <td><?php echo $row_ViewFeedback['comment']; ?></td>
        </tr>
        <?php } ?>
      </table>
</td>
  </tr>
</table>

[tuf-web_co_uk]

i would recommend putting th last

</table></td></tr></table>

before the end of the while statement <?php } ?>

[barand]

Like ?

Like line 9 where you free the result that you are trying to use on line 32.

[toddmarx]

barand,

I know where you are talking about. But you say something is wrong. So what is wrong? I looked at, and to me it looks like a normal query. I am not an sql guru. Please be a little more specific.

thanks

[toddmarx]

tuf-web.co.uk,

That fixed the error messages, but the table will only show 1 record. I was looking at making it repeat. I have never tried to make a formated table repeat so I dont know what I am doing wrong. 🙂😃

[barand]

tuf-web

Your revised code is fetching then ignoring the first record (immediately after the query)

Also, why repeat the table tags and header rows for every row of the query?

<?php require_once('../../Connections/mysqlconnect.php'); ?>
<?php
mysql_select_db($database_mysqlconnect, $mysqlconnect);
$query_ViewFeedback = "SELECT * FROM feedback ORDER BY id ASC";
$ViewFeedback = mysql_query($query_ViewFeedback, $mysqlconnect) or die(mysql_error());
// $row_ViewFeedback = mysql_fetch_assoc($ViewFeedback);
$totalRows_ViewFeedback = mysql_num_rows($ViewFeedback);


?>
<table width="752" border="0" cellpadding="0" cellspacing="0">
  <!--DWLayoutTable-->
  <tr> 
    <td width="752" height="145">&nbsp; <table border="1">
        <tr> 
          <td><div align="center">id</div></td>
          <td><div align="center">lname</div></td>
          <td><div align="center">fname</div></td>
          <td><div align="center">email</div></td>
          <td><div align="center">state</div></td>
          <td><div align="center">comment</div></td>
        </tr>
<?php  

while ($row_ViewFeedback = mysql_fetch_assoc($ViewFeedback)) {
?> 
        <tr> 
          <td><?php echo $row_ViewFeedback['id']; ?></td>
          <td><?php echo $row_ViewFeedback['lname']; ?></td>
          <td><?php echo $row_ViewFeedback['fname']; ?></td>
          <td><?php echo $row_ViewFeedback['email']; ?></td>
          <td><?php echo $row_ViewFeedback['state']; ?></td>
          <td><?php echo $row_ViewFeedback['comment']; ?></td>
        </tr>
        <?php } ?>
</table>
<?php mysql_free_result ($ViewFeedback) ?>

[barand]

toddmarx

How much more specific can I be?

On line 9 you have have freed the query resource that you are attempting you use on line 32. It doesn't exist anymore once you've freed it!

hth

[toddmarx]

I am sorry that I do not possess your infinite wisdom barand. I was simply asking for a little help when all you said was:

....Might have something to do with your

mysql_free_result($ViewFeedback)......

All I wanted to know is what was wrong. That is great that you told me where my code might be giving me problems. What kind of problems? But, if it were as obvious to me, as it is/was to you, then I would not have wasted time in this post to start with. I am new to php/mysql. When all you said was to take a look at a certain line of code. I did. I could not tell what was wrong with it.

All you seemed to accomplish was point out to me that I did not know what I was doing. Ok, good job! I dont know what I am doing. So what? That is why I am here asking questions trying to learn. I am doing this as a favor to a friend. My area of expertise is not php/mysql.

Thank you for your help,
Todd 😃