Hey! Having some more fun with htaccess again! - I have tired everyhting and cant get it to play... Below is my code: RewriteEngine On RewriteCond %{HTTPS} off RewriteRule ^(.*)$ https://%{HTTP_HOST}%{REQUEST_URI} [L,R=301] RewriteCond %{REQUEST_URI} !(/$|\.) RewriteRule (.*) %{REQUEST_URI}/ [R=301,L] RewriteBase /v2/ RewriteCond %{SCRIPT_FILENAME} !-d RewriteCond %{SCRIPT_FILENAME} !-f # Admin Pages RewriteRule ^admin/$ admin/index.php [NC,L,QSA] RewriteRule ^admin/([a-z0-9\-]+)/$ admin/$1.php [NC,L,QSA] # Super Admin Pages RewriteRule ^master/$ master/index.php [NC,L,QSA] RewriteRule ^master/([a-z0-9\-]+)/$ master/$1.php [NC,L,QSA] # If none just normal page RewriteRule ^([a-z0-9\-]+)/?$ users/$1.php [NC,L,QSA] ` https://domain.com/ - should load users/index.php https://domain.com/login/ (or any /XXX/) - should load users/$1.php https://domain.com/master/ - should load master/index.php https://domain.com/master/xxx/ …

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Magento 1.9 - How to include custom PHP Script into .PHTML file My magento tracking page: https://i.stack.imgur.com/3GIoN.png Back-end magento code : https://i.stack.imgur.com/ySgMA.png How can i add my PHP script into this tracking page. Tracking page code - trackorder.phtml [ https://i.stack.imgur.com/ySgMA.png ] <?php if(Mage::getStoreConfig('trackorder/trackorder_general/enabled')): ?> <div class="page-title"><h1><?php echo $this->__('Track Your Order ') ?></h1></div> <div class="form-list"> <form name="track_order" id="track_order" action="" method="post" onsubmit="sendAjax('track_order','<?php echo Mage::getUrl('*/*/track');?>'); return false;"> <!--<form name="track_order" method="post" id="track_order" action="<?php echo Mage::getUrl('*/*/view');?>">--> <ul class="form-list"> <li> <label for="order_id" class="required"><em>*</em><?php echo $this->__('Order Id') ?></label> <div class="input-box"> <input type="text" name="order_id" id="order_id" value="" title="" class="input-text required-entry" /> </div> </li> <li> <label for="email_address" class="required"><em>*</em><?php echo $this->__('Email …

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Heres a pic of my database table ![Screen_Shot_2019-11-16_at_2_35_54_AM.png](/attachments/large/4/0749479693675147a0ccc8219b89e87f.png) Im trying to create a search page with drop down selectors to filter through each of the last 8 columns. I want the the drop down selectors to be able to select multiple entries (I have this working already) I also want them to preload values from data already entered into my table columns. (I also got this working with the help of tutorials... although I admit I don't fully understand how this part works) Using these tutorials i've created a php page that contains 8 drop down selectors that automatically pull …

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I am skilled in Core PHP, JavaScript, Mysql, Ajax, HTML, CSS, jQuery and Codeigniter and having 4+ years of experience in Web Development. I have very vast knowledge in dynamic and static website development.

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How do I echo selected option? <?php session_start(); if(isset($_SESSION['cherianne'])){ }else{ header("Location:login.php"); } include("../includes/header.php"); // Retrieve our critical page setter variable. $char_id = $_GET['id']; //echo "<h1>$char_id</h1>"; // set this as a default value if (!isset($char_id)) { $result = mysqli_query($con, "SELECT bid FROM cde_blog LIMIT 1"); while ($row = mysqli_fetch_array($result)) { $char_id = $row['id']; } } // Step 3: UPDATE the DB with the new info from the form. Validate first. if(isset($_POST['mysubmit'])) { $title = strip_tags(trim($_POST['title'])); $message = strip_tags(trim($_POST['message'])); //echo "$title, $message"; $valid = 1; $msgPre = "<div class=\"alert alert-success\">"; $msgPreError = "<div class=\"alert alert-danger\">"; $msgPost = "</div>\n"; // Title Validation if((strlen($title) …

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Hello members, I am kind off new to working with self joins oe self referencing tables, I am wondering how one would update a self join table considering the table aliases naming. Your help will be appreciated. If some one can point me to a good resource, I will be highly grateful.

