The Slack system is comprised of a many different services and apps. We have 3 main clients: * Our web client is written in a mix of JavaScript and ES6, with React. We use Electron to ship it as a desktop app * Our Android client is written in a mix of Java and Kotlin * Our iOS client is written in a mix of Objective C and Swift On the backend, we have our core application which powers slack.com and our API, which is written in PHP/Hacklang running on HHVM. We store data in MySQL using Vitess. For caching, …

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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???

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**Meaning of Frontend verses Backend** Frontend is the a part of the internet site customers can see and engage with including the graphical user interface (GUI) and the command line together with the design, navigating menus, texts, pix, motion pictures, and so forth. Backend, at the contrary, is the part of the website customers can't see and have interaction with. It’s all approximately how the whole lot works. **Role of Frontend verses Backend** Both play a critical function in net improvement and despite the fact that they've their honest share of variations, they are like facets of the same coin. …

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i have manage to insert my checkbox array data into database. the problem is when i check multiple checkbox only the first data in array is inserted in database. i need when i check multiple checkbox, all the data i have check insert into database.can someone help me? this is my php code: <?php include ('connect.php'); $cat = implode(',', $_POST['check']); if(isset($_POST['submit'])) { for($i=0; $i<count($cat);$i++) { $p_id =$_GET ['sitter']; $price = $_POST ['price']; $pickup_date =$_POST ['pickup_date']; $dropoff_date =$_POST ['dropoff_date']; $numdays = $_POST ['numdays']; $total =$_POST ['test']; $sql2 = "INSERT INTO cat_sitter(sitter_fk,cat_fk, price, date_in, date_out,total_day, total)VALUES ('$p_id','" . $cat[$i] . "','$price','$pickup_date','$dropoff_date','$numdays', '$total')" …

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Parse error: syntax error, unexpected '||' (T_BOOLEAN_OR) in php pdo

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

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Im using MongoDB with PHP. Im making a check user exist using their email.It can be check, but when I enter the existing user and click the confirm button, it display the 'echo' and also display an error about converted document to int. Below is what the error say and my code. **ERROR** Fatal error: Uncaught Error: Call to a member function count() on null in C:\xampp\htdocs\project\OrgChart-master\OrgChart-master\resetPasswordSuccess.php:36 Stack trace: #0 {main} thrown in C:\xampp\htdocs\project\OrgChart-master\OrgChart-master\resetPasswordSuccess.php on line 36 **CODE** <!-- MongoDB Conn With Email Check Function V1 (FUNCTIONING) --> <!--Line 36 is at if($result==0)--> <?php if(isset($_POST['registration'])) { $registration = ($_POST['registration']); if(!empty($registration)) …

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already this phone number registration in alert msg how to display in php and mysql

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I have couple of shipping methods DHL & FedEx, if selected products exits in cart show only selected shipping method, how to archive using shopping cart pricing rule or any other way to do this?

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I have a form which I am using to filter a recordset. It all works as I want, but for one small annoying thing - there is an empty space at the top of the list of options (see screenshot). How can I get rid of this empty option? ![empty_option.jpg](/attachments/large/4/5ddcc6d6df75ecd1a8e0babbae2b31ba.jpg) I have checked the recordset which is used to generate the options for the dropdown list and this is OK, i.e. correct number of options. <?php $con=mysqli_connect("localhost","jtwiname"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $query_rs5 = "SELECT distinct class FROM classname"; $rs5 = mysqli_query($con, …

