[SOLVED] Listmenu value to variable

[biggieuk]

Hi,

 

Im having trouble assigning the value of a listmenu item to a variable.

 

the listmenu is populated dynamically from a database using 'username' as the label and 'email' as the value.

 

I need to assign the 'email' value of the selected item to a variable.

 

Can anybody please give me some guidance.

 

Thanks

[trq]

Can you post your code?

[biggieuk]

Your not going to like it, i used Dreamweaver to dynamically populate the listmenu 

 

<select name="select2" id="select2" disabled = "disabled">
    <?php
do {  
?>
    <option value="<?php echo $row_users['email']?>"<?php if (!(strcmp($row_users['email'], $row_users['email']))) {echo "selected=\"selected\"";} ?>><?php echo $row_users['username']?></option>
    <?php
} while ($row_users = mysql_fetch_assoc($users));
  $rows = mysql_num_rows($users);
  if($rows > 0) {
      mysql_data_seek($users, 0);
  $row_users = mysql_fetch_assoc($users);
  }
?>
  </select>

[trq]

Obviously your axpecting more than one email? If so, you'll need to store them in an array. eg;

 

<?php
do {  
?>
    <option value="<?php echo $row_users['email']?>"<?php if (!(strcmp($row_users['email'], $row_users['email']))) {echo "selected=\"selected\"";} ?>><?php echo $row_users['username']?></option>
    <?php
} while ($row_users = mysql_fetch_assoc($users));
  $rows = mysql_num_rows($users);
  if($rows > 0) {
      mysql_data_seek($users, 0);
      $row_users = mysql_fetch_assoc($users);
      $emails[] = $row_users['email'];
  }
?>

[biggieuk]

thanks, that worked great.