Forum Topics Tagged [php]

I am thinking of moving to ecommerce development. There is just so much out there so much so that I find myself at a loss as to where to start. I do front end development but I understand backend equally well. Ideally I would like to find a suitable e-commerce system that I can work with, tweak and customise. So if a client needs an ecommerce site my task would be to build a custom interface and hook it up with this system. If some changes are necessary to make this work that would be done. What are the best …

With this php code, I send a request for payment with Paypal sandbox account. However when I go to the submit page I cant see what amount I am paying: https://i.imgur.com/iTra1SQ.png -> here in the red shoulld be displayed - 10.00(amountPayable). How can I send it so it's showed? use PayPal\Api\Amount; use PayPal\Api\Payer; use PayPal\Api\Payment; use PayPal\Api\RedirectUrls; use PayPal\Api\Transaction; require 'bootstrap.php'; $payer = new Payer(); $payer->setPaymentMethod('paypal'); // Set some example data for the payment. $currency = 'GBP'; $amountPayable = 10; $invoiceNumber = uniqid(); $amount = new Amount(); $amount->setCurrency($currency) ->setTotal($amountPayable); $transaction = new Transaction(); $transaction->setAmount($amount) ->setDescription('10 GBP payment') ->setInvoiceNumber($invoiceNumber); $redirectUrls = …

Can anyone help me with my search engine. I need filter search in the system using PHP. The problem with the code: All the provided fields must be filled up..unless it will not work There were 6 fields such as Remark Status: Industry: Position: Location: Age Range: Min and Max I have to use these fields as the filters,but eventually all the fields must be filled up before it works..How if i provide just one field, let say i fill the position as HR Manager so i did not specify its age and location...Please Help me on this,,, if(isset($_GET['advance_searching'])){ if …

I am working on a project for drag and drop items in a shopping cart. For an item dropped into the cart, I am trying to implement jQuery $.ajax. The result back from the PHP is pre-pended with '`' and I can not figure out why. This is the javascript: $('#target').droppable({ drop: function(event,ui) { var id = $(ui.draggable).data('id'); var aisle = $(ui.draggable).data('aisle'); $.ajax({ url : base_url+'/index.php/groceries/jquery_append_to_list', cache : false, type : 'post', data : { 'id' : id, 'aisle' : aisle }, success : function(data) { $('#target').append('<p data-id="'+id+'" data-aisle="'+aisle+'" class="draggable">'+data.item+'</p>'); } }) } }); This is the PHP script: function …

i want to display the stored image in fdpf.

Receiving error - Warning: Missing argument 4 for Get_All_Orderlines_Range(), called in /home/download/domains/1ecommerce.com/public_html/dev/_cms/list_orders.php on line 184 and defined in /home/download/domains/1ecommerce.com/public_html/dev/_cms/list_orders.php on line 161 Line 161: function Get_All_Orderlines_Range($id, $all = "", $from_date, $to_date){ $sql = "SELECT * FROM orders"; $sql .= " INNER JOIN orderline on orders.OR_ORDER_NO = orderline.OL_ORDER_NO"; $sql .= " WHERE OR_CUSTOMER_ID = '$id'"; if($all != "ALL"){ if($from_date != "" and $to_date != ""){ $sql .=" AND (orders.OR_DATE_CREATED BETWEEN '" . $from_date . "' AND '" . $to_date . "')"; } } $sql .= " ORDER BY orderline.OL_ORDER_NO"; //echo $sql; return FetchSqlAsObjectArray($sql); } Line 184: $orderlines = Get_All_Orderlines_Range($printed, pack_calendar_date($from_date, "from"), …

How I can get current date and 15 days back date from current date in 'Y-m-d' format ?? e.g. if current date is 2011-06-21 then $currentdate=2011-06-21 and $dateback=2011-06-06;

hi, i have installed XAMPP, but i would like to run my localhost from another computer..