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Please suggest something ---- Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, bool given in C:\xampp\htdocs\tourtravel\index.php on line 14** <?php include 'header.php'; include 'slider.php'; ?> <div class="container" > <div class="row"> <div class="col-sm-6"> <?php include 'tab.php' ?> </div> <div class="col-sm-6"> <?php include("config.php"); $q = mysqli_query($con,"select * from packages"); while ($r = mysqli_fetch_array($q)) { echo "<table>"; echo "<tr>"; echo"<td style='padding-right: 20px;'> <div class='albumDetail'> <a href='find_holiday.php?goto=Goa'><img class='img-responsive imghover' src='upload/$r[1]' width='220' height='150' > <span class='albumtitle'>Goa</span> </a> </div> </div> </td>"; echo"<td style='padding-right: 20px;'> <div class='albumDetail'> <a href='find_holiday.php?goto=Leh Ladakh'><img class='img-responsive imghover' src='upload/$r[2]' width='220' height='150' > <span class='albumtitle'>Leh Ladakh</span> </a> </div> </div> </td>"; echo "</tr>"; echo "<br/><br/><br/>"; …

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** ## I want a Help to Intergrate Google Calendar API to create Events thure webisite ## ** Ihave tied some google documantation and some tutorials buth they acctually not working properly google documantions are not not working for me

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Hi, Hope this makes sense and hope someone can help me. I have a database table with the following columns: person_id, firstname, surname, email_address. Periodically I am given a set of files to send to some of the contacts in the database table. The filenames contain the person_id and they are always of the format: person_id.pdf or person_id-2.pdf. Sometimes there is only one of the files available or sometimes it is both of the files. The files are stored in a directory on our server Example. directory: letters files: 1111.pdf, 1111-2.pdf, 2222.pdf, 3333-2.pdf The user with person_id 1111: should receive …

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Hi, Is there a PHP script that doesn't allow the approver to be the same as the login e-mail? > $checkEmail = test_input($_POST["Approved_by_Email"]); > if (filter_var($checkEmail == session('session_staff_email'))) { > alert("Please do not use your e-mail as the approver."); > return false; > } I tried to solve the problem, but it's so difficult to do. Here is the code that I'm trying to do. "Approved_by_Email" is the text box.

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Why is $if (isset($_GET['edit']) returning empty value when it's inside if (isset($_POST['add_submit']) but it's returning the value that I want When it's not inside if (isset($_POST['add_submit']) and how to fix it. Thank you. if (isset($_POST['submit']) ) { $brand = $_POST['brand']); if (isset($_GET['edit']) ) { $sql = "UPDATE brand SET brand = '$brand' WHERE Bra_ID = '29'"; $stmt = $con->prepare($sql); $stmt->execute(); header("Location: brands.php"); exit(); } $sql = "INSERT INTO brand(brand) VALUES('$brand')"; $stmt = $con->prepare($sql); $stmt->execute(); header("Location: brands.php"); exit(); }

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I have read quite a few posts on how to do this (Retain Selected Value PHP Dynamic Drop Down List), but cannot quite understand what to do. What I have works, but it does not retain the selected. Any help appreciated. <form action="" method="POST" name="form1" id="form1"> <select name="selClass" size="1" id="selClass" onchange="form1.submit()"> <option value="">Select a class</option> <?php echo "<option value='" . "All records" . "' . >" . "all records" . "</option>"; while ($row1 = mysqli_fetch_array($rs5)) { echo "<option value='" . $row1["class"] . "'>" . $row1["class"] . "</option>"; } ?> </select> </form>

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Hello everyone, I am new to this daniweb community and I really hope you guys will help me with this as I am baffled at what I couldn't come up with or was it not possible to do that? I want to have a checkbox on either PHP page or on modal like I did below to be checked, then save (submit), and stored in the mysql database so users can have ability to hide or show the product's description every time we go there without it coming back when the page refresh or whatever...Is there any possible that it …

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Hi, I want to make online shop and I made most of it but I have problem on product page.I want for "customers" ( this is my personal project so ... ) to be able to choese and then what they choese to be store into table. I struggling with this more then 3 hour , I try everyhing , this is my last shoot. (I used the echo because I want to show it , so after that I would put insert into, but ... I was not able to do it ) <?php $con = mysqli_connect("localhost","root","","basic"); $query = …

