I made a little code to update record in database like this:
[PHP]
require_once('includes/config.php');

if ((isset($_POST["DO_update"])) && ($_POST["DO_update"] == "edit_product")) {

$ProductID = $_POST["txtProductID"];
$CategoryID = $_POST["txtCategoryID"];
$ProductName = $_POST["txtProductName"];
$ProductPrice = $_POST["txtProductPrice"];
$IsRebate = $_POST["selIsRebate"];
$RebatePrice = $_POST["txtRebatePrice"];
$ImageFileName = $_POST["txtImageFileName"];
$BigImageFileName = $_POST["txtBigImageFileName"];
$ProductDetails = $_POST["txtProductDetails"];

$query_update = "UPDATE ks_products SET `prod_name` = '$ProductName'";
$query_update .= ", `prod_price` = '$ProductPrice'";
$query_update .= ", `prod_is_rebate` = '$IsRebate'";
$query_update .= ", `prod_rebate_price` = '$RebatePrice'";
$query_update .= ", `prod_image` = '$ImageFileName'";
$query_update .= ", `prod_big_image` = '$BigImageFileName'";
$query_update .= ", `prod_details` = '$ProductDetails'";
$query_update .= ", `cat_id` = '$CategoryID'";
$query_update .= "WHERE `prod_id` = '$ProductID' LIMIT 1";

mysql_select_db($database_index, $index);
mysql_query($query_update, $index) or die(mysql_error());

}
[/PHP]
No error found but the record cannot be updated, plz help me!
Only prod_price, prod_is_rebate and prod_rebate_price column is mediumint, others is varchar.

PHP version 5.0.4, MySQL version 4.1

Remove the ' ' quote from your script will do
[PHP]$query_update = "UPDATE ks_products SET
prod_name = '$ProductName',
prod_price = '$ProductPrice',
prod_is_rebate = '$IsRebate',
prod_rebate_price = '$RebatePrice',
prod_image = '$ImageFileName',
prod_big_image = '$BigImageFileName',
prod_details = '$ProductDetails',
cat_id = '$CategoryID'
WHERE prod_id = '$ProductID' LIMIT 1";[/PHP]

Thanks for your help but it didn't work :cry:

I found the bug and fixed it. Thanks for your help :cheesy:

What error message you get?
Make sure you connect to the database properly.
Make sure your database got the variables you use.

You may want to add echo $ProductDetails; after
$ProductDetails = $_POST["txtProductDetails"];
to see if the POST is actually working.

Yes, you are right. I found that the $_POST['txtProductID'] didn't get value from submit form.

Thanks for your advice.