I'm a novice in PHP and I would like to know on how exactly could I define said row. Any helps/tips are appreciated. Thank you in advanced! function add_review() { if(isset($_POST['add_review'])){ (THE UNDEFINED ROW) **$product_id = $row['product_id'];** $review_name = escape_string($_POST['review_name']); $review_email = escape_string($_POST['review_email']); $review_content = escape_string($_POST['review_content']); $review_created = date("Y-m-d H:i:s"); $rating = escape_string($_POST['rating']); $review_query = query("INSERT INTO reviews(review_id, product_id, review_name, review_email, review_content, review_created, rating) VALUES('NULL', '$product_id', '$review_name', '$review_email' , '$review_content' , '$review_created' , '$rating')"); confirm_query($review_query); set_message("Thank you for the feedback!"); }}

Hi all, I am very new to PHP and MYSQL and have a class assignment I need help with. I am trying to make a page with an HTML form that will update my MYSQL database. You can view my pages online here: [URL="http://baileyjumper.aisites.com/Scripting-Week7/homework4/exercise-five.php"]http://baileyjumper.aisites.com/Scripting-Week7/homework4/exercise-five.php[/URL] I need help with the "Edit" section, if you look online. Here is the code for my Edit page: [CODE=PHP] <?php $hostname = "localhost";//host name $dbname = "baileyjumper_imd203";//database name $username = "baileyjumper_imd";//username you use to login to php my admin $password = "chris4ever";//password you use to login //CONNECTION OBJECT //This Keeps the Connection to the Databade …

Got Client with need of **"online print shop"** Client is providing paper printing (like business cards, flyers), printing on cloths, office stationary, logo embroidery on any type of clothing. Client provided some examples. **Links** * [rockdesign](https://www.rockdesign.com/) * [moo](https://www.moo.com/ca/) * [readyprint](https://www.readyprint.ca/) * [vistaprint](https://www.vistaprint.ca/) * [printshop](https://printshop.ca/) * [printvenue](https://www.printvenue.com/) **Where do I start?** Which platform should I choose? 1. CorePHP 2. Wordpress 3. WooCommerce 4. Magento 5. Prestashop 6. WordPress Is there any plugin or script that should I go for???

I have predefined value like shipping_weight = 10$, my form consisting two dropdown list, if the dropdown selected based on the selection how to return my ajax value, Index.php : <tr> <th>I have : </th> <td> <select id="old" name="i_have"> <option value = "select_option">Select Option</option> <option value = "three_compact">3 Compact</option> <option value = "three_regular">3 Regular</option> <option value = "three_triple">3 Triple</option> <option value = "five_compact">5 Compact</option> <option value = "five_regular">5 Regular</option> <option value = "five_triple">5 Triple</option> <option value = "seven_compact">7 Compact</option> <option value = "seven_regular">7 Regular</option> <option value = "seven_triple">7 Triple</option> <option value = "nine_compact">9 Compact</option> <option value = "nine_regular">9 Regular</option> <option …

I have a search form, and I want to change the url from: https://www.example.com/search?search=term+term to: https://www.example.com/search/term+term I know how to use htaccess, but I don't know how to send user to friendly URL after he submit the form. Here is my form: <form action="search" method="get"> <input type="text" name="search" id="Searchx" placeholder="Search Marketplace" /> </form>

Hello i have this form and my php code skills not very good if someone make change password I would be glad thank you <div class="col-lg-9"> <div class="nk-box-3 bg-dark-1"> <form class="change-password" method="post" action=""> <input type="password" class="form-control input-change-password" name="opass" id="opass" placeholder="Old Password"> <div class="nk-gap-1"></div> <input type="password" class="form-control input-change-password" name="npass" id="npass" placeholder="New Password"> <div class="nk-gap-1"></div> <input type="password" class="form-control input-change-password" name="cnpass" id="cnpass" placeholder="Repeat New Password"> <div class="nk-gap-1"></div> <button type="Update" class="nk-btn link-effect-4 ready"><span class="link-effect-inner"><span class="link-effect-l"><span>Change password</span></span><span class="link-effect-r"><span>Change password</span></span><span class="link-effect-shade"><span>Change password</span></span></span></button> </form> </div> </div>

I am required to change ip adress everytime this function gets executed I am trying this code below .But I am completly confused.How can I do this function getData($domainName, $ext) { $proxy = array( 1 => array( '88.255.101.247', '8080' ), 2 => array( '176.53.2.122', '8080' ), 3 => array( '37.123.96.237', '8080' ) ); shuffle($proxy); $servers = array( "biz" => "whois.neulevel.biz", "com" => "whois.internic.net", "us" => "whois.nic.us", "info" => "whois.nic.info", "name" => "whois.nic.name", "net" => "whois.internic.net", "tv" => "whois.nic.tv", "ru" => "whois.ripn.net", "org" => "whois.pir.org", "com.tr" => "whois.nic.tr", "gen.tr" => "whois.nic.tr", "web.tr" => "whois.nic.tr", "k12.tr" => "whois.nic.tr", "org.tr" => "whois.nic.tr" ); …