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Here my form auto responce based on drop-down list, <tr> <th>Shipping Cost (Rs) : </th> <td id="findata"></td> <input type="hidden" name="shipping_cost" id="shipping_cost"/> </tr> <tr> Same page Ajax : <script> $(document).ready(function(){ $('#new').on('change',function(){ var zip = $("#zip_postal_code").val(); var country = $("#country").val(); $.ajax({ type: "POST", // url: "ajax_ship_cost_data.php", url: "sp_cost.php", dataType: "text", data: { zip_postal_code: zip, country: country}, success: function(data) { // Check the output of ajax call on firebug console //console.log(data); $('#findata').html(data); $('#shipping_cost').val(data); } }); }); }); </script> The field **Shipping Cost** is auto responce based following field, <tr> <th>I have : </th> <td> <select id="old" name="i_have"> <option value = "select_option">Select Option</option> <option …

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hello every one please can any body do this task for me please becouse our instructor in the university uploaded it and I dont have enough time becouse im studying for my final exams please any one and i will be helpfull for him/her :) this is the task Create a car rental system where customers rent cars for a given period of time. Car – a class for all cars. Holds car information and an array of rents. Implement several types of cars as subclasses: Small car – 4 passengers, Motor 1200, air-conditioner Medium car – 4 passengers, Motor …

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I need to add couple of php page as from action like, <form action=“one.php”, “two.php” method=“POST”> The code above is not working, how can i use more than one form action url?

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HI All I am displaying 10 questions randomly , if user play quizz multiple times - i want to display updated score of the users's session. demo url is : http://telugumirchi.com/quiz/ **index.php** <?php session_start(); ?> <?php require 'config.php'; ?> <!DOCTYPE html> <html> <head> <title>ABCD COMPANY</title> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <!-- Bootstrap --> <link href="css/bootstrap.min.css" rel="stylesheet" media="screen"> <link href="css/style.css" rel="stylesheet" media="screen"> <!-- HTML5 shim and Respond.js IE8 support of HTML5 elements and media queries --> <!--[if lt IE 9]> <script src="../../assets/js/html5shiv.js"></script> <script src="../../assets/js/respond.min.js"></script> <![endif]--> </head> <body> <header> <p class="text-center"> Welcome <?php if(!empty($_SESSION['name'])){echo $_SESSION['name'];}?> </p> </header> <div class="container"> <div class="row"> <div class="col-xs-14 …

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Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, bool given in C:\Users\USER\Desktop\Final Year Project\XAMPP\htdocs\littleavenue\shop.php on line 320 <?PHP if(!isset($_GET['product_category'])){ if(!isset($_GET['category'])){ $per_page = 6; if(isset($_GET['page'])){ $page = $_GET['page']; } else { $page = 1; } $startfrom = ($page-1) * $per_page; $getproducts = " SELECT * FROM products OREDR BY 1 DESC LIMIT $startfrom, $per_page"; $runproducts = mysqli_query($connection,$getproducts); ////LINE 320 //// while($row = mysqli_fetch_array($runproducts)){ ////LINE 320//// $productid = $row['product_id']; $producttitle = $row['product_title']; $productprice = $row['product_price']; $productimgone = $row['product_img1']; echo "<div class='col-md-4 col-sm-6 center-responsive' > <div class='product' > <a href='productdetails.php?product_id=$productid' > <img src='admin/product_images/$proproductimgone' class='img-responsive' > </a> <div class='text'> <h3><a href='productdetails.php?product_id=$productid'>$producttitle</a></h3> <p class='price'>$ $productprice</p> …

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I am doing this assignment i have to redirect the login page to the welcome back page to authenticate the use identity but i am getting this error in the welcome_back page in page. i have look everywhere to find the error. the error is the and now here is my code for it. <div id="contentArea"> <?php session_start(); if(!session_is_registered("username")) { header("location:login_user.php") } <--the error is right here Parse error: syntax error, unexpected '}' in C:\wamp64\www\Taylor_Luana_ITEC244_Assignment\php\welcome_back.php on line 43 ?> <?php echo "Welcome back " . $_SESSION['username']; ?> <div id="footerbox"> <div id="footerholder"> <div id=""> <a href="browsenationalwonders.php">Browse National Wonders</a> <a href="updateprofile.php">Update Profile</a> …

