error in displaying results.
 

hi,

this is a snippet of my codes,

<td><a href="editFunction.php?ID=<?php echo $contact['ID']; ?>">update</a></td>
after clicking "update", the ID will be passed to the editFunction.php page.

editFunction.php

<?php 

                //get variables from previous page

                $ID = $_GET['ID'];

                

                

                //sql statement to retrieve the commands 

                $sql = "SELECT * FROM contacts WHERE ID LIKE '".$ID."'";

                

                //execute query 

                $result = mysql_query($sql) or die (mysql_error());

                

                //display on table 

                $contact = mysql_num_rows($result);  {

?> 

.

......

<td width="140"><span class="style2">ID : </span></td>

    <td width="384">

          <input name="ID" type="ID" id="ID" value="<? echo $_GET['ID'] ?>" size="50" readonly />        </td>

  </tr>

  <tr>

    <td width="140"><span class="style2">First Name : </span></td>

    <td width="384">

      <input name="firstName" type="text" id="firstName" value="<? echo $contact['firstName'] ?>" size="50" />        </td>

  </tr>

</table>



<?php 

                        

            }

?>

</body>

</html>

why aren't my results displaying out in the respective fields except for ID?
where do i start debugging? Thanks alot.

Re: error in displaying results.
 

because contact contains the number of rows returned by the query.

change:

$contact = mysql_num_rows($result);

to

$contact = mysql_fetch_assoc($result);

Re: error in displaying results.
 

using this wll help you.
actually mysql_assoc is used to retrive resuts form

while($row = mysql_fetch_array($result, MYSQL_ASSOC))

{

    echo "Name :{$row['name']} <br>" .

         "Subject : {$row['subject']} <br>" . 

         "Message : {$row['message']} <br><br>";

}

Re: error in displaying results.
 

i don't think he needs the while loop because he is updating a specific row.

i use mysql_fetch_assoc() all the time. works perfectly. you can also use mysql_fetch_array, they complish the same thing.

Re: error in displaying results.
 

use mysql_fetch_array ..........
for mysql_num_rows........

Re: error in displaying results.
 

ah, thanks alot. PROBLEM SOLVED :)