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i wish to send another field together with my FormData, so after i googled it, they said to add append to my form data, done it. but it did not send anything to db including the content of the file. here is my code:- $('#upload_csv').on("submit", function(e){ var data = new FormData(this); data.append('test_filename', file); }); $.ajax({ url:"uploadfileprocess", method:"POST", data:{data,data}, contentType:false, cache:false, processData:false, success: function(data){ alert(data); if(data=='Error1') { alert("Invalid File"); } else if(data == "Error2") { alert("Please Select File"); } else { $('#importfiletodb').html(data); } } }) all the content and the file name should be able to save together when user click …

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Would appreciate some kind help here as I'm experiencing some issue with regards to sorting of result from my map database. I trying to use onchange="this.form.submit() to activate my dropdown sort function, but somehow I kept getting plain text display of the sorted result. However, it would work fine if I were to click on search & preset the sort(for example, if I set it to sort according to lowest price value & click on the search button, the sort result would show out as normal, as seen in the below screenshot). Would really appreciate if anyone of you could …

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Hi everyone, I am trying to connect my html form to database using php pdo but everytime i try to submit my data it shows undefined

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Please I need your help, I am new to PHP and I need your help in inserting multiple select array into the database in the form of this. This is what I want to achieve. My table has 3 columns (id, examno, results). I want results column to be inform of subjects grades, subjects grades.........depending on the numbers of subjects and grades users select (e.g English C6, Mathematics C6) all in one column results. ## This is my html codes ## <form action="insert.php"> <div class="form-group"> <label>Exam Number</label> <input type="text" class="form-control" name="examno" id="examno"> </div> <table width="100%" cellpadding="0" cellspacing="0" border="0" class="table table-borderless" …

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Please I need your help, I am new to PHP and I need your help in inserting multiple select array into the database in the form of this. This is what I want to achieve. My table has 3 columns (id, examno, results). I want results column to be inform of subjects grades, subjects grades.........depending on the numbers of subjects and grades users select (e.g English C6, Mathematics C6) all in one column results. ## This is my html codes ## <form action="insert.php"> <div class="form-group"> <label>Exam Number</label> <input type="text" class="form-control" name="examno" id="examno"> </div> <table width="100%" cellpadding="0" cellspacing="0" border="0" class="table table-borderless" …

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Hello. I am having a problem here. So I have this slider which display the pictures with status "1" which means displayed. Here comes the problem, when there's no pictures with the status AKA "0" (not-displayed), the default image doesn't show up. Here's the script. <div class="carousel-inner"> <?php $no=1; $default=$this->db->get_where("penyaluran", array("status" => "1"))->result(); $g='image.jpg'; $n=''; foreach($default as $pict) { if ($g=$pict->foto_dokumentasi) { $n=$pict->nama_penyaluran; } else { $g='image.jpg'; $n= ''; } if ($no == 1) { $status = 'active'; } else { $status = ''; } ?> <div class="carousel-item <?php echo $status; ?>"> <img src="<?php echo base_url('uploads/'.$g); ?>" alt="" style="max-height: 300px; …

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Greetings Digital Natives. I'm excited join this platform. I am a sofware developer at [Timbu](http://www.timbu.com/south-africa) Hoping to learning and connecting with you guys. I want to really upgrade my tech skills. I have stayed along in front end. I need to do little of backend and move with likeminded fellow.

