I cleaned up the code a bit and although I don't know much about setting up the arrays in a html form I believe the complete code should look more like the following:
<strong>Update multiple rows in...

That code in it's design is flawed. The following line will cause a lot of trouble:
$sql1="UPDATE $tbl_name SET AcMake='$AcMake[$i]', AcCapacity='$AcCapacity[$i]', AMP='$AMP[$i]', qty='$qty[$i]'...

The following should do the trick the the result being in the $res array
<?php
$var=$_POST['box'];
$i=1;
$res=array();
$var=(empty($var))?array():$var;
foreach ($var AS $val) {
...

I see how that can be handy but another thing I have noticed with sites like google and youtube is that they seem to redirect users to their country code domain meaning there is a separate database...

Just a note on that. Wikipedia receives 1,000 new articles every day and many more edited articles so it is possible to compare Wikipedia to ever changing websites such as Facebook. I guess it is a...

Are you at all familiar with the mediawiki cms. That is the same cms wikipedia uses and it does not spread into millions of tables. Instead it has hundreds of computers that just process the...

Well actually I have been told in the mysql forum that if you design a table so that it has indexing you shouldn't ever need multiple tables for the one job. That is when designed properly because...

Maybe my explanation was too short. Sometimes people will do the following:
<?php
$connection = mysql_connect("myserver.com","myuser","mypassword") or die ("Couldn't connect to server.");
$db =...

So it displayed something... So we know that you were using the right names but the normal cause of this is an extra line of code. Try the following script.
<?php
$connection =...

To test if no rows are found try the following:
while( $data2 = mysql_fetch_array($query)) {
echo $data2[0];
echo $data2[1];
echo $data2[2];
}
The nothing is displayed then that means there...

Replace lines 11 to 13 with the following
echo $data2['firstname'];
echo $data2['lastname'];
echo $data2['task'];

Just to make it clear - indeed there should be no echos before the header().

That is correct from what I have read in the past.

Perhaps this:
<?php
mysql_connect(HOST,USER,PASS);
@mysql_select_db("db303278079") or die( "Unable to select database");
$query="SELECT * FROM restaurants";
$result=mysql_query($query);...

Well actually the entire page will be process by php unless the exit; statement is placed there. So for example if you have a mysql insert it will insert before redirecting unless the exit; is placed...

If you want wildcards then the following contains the wildcards:
$sql = 'SELECT * FROM device WHERE `device_num` LIKE "%'.mysql_real_escape_string($val_d).'%" OR `dib` LIKE...

<?php
header("Location: http://yoursite.com/index.php");
exit;
//above must be at very top before any browser output

Try adding the following to the top of your php file(s)
setcookie ('PHPSESSID', $_COOKIE['PHPSESSID'], time()+172800);

Try replacing it with the following:
$sql = 'SELECT * FROM device WHERE `device_num` LIKE "'.mysql_real_escape_string($val_d).'" OR `dib` LIKE "'.mysql_real_escape_string($val_dp).'"';

I would have to say if you managed to get XAMPP working then good, keep it as XAMPP has a few minor additional features that WAMP doesn't have (eg. mailserver) although the installation for XAMPP is...

Try the following:
<?php
require_once('config.php');

for ($i=1;$i<11;$i++) {
$aid='aid'.$i;
$$aid=$_POST['Question-'.$i];

$qid='qid'.$i;
$$qid=$_POST['qid'.$i];

You may want to note that for assigning a possibly undefined variable or array index/key the following is the correct method.
$PK = (isset($_POST['PK']))?$_POST['PK']:'';
The above is what line 24...

Your links are all pointing to the wrong location and should be something like as follows: <?php
//REMEMBER TO CONNECT TO DATABASE!

$show_all=$_POST['rg01']; // Show All
//echo $show_all;
...

Why won't that work. Is there something in the included library that prevents it working. The only thing I would suggest is escaping the variable $key like the following:$zips =...

From my experience yes it is.

I entered in the specs of the plan I'm on with my host and it's over double the price. If there was something like that where you can decide cpu, storage space and bandwidth for a third of the price...

To retrieve the variable param from the url simply use $_GET['param'] to get its value.

I did a google search and the following link might be of great interest to you. http://www.m6.net/ Enjoy.

That part of the syntax is totally wrong. I don't think that an include is not meant to be in an if statement. Also replace the include with the require function which does the same thing but with...

Try rebooting the apache httpd service. I've noticed on windows rebooting the computer doesn't restart the httpd service so you will need to go into your controll panel to restart the apache httpd...

Well a server can have several php.ini files and only one of them are really used. So make sure the php.ini file you changed is the same one as listed when you use the following code:
<?php...

Perhaps the following:
//get username
$username='Bill Gates';
//validate
$username=mysql_real_escape_string($username);
//set the sql query
$result=mysql_query('SELECT * FROM `table` WHERE...

What's the problem or is this solved?

In future please use code tags as this is not your first post. Also the following code should do the trick:
$category_id=mysql_real_escape_string(stripslashes($_POST['category_id']));...

In Ubuntu. ok. I found this problem when I first got a Virtual Private Server. The server had a different file structure than what my Ubuntu did. So try to browse around your server to make sure...

No error. That's because you didn't echo the exec function. Line 18 should be as follows:
echo exec($createBackup);

Try replacing line 11 with the following:
$file[] = str_replace(" ","_",$_POST[$files1]);

Try on line 34 the following code and tell us the results:
echo '<xmp>'; print_r($file_names); echo '</xmp>';
That should dump the keys and values of the array. They don't always have numbers...

I believe it is the break; code in the switch statement. That is why I suggested replacing it with an if statement. But obviousley there is something fishy going on with the $returnCode variable.

Weird. Should have worked. :?:
Below is another piece of code you can try and hopefully it should work.
$returnCode = (int)curl_getinfo($ch, CURLINFO_HTTP_CODE);
if ($returnCode!=='200' ||...