I am attempting to print out mySQL database into a html table. I have watched many tutorials on how to do this but am unsure as to how I refer to the html table in my php code. The information gets printed fine and connects to the database but for some reason it isn't output in the table format.
<?php
$conn = mysqli_connect('localhost', 'Admin', 'admin1', 'info');
if (!$conn) {
echo "Connection failed:" . mysqli_connect_error();
}
//Writing query for database.
$sql = "SELECT `First Name`,`Last Name`,Emails,`Date Created` FROM clientinfo ORDER BY `Date Created`";
//Querying and getting results
$result = mysqli_query($conn, $sql);
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo "<tr><td>" . $row["First Name"] . "</td></tr>" . $row["Last Name"] . "</td></tr>"
. $row["Emails"] . "</td></tr>" . $row["Date Created"] . "</td></tr>";
}
echo "</table>";
} else {
echo "0 result";
}
//Fetch resulting rows as an array
$informed = mysqli_fetch_all($result, MYSQLI_ASSOC);
// Freeing result from the memory.
mysqli_free_result($result);
mysqli_close($conn);
?>
<!DOCTYPE html>
<html lang="en-US">
<head>
<div class="Contained">
<div class="row">
<?php foreach ($informed as $inform) { ?>
<div class="col s6 medium-3">
<div class="card z-depth-0">
<div class="card-content center">
<h6><?php echo htmlspecialchars($inform['First Name']); ?></h6>
<div><?php echo htmlspecialchars($inform['Last Name']); ?></div>
</div>
<div class="card-action right-align">
<a class="brand-text" href="#">More Info
</div>
</div>
</div>
<?php } ?>
</div>
</div>
<title> Email and Name List </title>
</head>
<body>
<table>
<tr>
<th>First Name</th>
<th>Last Name</th>
<th>Emails</th>
<th>Date Created</th>
</tr>
</table>
</body>
</html>
Output in browser gyazo:
You must change code for output all table in php like:
<body>
<?php
$conn = mysqli_connect('localhost', 'Admin', 'admin1', 'info');
if (!$conn){
echo "Connection failed:" . mysqli_connect_error();
}
//Writing query for database.
$sql = "SELECT `First Name`,`Last Name`,Emails,`Date Created` FROM clientinfo ORDER BY `Date
Created`";
//Querying and getting results
$result = mysqli_query($conn,$sql);
if ($result->num_rows>0){
echo '
<table>
<tr>
<th>First Name</th>
<th>Last Name</th>
<th>Emails</th>
<th>Date Created</th>
</tr>';
while($row = $result->fetch_assoc()){
echo "<tr> ";
echo "<td>" . $row["First Name"] . "</td>";
echo "<td>" . $row["Last Name"] . "</td>";
echo "<td>" . $row["Date Created"] . "</td>";
echo "</tr> ";
}
echo"</table>";
}
else{
echo "0 result";
}
//Fetch resulting rows as an array
$informed = mysqli_fetch_all($result, MYSQLI_ASSOC);
// Freeing result from the memory.
mysqli_free_result($result);
mysqli_close($conn);
?>
</body>
Another question are you sure is $row["First Name"] and not $row["First_Name"]?
Last tip learn how prepare stm for prevent sql inject
Based on your code, you are trying to print the table before the actual table is defined below. You can try something like this:
<?php
$conn = mysqli_connect('localhost', 'Admin', 'admin1', 'info');
if (!$conn){
echo "Connection failed:" . mysqli_connect_error();
}
//Writing query for database.
$sql = "SELECT `First Name`,`Last Name`,Emails,`Date Created` FROM clientinfo ORDER BY `Date Created`";
//Querying and getting results
$result = mysqli_query($conn,$sql);
$informed = mysqli_fetch_all($result, MYSQLI_ASSOC);
?>
<!DOCTYPE html>
<html lang="en-US">
<head>
<div class="Contained">
<div class="row">
<?php foreach($informed as $inform){?>
<div class="col s6 medium-3">
<div class="card z-depth-0">
<div class="card-content center">
<h6><?php echo htmlspecialchars($inform['First Name']); ?></h6>
<div><?php echo htmlspecialchars($inform['Last Name']);?></div>
</div>
<div class="card-action right-align">
<a class="brand-text" href="#">More Info</div>
</div>
</div>
<?php }?>
</div>
</div>
<title> Email and Name List </title>
</head>
<body>
<table>
<tr>
<th>First Name</th>
<th>Last Name</th>
<th>Emails</th>
<th>Date Created</th>
</tr>
<?php
if ($result->num_rows>0) {
while($row = $result->fetch_assoc()){
echo "<tr><td>" . $row["First Name"] . "</td><td>" . $row["Last Name"] . "</td><td>" . $row["Emails"] . "</td><td>" . $row["Date Created"] . "</td></tr>";
}
} else {
echo "<tr><td rowspan=\"5\">0 result</td></tr>";
}
?>
</table>
</body>
<?php
// Freeing result from the memory.
mysqli_free_result($result);
mysqli_close($conn);
?>
Related
I want to know where to put the condition to display data in a table after I select the value from a dropdown list.
