How to update a database value with an imagebutton in php - php

Question

I am working on an php file, winch shows an table with values of the database. every row has an imagebutton at the end, which should change the status, displayed with an image.
so I think short function of it all must be, on click go to script go to php or jQuery?
the buttons also have to give the id of the user to the next script or function, or os there a better way?
here is my basic approach, with no functionality, for the moment
in index.php
include("function.php");
while ($row_user = mysql_fetch_assoc($result_user)) {
$sqle = mysql_query("
SELECT COUNT(*) FROM
places2user
WHERE
user = '".mysql_real_escape_string($row_user['ID'])."'
");
$res23 = mysql_fetch_array($sqle);
echo <table>
echo <tr>
echo "<td>".$row_user['ID']." </td><td><input type=\"image\" src=\"images/".htmlentities($row_user['receive'], ENT_QUOTES)."\" id=\"status\" name=\"status\" ></td>
echo </tr>
echo </table>
}
there is also a function.php, with a changestatus($id) function, which updates the database values, but I think, this is not the right way to code this.

Answer 1

While programming, don't rely on javascript only. I always create form using just html and css, make sure it works, and than add fancy stuff with jQuery or so. This way you know the form will also work without javascript (great for users without it).
But to come to your question:
if you use this, it should do what you want:
<?php echo "<form action='?' method='post'>
<table>
<tr>
<td>".$row_user['ID']." </td>
<td><input type='hidden' name='userId' value='".$row_user['ID']."'/>
<input type=\"image\" src=\"images/".htmlentities($row_user['receive'], ENT_QUOTES)."\" id=\"status\" name=\"status\" ></td>
</tr>
</table></form>";