如何将ID作为值和名称显示在dropdownlist php中

Question

i have a problem with that i need to get the value from the dropdown list to be a number and the name for a kategory to be the name that the user picks.

<select name="kategori">
<?php
    $query=mysql_query("SELECT KategoriID from Kategori");
    $second=mysql_query("SELECT KategoriNavn from Kategori");
    while($r=mysql_fetch_row($query) && $v=mysql_fetch_row($second)){
        echo "<option value='$r[0]>$v[0]</option>";
    }
?>

This is the code i have, but i cant make it to work. Im kinda new to PHP. Thanks!

Answer 1

There's no need to write two different queries. You could have written just a single one. I think mysql fetch_assoc is a tad easier to understand.

You can try something like this:

  <?php

     $query = mysql_query("SELECT KategoriID, KategoriNavn  from Kategori") or die(mysql_error()); // Debugging displays SQL syntax errors, if any.

     echo "<pre>";
     print_r($query);
     exit;              // Let me know what the array looks like.

     while ($r= mysql_fetch_assoc($query)) { ?>

   <option value=<?php echo $r['KategoriID']; ?> > 
      <?php echo $r['KategoriNavn']; ?>
   </option>

   <?php } ?>

   <?php 

         echo "<pre>";
         print_r($_POST); // Do this where you're checking your POST data
         exit;
     ?>

Assuming you want option value to be KategoriNavn and the option to display to be KategoriID.

Hope this helps.

Peace! xD