Both have the same id (dropdown and table).
php table
<html>
<head>
</head>
<body>
<?php
$con=mysqli_connect("localhost","root","root","company");
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT employees.id,employees.jobs FROM employees WHERE employees.jobs in ("programmer","hr","qa")";
if ($result=mysqli_query($con,$sql)){
?>
<label for="y">Select the job:</label>
<select name="loads" id="loads" onchange="">
<?php while($ri = mysqli_fetch_array($result)) {
?>
<option value="<?php echo $ri['id'];?>" > <?php echo $ri['jobs']; ?> </option>
<?php
}
}
?>
</select>
<table class="striped" border="1" align="center" id="demo">
<tr class="header">
<td align="center"><b>Name</b></td>
</tr>
<?php
$con=mysqli_connect("localhost","root","root","company");
// Check connection
if (mysqli_connect_errno()){
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql2="SELECT employees.id,employees.name FROM employees WHERE employees.jobs in ("programmer","hr","qa")";
if ($result=mysqli_query($con,$sql2)){
// Fetch one and one row
while ($row=mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row["name"] . " " . "</td>";
echo "</tr>";
}
}
mysqli_close($con);
?>
</table>
</body>
</html>
If you mean: you want to something happens to your table when an item get selected in dropdown list(select tag). then it is not possible through php because php codes compiled once after each load on a page and it doesnt work live!
so you have to use JQUERY and AJAX to do that.
if this is what you searching for, reply me so i can help you.
by the way you dont need to connect 2 times into database and run the same query, i just edited your code a little bit:
<html>
<head>
<title></title>
</head>
<body>
<?php $con = mysqli_connect("localhost","root","root","company");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql="SELECT * FROM employees WHERE employees.jobs in ("programmer","hr","qa")";
$result = mysqli_query($con, $sql);
if ($result) { ?>
<label for="y">Select the job:</label>
<select name="loads" id="loads" onchange="">
<?php while($ri = mysqli_fetch_array($result)) { ?>
<option value="<?php echo $ri['id'];?>" > <?php echo $ri['jobs']; ?> </option>
<?php
}
}
?>
</select>
<table class="striped" border="1" align="center" id="demo">
<tr class="header">
<td align="center"><b>Name</b></td>
</tr>
<?php
// Fetch one and one row
while ($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row["name"] . " " . "</td>";
echo "</tr>";
}
mysqli_close($con);
?>
</table>
</body>
</html>
Am really confused about this error
i have done everything to solve this error but i cant do anything
can someone help me
am trying to display all value store in SQL database in php
<?php
require('includes/config.php');
//if not logged in redirect to login page
if (!$user->is_logged_in()) {
header('Location: login.php');
}
//define page title
$title = 'Members Page';
//include header template
require('layout/header.php');
?>
<div class="container">
<div class="row">
<div class="col-xs-12 col-sm-8 col-md-6 col-sm-offset-2 col-md-offset-3">
<h2>Member only page - Welcome <?php echo $_SESSION['username']; ?></h2>
<p><a href='logout.php'>Logout</a></p>
<hr>
</div>
</div>
<?php
$result = $db->prepare("SELECT * FROM password");
$result->execute();
while ($row = $db->fetch(PDO::FETCH_ASSOC))
{
$title = $row['email'];
$body = $row['pass'];
}
echo "<table>
<tr>
<th>Usernamw</th>
<th>Password</th>
</tr>
<tr>
<td>" . $title . "</td>
<td>" . $body . "</td>
</tr>
</table>" ?>
</div>
<?php
//include header template
require('layout/footer.php');
?>
this is were i get error
while ($row = $db->fetch(PDO::FETCH_ASSOC))
You're mixing up your variables/objects:
while ($row = $db->fetch(PDO::FETCH_ASSOC))
should be
while ($row = $result->fetch(PDO::FETCH_ASSOC))
You have to fetch your result, not your connection.
EDIT: To get all of your rows to echo, you have to include them in the while loop:
echo "<table>
<tr>
<th>Usernamw</th>
<th>Password</th>
</tr>";
while ($row = $result->fetch(PDO::FETCH_ASSOC))
{
$title = $row['email'];
$body = $row['pass'];
echo "<tr><td>" . $title . "</td><td>" . $body . "</td></tr>";
}
echo "</table>";
Try to call fetch on $result $result->fetch().
prepare() is a method of a PDO object, while fetch() is a PDOStatement method. $result->fetch should work.