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Hi, I do need help with my solo project called Salary Expenses project in PHP and Laravel and I don't see any data for the current month in the **Monthly Expenses page**, even the data is recorded during the current matching month. Can help? My project file: https://drive.google.com/open?id=11qrmVHOwdgiiYQbw981o4y_CNp9DrTUM

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I apologize in advance if the title in unclear. I am fairly new to PHP/MySQL programming. I have a script I am working on - I have the login, register setup correctly and working fine. I have the pages restricted so that you're required to login. I am now trying to work on an administration panel, my users table in the database is currently setup as follows; id username password created_at I want to add another column called `is_admin` and the default value set to 0, where if `is_admin` is set to 1 then user is admin. I know how …

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Hey guys, So I can't seem to figure out why my live server is throwing this error. This works perfectly on my local server. Can anyone help me figure out how to fix this, or lead me in the right direction as to fixing this so my login and registration scripts work? This is the error that shows: "**Notice: Undefined variable: mysqli in /home/public_html/login.php on line 44 Fatal error: Uncaught Error: Call to a member function prepare() on null in /home/public_html/login.php:44 Stack trace: #0 {main} thrown in /home/public_html/login.php on line 44**" My code is below. Login.php <?php // Initialize the …

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validation is working fine for me but i cant submit please help me thanks here is my php

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<center><div id="piechart"></div></center> <script type="text/javascript" src="https://www.gstatic.com/charts/loader.js"></script> <script type="text/javascript"> google.charts.load('current', {'packages':['corechart']}); google.charts.setOnLoadCallback(drawChart); function drawChart() { var data = google.visualization.arrayToDataTable([ ['A', 'B'], <?php $query = "SELECT COUNT (*) AS a FROM TABLE WHERE A = '1'"; $results = sqlsrv_query($conn, $query); $row = sqlsrv_fetch_array($results); $A = $row['a']; ?> <?php $query = "SELECT COUNT (*) AS b FROM TABLE WHERE B = '1'"; $results = sqlsrv_query($conn, $query); $row = sqlsrv_fetch_array($results); $b = $row['b']; ?> ]); var options = {'title':'Survei Harian ', 'width':850, 'height':550}; var chart = new google.visualization.PieChart(document.getElementById('piechart')); chart.draw(data, options); } </script> help I wear script on why not play please grafik assistance

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I have a problem with laravel and vuejs the code below is in my resources in test.blade.php, I don't see anything in the internet browsers <!doctype html> <html lang="en"> <!doctype html> <head>` <meta charset="utf-8"> <meta name="viewport" content="width=device-width, initial-scale=1"> <title>Laravel</title> <!-- Fonts --> <link href="https://fonts.googleapis.com/css?family=Nunito:200,600" rel="stylesheet"> <!-- Style --> </head> <body> <div id="app"> <test></test> </div> <script src="http://192.168.3.129/js/app.js"></script> </body> </html>

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Hi, I am trying to insert a data from the database to select drop down list in a table where this comes in the 4th column. But it does not work. There is already data from the database to other columns and it is working , but this is not working for the 4th column in the drop down list. How do you add in a table 4 th column - This is not working - echo "<td> <select> <?php $sql = "SELECT name FROM categories"; $result=mysql_query($sql); while ($row = mysql_fetch_array($result)); <option>". $row[name] ." </option> } ?> </select></td>"; echo "<td><a …

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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 …

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I am under developing PHP add to cart without DB, so that i am using SESSION, actually get data from fetch_data.php data to my_cart.php using the POST method, successfully retun the values, After receiving the post data how can i convert to display like a table. workout: fetch_data.php return values. https://snag.gy/IASCMZ.jpg & values received https://snag.gy/ojWxHe.jpg Here is my my_cart.php : // FYI -> Here i am using only two fields : voice_sku & voice_name <table width="100%" cellpadding="6" cellspacing="0"> <thead> <tr> <th>Voice Sku</th> <th>Voice Name</th> <th>Remove</th> </tr> </thead> <tbody> <?php session_start(); $voice_sku = ''; $voice_name = ''; if(isset($_POST['voice_sku'])&& isset($_POST['voice_name'])) { // …

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I have One Problem in my coding ..I want to upload multiple images in the database in one row with using comma ... for($i=0;$i<count($_FILES['eventpic']['name']);$i++) $pic = $_FILES['eventpic']['name'][$i]; $tmp =$_FILES['eventpic']['tmp_name'][$i]; $location="pics/".$pic; move_uploaded_file($tmp,$location); $q="insert into tbl1(title,tdate,eventpic)values('','','$pic')"; echo $q; through this code i can upload tthe multiple images but not in a single row ....

