Question

What I am trying to do is change the image when an item is selected from the drop down. This is part of a form so I cant have the value change. However the option value is the row id, that row would also contain the target for the image. But because the target ‘file’ is called outside the loop it isn’t firing.

I read I have to call it within the loop first but can’t get it to work. Could you look at the code below and throw me a hint?

Thanks

    <?php

include ("conned-db.php");

$result = mysql_query("SELECT * FROM gallery") 
or die(mysql_error());  

echo "<select id='gallery_id' name='gallery_id' style='width:200px;' >";


while($row = mysql_fetch_array( $result ))
{
echo '<option value=' . $row['id'] . '>';
echo $row['gallery_name'];
echo '</option/>';
}
echo "</select>";
echo "</td>";
echo "<td colspan='2' rowspan='2'>";
echo '<img src=' .$row['file']. '/></td>';

    ?>

Answer 1

This should work too. If the file locations of the images are available at the time you load the page using ajax is not a must. You have to use ajax if you need to query the server again to retrieve the required file location. The following code assumes that you have the location of the images for each item of the dropdown list at the time you load the page.

<select id='gallery_id' name='gallery_id' style='width:200px;' 
  onchange='document.getElementById("image").src=this.options[this.selectedIndex].title' >
 <?php
 while($row = mysql_fetch_array( $result ))
 {
  ?>
   <option value='<?php echo $row["id"]; ?>' title='<?php echo $row["file"]; ?>'> 
        <?php echo $row["gallery_name"]; ?>
   </option>
  <?php
  }
  ?>
<img id="image" />