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Hi, I want to display the avearage row as images. Is it possible? function reviews_average(){ $avgproduct_id = escape_string($_GET['id']); $avgquery = "SELECT avg(rating) as average_rating FROM reviews WHERE product_id = $avgproduct_id"; $avgresults = query($avgquery); confirm_query($avgquery); while($row = fetch_array($avgresults)){ $avgratings = **--THIS IS THE ONE I WANT TO DISPLAY AS IMAGES--** "{$row['average_rating']}"; echo $avgratings; } } This is the image `<img src='../public/images/star.png' height='20px' width='20px'/>`

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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!"); }}

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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 …

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+0 forum 21

I have magento shipping charge code : <?php ob_start(); // require_once(__DIR__ . '/app/Mage.php'); require_once('./../app/Mage.php'); umask(0); ini_set('display_errors',true); ini_set('memory_limit', '1024M'); Mage::app()->loadArea('frontend'); function getShippingEstimate($productId,$productQty,$countryId,$postcode ) { // $quote = Mage::getModel('sales/quote')->setStoreId(Mage::app()->getStore('default')->getId()); $quote = Mage::getModel('sales/quote')->setStoreId(Mage::app()->getStore('english')->getId()); $_product = Mage::getModel('catalog/product')->load($productId); $_product->getStockItem()->setUseConfigManageStock(false); $_product->getStockItem()->setManageStock(false); $quote->addProduct($_product, $productQty); $quote->getShippingAddress()->setCountryId($countryId)->setPostcode($postcode); $quote->getShippingAddress()->collectTotals(); $quote->getShippingAddress()->setCollectShippingRates(true); $quote->getShippingAddress()->collectShippingRates(); $_rates = $quote->getShippingAddress()->getShippingRatesCollection(); $shippingRates = array(); foreach ($_rates as $_rate): if($_rate->getPrice() > 0) { $shippingRates[] = array("Title" => $_rate->getMethodTitle(), "Price" => $_rate->getPrice()); } endforeach; return $shippingRates; } // echo "<pre>"; // product id, quantity, country, postcode // print_r(getShippingEstimate('14419','1',"IN","642001")); // echo "</pre>"; $results = getShippingEstimate('14419','1',"IN","642001"); // $results = getShippingEstimate('14419','1',"US","99501"); $count = -1; echo "<select>"; foreach ($results as $result): $count++; ?> …

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Hi i am trying to fetch data from database into my dropdown menu in codeigniter, but havind error of 'undefined variable: unit_name' . this is my model file `public function get_unit_value() { $result = $this ->db ->select('id, unit_name') -> get('units') -> result_array(); $unit_name = array(); foreach($result as $r) { $unit_name[$r['id']] = $r['unit_name']; } $unit_name[''] = 'Select unit...'; return $unit_name; }` this is my controller `public function unit(){ $this->load->database(); $this->load->model('client_model'); $data['unit_name'] = $this ->client_model-> get_unit_value(); $this -> load -> view('add_quote', $data); }` this s my view file `<div class="panel panel-default"> <div class="panel-heading">The form</div> <div class="panel-body"> <!--dropdown input--> <?php echo form_dropdown('unit_name',$data['unit_name'], '', …

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Hello i am getting an error on a CASE WHEN THEN syntax on my mySql SELECT DISTINCT u.uid,c.c_id,u.name,u.profile_pic,u.username,u.email,c.time FROM conversation c, users u, conversation_reply r CASE WHEN c.user_one = '99' THEN c.user_two = u.uid WHEN c.user_two = '99' THEN c.user_one= u.uid END AND ( c.user_one ='99' OR c.user_two ='99' ) AND u.status='1' AND c.c_id=r.c_id_fk AND u.uid<>'1'ORDER BY c.time DESC LIMIT 15 #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'CASE WHEN c.user_one = '99' THEN c.user_two = u.uid WHEN c.user_t' at …

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The End.