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Hello I am using Crystal report version 11.5.8.826, php version 7.0.10, Apache version 42.4.33 and MySql Version 5.7.14. I have created the sampel rptone.rpt file that contents sample report. I want to load that rptone.rpt file using php. With Crystal report 11.5.8.826, php 7.0.10, apache 42.4.33 and MySql 5.7.14 i have tried to connect/load crystal report file that is .rpt file using COM Object but it gives error $my_report = "report/rptone.rpt"; $COM_Object = "CrystalReports115.ObjectFactory.1"; $objCom = new COM($COM_Object, "localhost", 0, ""); $crapp = $objCom->CreateObject("CrystalDesignRunTime.Application"); $creport = $crapp->OpenReport($my_report, 1); This expected to be open rptone.rpt file that must show the report …

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I am following the below module, https://www.webslesson.info/2018/04/shopping-cart-by-using-bootstrap-popover-with-ajax-php.html How can i use without quantity [Product Name, Price, total, Action] i tried lot fail again and again. I am learning stage in PHP, any help pls welcome.

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I am having problem in submitting this form when I select "1" on the number of setting in html form. The submit button is working if I select 2 in the number of sitting from html form as it will show all the form fields but when I select 1, submit form is not working. I am new to PHP This is my html form <form name="register" method="POST" action="process.php"> <label>Select Number of Sitting(s)</label> <select name="sitting" id="sitting" class="form-control" onchange="showDiv(this)"> <option value="1">1</option> <option value="2">2</option> </select> <div class="output1" id="div1" name="onesitting"> <label>School Name</label> <input type="text" name="schoolname[]" id="schoolname1" class="form-control" placeholder="School Name" required> <label>Exam Type</label> <select …

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Hi i'm Anis. I want to ask how i can attach image from database in FPDF?

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My code about filterable audio searching, here how can i add Add-To-Cart functionality? code : index.php <?php //index.php include('database_connection.php'); ?> <!DOCTYPE html> <html lang="en"> <head> <meta charset="utf-8"> <meta http-equiv="X-UA-Compatible" content="IE=edge"> <meta name="viewport" content="width=device-width, initial-scale=1"> <meta name="description" content=""> <meta name="author" content=""> <title>Voice Repository</title> <script src="js/jquery-1.10.2.min.js"></script> <script src="js/jquery-ui.js"></script> <script src="js/bootstrap.min.js"></script> <link rel="stylesheet" href="css/bootstrap.min.css"> <link href = "css/jquery-ui.css" rel = "stylesheet"> <!-- Custom CSS --> <link href="css/style.css" rel="stylesheet"> </head> <body> <!-- Page Content --> <div class="container"> <div class="row"> <br /> <div class="col-md-12"> <h2 align="center">Voice Repository Bank</h2> </div> <div class="col-md-3"> <!-- Price --> <!-- <div class="list-group"> <h3>Price</h3> <input type="hidden" id="hidden_minimum_price" value="0" /> <input type="hidden" …

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*Multiple database select - but print ids separately* $sql = "SELECT * FROM peron, peronmedia WHERE peronmedia.PID=5" and peron.ID=5"; $xc = mysqli_query($baglanti, $sql); $rs=mysqli_fetch_array($xc); echo $rs["peron.PID"]; echo $rs["peronmedia.ID"]; **but the codes don't work** ## **How can I do it?** ## Thank you.

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Magento how to apply discount for new customer like 1st order 5% discount and 3rd order 10% discount?

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The End.