Question

G’day,
I’m having a problem getting the right value to post on a form I’ve made. It is passing the ID but not the name.
Here’s my code:

<td name="tech_userlogin" id="tech_userlogin" align="right" valign="top"><select name="tech_userlogin" id="tech_userlogin" class="db_field_name">

        <?php
        $sql="SELECT techID, tech_userlogin FROM technicians";
    $result=mysql_query($sql);

    $options="";

    while ($row=mysql_fetch_array($result)) {

        $techID=$row["techID"];
        $tech_userlogin=$row["tech_userlogin"];
        $options.="<OPTION VALUE=\"$techID\">$tech_userlogin</option>";
    }
    ?>

    <OPTION VALUE="">---Select---
    <?php echo $options; ?>
    </SELECT> 

The form posts to this page:

<?php
include 'sql_connect_R.inc.php';

$id = mysql_real_escape_string($_POST['jobID']); 
$equip = mysql_real_escape_string($_POST['wo_equip']);
$techID = mysql_real_escape_string($_POST['techID']);
$tech_login = mysql_real_escape_string($_POST['tech_userlogin']);

mysql_query("UPDATE work_orders SET wo_equip = UCASE('$equip'), wo_techID = '$techID', tech_userlogin = '$tech_login'
WHERE jobID = '$id'");

mysql_close($con);

I’m very new to PHP/MySQL and it took me a couple of days to get something that would put the tech_userlogin into a dropdown box. That is working, but when it posts the techID comes through on both $techID and $tech_userlogin.
Could someone please help me sort this out? Any help would be greatly appreciated.
Cheers, Spud

Answer 1

What seems to be happening is that you’ve named the select box “tech_userlogin”, but in the option tags the value is set to the techid. Since the value is what’s sent back in the $_REQUEST array, try something like this instead:

while ($row=mysql_fetch_array($result)) {
    $tech_userlogin=$row["tech_userlogin"];
    $options.="<OPTION VALUE=\"$tech_userlogin\">$tech_userlogin</option>";
}

I believe that will give you what you want.