Hi all, this is my first post and would appreciate some insight as to where im going wrong. I have the following form that successfully registers a new user to my mysqsl database via PHP - with the following $sql INSERT Function The problem im having is sending the activation \ confirmation email to the new users email address. [code] //email function ini_set("SMTP", "smtp.server.com");//confirm smtp $to=$email; // Your subject $subject="activation required"; // From $header="from: test email <Test@test.com>"; // Your message $message="Your Comfirmation link\r\n"; $message.="Click on this link to activate your account\r\n"; $message.="You can not login to your new account until …

I have to take over a php project and I am totally new to php. I have this webpage that displays all packages sent to a customer and I want them to be able to click on the tracking number to do UPS website tracking. The code: <?php $result = mysqli_query($conn, $sql); if (!$result) { exit(); } else { if (mysqli_num_rows($result) > 0) { while ($row = mysqli_fetch_assoc($result)) { echo "<tr>"; echo "<td><strong>"; echo $row['trackingno']; echo "</td>"; echo "<td>"; echo $row['date_shipped']; echo "</td>"; echo "<td>"; echo $row['desc_reference2']; echo "</td>"; echo "<td>"; echo $row['status']; echo "</td>"; echo "<td>"; echo $row['date_delivered']; echo …

So here's a little background to help with what I'm trying to accomplish (be mindful I'm also fairly new to MySQL): I'm creating a dynamically built navigation for a website using PHP and MySQLi. I've created a class called *Database* where I've got helper functions, such as connect() for connecting to the database, and the like. Then I have another class called *Navigation* which extends *Database*, and there I'm creating helper functions to do specific things for the database that holds my navigation links, such as selecting and inserting and so on. Now, the following code works flawlessly when working …

Hi, I want to replace ticks in text with other text. This method usually works, but when adding three dots, I get php warning ltrim(): Invalid '..'-range, no character to the right of '..' or invalid range needs to be incrementing. I don't know where to look for error, except deleting dots. Here is the code: $code = "some text before block ´´´´code block1´´´´ text after block ´´´´code block2...´´´´"; // I want ticks to be replaced to this: <pre><code> code block1 </code></pre> $pattern = "´´´´"; // we are using ticks $countCodeBlock = substr_count($code, $pattern); // count how many times pattern …

Hello, Recently I am using phpexcel and afer many tries I wasn't able to find out how to make a document with a image in its header. Here is a simplified code that I am using. Can someone helps me with this issue?! <?php require_once 'PHPExcel.php'; $objPHPExcel = new PHPExcel(); $objDrawing = new PHPExcel_Worksheet_HeaderFooterDrawing(); $objDrawing->setPath('header.jpg'); $objPHPExcel->getActiveSheet()->getHeaderFooter()->addImage($objDrawing, PHPExcel_Worksheet_HeaderFooter::IMAGE_HEADER_RIGHT); $objPHPExcel->getActiveSheet()->getHeaderFooter()->setOddHeader("&C&G"); $objPHPExcel->setActiveSheetIndex(0)->setCellValueExplicit('A1', 'CODICE PRATICA'); $objWriter = PHPExcel_IOFactory::createWriter($objPHPExcel, 'Excel2007'); $objWriter = new PHPExcel_Writer_Excel2007($objPHPExcel); $objWriter->save('test.xlsx'); ?>

Write a PHP code to: 1- Create a 2-dimensional array, with a random number of rows and columns. 2- Fill the array with random numbers. 3- Send the array to a function that returns the number of the highest frequency in the array. Who can help me to solve it?

inside mysql db which is table employe in coloumn *id_employee*, some data have a leading zero, etc : **id_employee : 0378192918 empy_name : Daniweb** i need to get id_employee for delete employe in my aplication, then i using ajax method like this : if(id) { alert(id); $("#deleteBtn").unbind('click').bind('click',function(){ $.ajax({ url: 'pegawai/delpeg.php', type:'post', data: {emp_id : id}, datatype: 'json', success:function(response) { return value is **378192918**, the first number **0** not showing in alert box. why ? then how to get entire value id_employe from db ?

