PHP While Loop - Variable will passed on the last entry only - php

Question

When I press the Submit Button the Variable "rating" will only passed from the last line. The others are simply empty.
<form method="POST" action="<?php echo $_SERVER["PHP_SELF"]; ?>" enctype="multipart/form-data">
<table width="800" border="0" align="center" class="zui-table">
<thead>
<tr>
<td colspan="4">
</td>
</tr>
<tr>
<th width="96">Kategorie</th>
<th width="177">Unterkategorie</th>
<th width="299">Beitrag</th>
<th width="210"><? if ($rolle == '1') { echo 'Bewertung'; } elseif ($rolle== '9') { echo 'Freischalten';} else { echo 'Status';} ?></th>
<? while ($show = mysql_fetch_array($abfrage)) { ?>
</tr>
</thead>
<tbody>
<tr>
<td><? echo $show['category']; ?></td>
<td><? echo $show['subcategory']; ?></td>
<td><img src="timthumb.php?src=<?php echo $show['url'] ?>&w=300&h=150" alt="resized image" BORDER="2"/> </td>
<td><? if ($rolle == '1') { ?>
<select name="rating" class="style28" style="width: 120px" id="rating">
<option value="0">0</option>
<option value="5">5</option>
<option value="15">15</option>
<option value="20">20</option>
<option value="25">25</option>
<option value="30">30</option>
<option value="35">35</option>
<option value="40">40</option>
<option value="45">45</option>
<option value="50">50</option>
<option value="55">55</option>
<option value="60">60</option>
<option value="65">65</option>
<option value="70">70</option>
<option value="75">75</option>
<option value="80">80</option>
<option value="85">85</option>
<option value="90">90</option>
<option value="95">95</option>
<option value="100">100</option>
</select> <? echo '<button type="submit" value="' . $show['id']. '" name="submit">Bewerten?</button>'; }
elseif ($rolle == 9) {
echo '<button type="freigeben" value="' . $show['id']. '" name="freigeben">Freigeben?</button>';
}?></td>
</tr>
</tbody>
<? }
}?>
</table>
Only the "rating" in the last line will be passed to
if (isset($_POST['submit'])) {
echo $_POST['rating'];
echo $_POST['submit'];
//mysql_query("SET CHARACTER SET 'utf8'");
$db = mysql_connect("localhost", "xx", "xx");
mysql_select_db("cx",$db);
//Rating eingeben
$sql = "UPDATE wp_awa_upload SET count = count + '$_POST[rating]' WHERE id = '$_POST[submit]'";
$result = mysql_query($sql,$db);
//Session eintragen
$sql_insert = "INSERT into wp_awa_session (upload_id, user_id) VALUES('$_POST[submit]', '$_SESSION[id]')";
$result = mysql_query($sql_insert,$db);
echo $sql_insert;
echo $sql;
}
All the lines will be displayed, but when I press on the Submit button, then the variable from the rating dropdown will only be passed, when the button of the last line has been pressed.
Thanks for any help!
Kind Regards,
Stefan

Answer 1

If I understand what you try to achieve, you want the rating variable to contain the rating that corresponds to the button that was pressed. But if you think about it, there's no way for the browser to figure out which drop down the button corresponds to. It makes sense for you (because they appear to be next to each other), but HTML has no concept of "being next to each other", from perspective of HTML you have multiple drop downs, for some reason all with the same name, and multiple buttons. When you press one of the buttons, it needs to decide which value to send as the rating. It chooses to use the last one, because it has no way to guess better.
The simplest solution for you would be to actually make each drop down and a button to be its own form. Currently you have one huge form (see the <form> tag at the beginning of your script). Instead, open the form at the beginning of your while loop body and close at the end, along the lines of
<? while ($show = mysql_fetch_array($abfrage)) { ?>
</tr> <form method="POST" action="<?php echo $_SERVER["PHP_SELF"]; ?>" enctype="multipart/form-data">
...
This way the browser can figure out which drop down corresponds to which button, because for each button there's only one drop down in the corresponding form.