issaru07

i am trying to do credit card validation. the card details are entered n creditcard.php and process in creditcard-process.php and this is the error for creditcard-process.php.. can i check whats wrong with creditcard-process.php ? Notice: Use of undefined constant ccnum - assumed 'ccnum' in D:\xampp\htdocs\book\creditcard-process.php on line 109 Notice: Undefined …

can i check is there anything wrong with my if else, if seems like it will only go banktransfer.php instead. [CODE] if ($_POST[Payment] = "Bank Transfer") { $btpage = "banktransfer.php"; echo '<script language="javascript" type="text/javascript">window.location.href="'.$btpage.'";</script>'; } else if ($_POST[Payment] = "Credit Card") { $ccpage = "creditcard.php"; echo '<script language="javascript" type="text/javascript">window.location.href="'.$ccpage.'";</script>'; } …

i know this is a simple error but i just couldnt spot the error and i run the query seems fine. really appreciated... pleasee help Parse error: syntax error, unexpected '=' in D:\xampp\htdocs\book\admin-order-details.php on line 16 [CODE] <?php $objConnect = mysql_connect("localhost","root","boon84") or die(mysql_error()); $objDB = mysql_select_db("db-test"); $id = $_GET['id']; strSQL …

i got this error when i trying to submit a form.. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'order(firstname,lastname,email,address,city,postalcode,country,totalprice,paymen' at line 1 appreciate if you could assist me.. [CODE] $sql2= mysql_query("INSERT INTO …

i have to run 2 queries for inserting 2 different tables userorder table contains orderid , title order_details containts orderid, qty so how can i get orderid for my order_details? the code below donts seems able to get last orderid [CODE] $query = mysql_query("SELECT last_insert_id()"); list($oid)=mysql_fetch_row($query); [/CODE]

i actually trying to run 2 select queries and get user id from members3 table and productid, title,price, categoryname and img from product4 tables. i quite noob in php so not too sure whats went wrong. there is no error msg show just that my database data only have userid …

can i check is there any wrong with this? it prompted me error in line 5 Parse error: syntax error, unexpected T_IF [CODE] <?php session_start() if (!(isset($_SESSION['email']) && $_SESSION['email'] != '')) { header ("Location: login.php"); } [/CODE]

hi everyone, keep getting this error although it shows the result perfectly..i cant do a $cart= ''; as it will show the incorrect result Notice: Undefined variable: cart in D:\xampp\htdocs\book\includes\functions.inc.php on line 8 [CODE] <?php function writeShoppingCart() { if(isset($_SESSION['cart']) ) { $cart = $_SESSION['cart']; } if (!$cart) { return '<p>You …

the output of my page is correct yet it shows Notice: Undefined variable: total in D:\xampp\htdocs\book\includes\functions.inc.php on line 36 is there any ways to resolve this? thanks [CODE] <?php function writeShoppingCart() { $cart = $_SESSION['cart']; if (!$cart) { return '<p>You have no items in your shopping cart</p>'; } else { …

i having problem doing a echo with <img> , the image do not seems to able to display although there is not error message. please advice... thanks for your help. [CODE] $pic = $row["img"]; echo "$count.)&nbsp; $title, &nbsp;$price" ; echo '<img src="site_images/$pic" border=0>'; [/CODE]

i have search page to search by title, author, isbn from the dropdown list. But when they display the result for my search, it unable to retrieve my other data from my table (product4) like the price. it shows this error below Notice: Undefined variable: price in D:\xampp\htdocs\book\search.php on line …

been stuck this for the whole day.. need to seek advice from u guys.. it dont seems to work if i login as normal user... but if i log in as admin, it will redirect me to admin-index.php correctly. [CODE] <?php // Array for recording errors: $login_errors = array(); // …

i having problem and stucked with retrieving like lastnametname. and i am stucked with how do i check if admin=1 will redirect to admin.php if admin=0 will redirect to welcome.php instead. this is the part of my login process script [CODE] $q = "SELECT * FROM members2 WHERE (email='$e' AND …

i have a login page and it works well. but how can i make like after the log in successfully and all my other pages will have Welcome (lastname) of the user ? as currently, after log in succesfully will redirect to welcome.php and it seems like it only able …

i have a page which list all the products and there are checkbox to do muiltiple deletion. but there is an error on line 62 for the $delete [CODE]<html> <head> <title>delete</title> </head> <body> <? $objConnect = mysql_connect("localhost","root","booo33384") or die(mysql_error()); $objDB = mysql_select_db("db-test"); //$strSQL = "SELECT * FROM products3"; $strSQL = …

hi everyone, i encounter problem when trying to create a form page to upload images to images folder and imagelocation will store information for the pathname where the image will be located ( will be used for display image in future) the code below does not have any error when …

hi everyone, i am new here and i also quite new to php. i know this issue is quite common, been spending hours try to troubleshoot for this issue earlier. i have to give up and decide to seek advice from you guys, thanks for your time. The problem actually …