change an image via php dropdown - php

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

Try this, I think this is what you are looking for
If you want to do something like this you must use Ajax. Here you go for the link that helps you to understand about Ajax.
http://www.w3schools.com/php/php_ajax_database.asp
Note:
If you wanted it to be only PHP without Javascript, you would have to sacrifice the 'must not refresh' constraint, as the only way to submit the form is by pressing the button, and submitting the content.

Answer 2

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" />

Answer 3

Here is a short example which implements java scrip and php where i update src of an image based on the id from the select you might want to change with specific src based on that id
<?php
include ("conned-db.php");
$item = $_GET["imageid"];
if ($item == "")
{
$item = 1;
}
$result = mysql_query("SELECT * FROM gallery")
or die(mysql_error());
?>
<select id='gallery_id' name='gallery_id' onChange="window.location='file.php?imageid='+this.value" style='width:200px;' >
<?
while($row = mysql_fetch_array( $result ))
{
echo '<option value=' . $row['id'] . '>';
echo $row['gallery_name'];
echo '</option/>';
}
?>
</select>
</td>
<?
echo "<td colspan='2' rowspan='2'>";
?>
<img src=' <?=$item?> '/></td>