PHP/MYSQL. Updating multiple rows failing in the `for loop` - php

Question

I have following HTML table:
<form method="post" action="update-table.php">
<table class="table-data" style="width: 960px;">
<thead>
<tr>
<td class="sorting" rowspan="1" colspan="1" style="width: 193px;">ID</td>
<td class="sorting" rowspan="1" colspan="1" style="width: 54px;">Live</td>
</tr>
</thead>
<tbody>
<tr class="odd">
<td class="nth-1"><input type="text" value="12" name="id"></td>
<td class="nth-2"><input type="checkbox" checked="checked" name="live"></td>
</tr>
<tr class="even">
<td class="nth-1"><input type="text" value="11" name="id"></td>
<td class="nth-2"><input type="checkbox" checked="checked" name="live"></td>
</tr>
<tr class="odd">
<td class="nth-1"><input type="text" value="10" name="id"></td>
<td class="nth-2"><input type="checkbox" checked="checked" name="live"></td>
</tr>
</tbody>
</table>
<input type="submit" />
and I'm trying to update live values accordingly to the ids with this file:
<?php
# update
session_name('users');
session_set_cookie_params(2*7*24*60*60);
session_start();
define('INCLUDE_CHECK',true);
require 'connect.php';
require 'functions.php';
if(!$_SESSION['id']) {
header ("Location: index.php");
}
//Function to sanitize values received from the form. Prevents SQL injection
function clean($str) {
$str = #trim($str);
if(get_magic_quotes_gpc()) {
$str = stripslashes($str);
}
return mysql_real_escape_string($str);
}
//Sanitize the POST values
$id = clean($_POST['id']);
//$usr2 = $_SESSION['usr'];
$live = (isset($_POST['live']))?1:0;
//$updated = date("F j, Y, g:i a",time()+60*60);
//Create INSERT query
foreach ($display_order as $id => $live) {
$sql = "UPDATE news SET display_order = $live WHERE id = $id";
$result = mysql_query($sql);
if($result) {
//header("location: notes.php");
//exit();
print_r($sql);
print_r($display_order);
}else {
die("Query failed");
}
}
?>
But I'm getting an error:
Warning: Invalid argument supplied for foreach() in C:\Documents and Settings\USER\Desktop\Dropbox\wamp_at_work\update-table.php on line 33
Line 33 is: foreach ($display_order as $id => $live) {
What the problem? Any suggestions much appreciated.

Answer 1

The problem is that you're duplicating the 'name' attributes of the input elements. What you want to do is send arrays back to the server, ie:
Updated with extended answer:
<?php
$inp_orders = !empty($_POST['live']) ? $_POST['live'] : array();
// proccess post
if ($inp_orders)
{
foreach ($inp_orders as $id => $live) {
$id = (int) $id;
$live = (bool) $live;
// update orders
$sql = "UPDATE news SET display_order = $live WHERE id = $id";
$result = mysql_query($sql);
if($result) {
echo 'News updated';
}else {
die("Query failed");
}
}
}
// get orders from db
$res = mysql_select("SELECT * FROM news");
$view_orders = array();
while ($row = mysql_fetch_assoc($res))
{
$view_orders['id'] = $row['id'];
$view_orders['live'] = $row['live'];
}
?>
<form method="post" action="">
<table class="table-data" style="width: 960px;">
<thead>
<tr>
<td class="sorting" rowspan="1" colspan="1" style="width: 193px;">ID</td>
<td class="sorting" rowspan="1" colspan="1" style="width: 54px;">Live</td>
</tr>
</thead>
<tbody>
<?php foreach ($view_orders as $order): $id = (int) $order['id']; $odd = true; ?>
<tr class="<?php echo $odd ? 'odd' : 'even' ?>">
<td class="nth-1"><?php echo $id ?></td>
<td class="nth-2"><input type="checkbox" <?php echo $order['live'] ? 'checked="checked"' : '' ?> name="live[<?php echo $id ?>]" /></td>
</tr>
<?php $odd = $odd ? false : true; endforeach; ?>
</tbody>
</table>
<input type="submit" />
</form>

Answer 2

I'm going to assume people can not change the ID. So do this:
<td class="nth-2">
<input type="hidden" name="live[12]" value="0">
<input type="checkbox" name="live[12]" value="1" checked>
</td>
That will get you an array called $_POST['live'] with either 0 or 1 in it depending on if they clicked it.
The hidden field is because if they don't click it nothing at all is sent, which can be more difficult to parse, so first I send a 0, then overwrite it if the checkbox is checked.
If people can change the ID you'll need to modify it a bit. Leave a comment if so.