Hi, I have a problem with retrieving uploaded images from a database. It appears to upload fine, but when i try to display the image i get loads of binary jargon? I have tried looking everywhere for a solution, but unable to, any help would be greatly appreciated. The code for uploading is [code] include ("database.php"); if(isset($_POST['Submit'])) { if(isset($_FILES['image'])) { /* Upload file from anywhere and put in a set temp dir */ $image = file_get_contents($_FILES['image']['tmp_name']); $image = mysql_real_escape_string($image); $handle = fopen($image, 'rb'); // read binary $contents = fread ($handle, filesize ($image)); //$contents = addslashes($contents); fclose ($handle); $description = $_POST['description']; …

Hello, So I have buttons on my site connected to a PHP code that updates the column "status" in "users" table with a value based on what button you press. This is the PHP script for that: <?php include('functions.php'); $servername = "localhost"; $username = "root"; $password = ""; $dbname = "emildeveloping"; // Create connection $conn = mysqli_connect($servername, $username, $password, $dbname); // Check connection if (!$conn) { die("Connection failed: " . mysqli_connect_error()); } $id = $_SESSION['user']['id']; if (isset($_POST['108'])) { $sql = "UPDATE users SET status = '10-8' WHERE id='".$id."'"; } if (isset($_POST['107'])) { $sql = "UPDATE users SET status = '10-7' …

I'm somewhat new to coding websites. I'm trying to have a code where it pulls multiple images from a single database and displays it in a grid, not a table. At the top of my page I have a normal connection to the database. (very sorry about not having a code snippet, I'm new to daniweb and keep having errors when trying to use code snippets.) <div id="grid" class="grid-container"> <?php while ($row = mysqli_fetch_array($query)) echo '<div class="grid-item"> <p>'.$row['images'].'</p> <p>'.$row['project'].'</p> <p>'.$row['summary'].'</p> </div>'; ?> </div>

Hello i have a mysql statement and i want to add more parameters to get different results. SELECT DISTINCT M.message,F.friend_one,F.friend_two, F.role,U.uid,U.username FROM users U,friends F, messages M WHERE ( U.status='1' AND F.friend_one = '88') OR (F.friend_two='88' ) AND F.role='friend' 88 is the users id so i want to show M.message from friends of 88 but not 88 itself hints: when one is becoming friend with other friend_one is 88 -for example- and friend_two another user id or vice versa M table has id of user as a column of course

I realize that this question has been asked many times, but I've yet to find a flatfile version. I'm looking for a way to achieve this: **Scenario:** A person submits their email address via a form. The email is then stored in flatfile database (.txt) The person then gets an auto-reply email with a link to confirm their email address. Upon confirming after clicking the link, their email address is saved to a new file confirmed_emails.txt for example. Is this possible without the use of a SQL/similar datase, and how can I achieve this? Note: SQL or other database option, …

I have a form that uploads a news article with it's picture and then stores all the info correctly. I am using codeigniter with Active Record and would like to know how to display the image with the post it was uploaded with. here is my upload script: public function create(){ $data = array('error' => ''); $config = array( 'upload_path' => './uploads', 'allowed_types' => 'gif|jpg|png', 'max_size' => '204800', 'max_width' => '1920', 'max_height' => '1080', 'encrypt_name' => true ); $this->load->library('upload', $config); if(!$this->upload->do_upload()) { $error = array('error' => $this->upload->display_errors()); $data['title'] = 'Create a news item'; $data['main_content'] = 'news/create'; $this->load->view('templates/template', $data, $error); } …

$query = "INSERT INTO 'user-profiles' ('name', 'avatar', 'description', 'votes', 'pay_methods', 'location', 'user_id') VALUES ('sadj', 'qwesad', 'sadqwdqw', '0','sadsad', 'sadasdsad''4')"; $resultprof = $this -> conn -> query($query); votes and user_id are INT columns, and user_id is a foreign key. Making this insert on mySQL directly it works perfectly, but on my PHP the "success" is false....

whats the coding of combo box in php????????? can any one tell meplzzzzzz