I finally get reviews to work, which will allow user to post their review about the product and display them in a review page with product details, however I can't get it to work perfectly. When a user posts their review, it will update in the table but only the second post will show up.
The following image is from the test i was running,
Before posting a review
After posting the first review
After posting the second review
As the images display, the first review will never show up, only starting from the second and above,
Here my code for review page updated with full code
<?php
if (!isset($_SESSION)) {session_start();} //start session
if (!isset($_SESSION['client_ID'])) {
//echo "<script>alert('not logged in');</script>";
header("Location: index.html" );
}
?>
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
<meta name="keywords" content="Games, Gaming, PS4, PS3, XBOX, Video games">
<meta name="description" content="Games 4 You">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<title>Games 4 You</title>
<link rel="stylesheet" type="text/css" href="Styles/ProductsStyle.css">
<!-- javascript/jQuery -->
<script src="http://code.jquery.com/jquery-1.11.0.min.js"></script>
</head>
<body>
<!--Add the following script at the bottom of the web page (before </body></html>)
Support system using MyLiveChat.com
-->
<script type="text/javascript" async="async" defer="defer" data-cfasync="false" src="https://mylivechat.com/chatinline.aspx?hccid=42206151"></script>
<script>// disable zoom to keep image fit and always in position
document.firstElementChild.style.zoom = "reset";
// the above script will disable zoom in and out
</script>
<script type="text/javascript">
// this will auto change the background image to the following 7 images which are in the root Images/
// this is set to change every five second
// declare list of backgrounds
var images = ['bg-01.jpg', 'bg-02.jpg', 'bg-03.jpg', 'bg-04.jpg', 'bg-05.jpg', 'bg-06.jpg', 'bg-07.jpg'];
// declare function that changes the background
function setRandomBackground() {
// choose random background
var randomBackground = images[Math.floor(Math.random() * images.length)];
// set background with jQuery
$('body').css('background-image', 'url("Images/' + randomBackground + '")');
}
// declare function that sets the initial background, and starts the loop.
function startLoop() {
// Set initial background.
setRandomBackground();
// Tell browser to execute the setRandomBackground every 5 seconds.
setInterval(setRandomBackground, 5 * 1000);
}
// One the page has finished loading, execute the startLoop function
$(document).ready(startLoop);
</script>
<header id="header">
<div class="container">
<center><img src="Images/Title.png" alt="Title"></div>
</center>
</header>
<center>
<nav>
<?php
echo "<p> Welcome ".$_SESSION['client_name']."</p>";
//create connection
$con = new mysqli("localhost", "student", "student", "cib4003_h00233671_at");
if ($con->connect_errno) { //failed
echo "Failed to connect to MySQL: (" . $con->connect_errno . ") " . $con->connect_error;
}?>
<div class="wrapper">
<ul id="category" >
<li>Home</li>
<li>Products</li>
<li>View Cart</li>
<li>About</li>
<li>Settings</li>
<li>Logoff</li>
</ul>
</nav>
</div>
</center>
<main>
<h3>Available Products</h3>
<?php
$product = $_GET["RID"];
$_SESSION["product_name_RID"] = $_GET["RID"];
$sql="SELECT * FROM products,reviews WHERE products.Product_Name = '$product' AND reviews.Product_Name = '$product'";
//$sql="SELECT * FROM reviews WHERE Product_Name = '$product'";
// $sql="SELECT * FROM pizza,pizzacart WHERE pizza.Pizza_ID=pizzacart.Pizza_ID AND pizzacart.client_ID=".$_SESSION['client_ID'];
//echo "connected to DB";
//run SQL query
$result = mysqli_query($con,$sql);
//output result
if(mysqli_num_rows($result)==0) //no records found
{
$sql="SELECT * FROM products WHERE Product_Name = '$product'";
$result = mysqli_query($con,$sql);
// echo "<p>no records in DB".mysqli_num_rows($result)."</p>";
// echo "<p><a href=products.php></a>";
// link has been disable because i am using the <a for something else so i can't force the image to be in the center when using <a
// so the result will only be image that tell the customers no products found click all or search with different data
?>
<table class="table-style-one">
<tr>
<th>Product Image</th>
<th>Product Name</th>
<th>Description</th>
<th>Product Type</th>
<th>Console Type</th>
</tr>
<?php
while($row = mysqli_fetch_array($result)) { //loops through records
echo "<tr>";
echo "<td><img src='".$row['picture']."'/>";
echo "<td>".$row['Product_Name']."</td>";
echo "<td>".$row['Description']." <center><b><br>".$row['Trailer']."<br></b></center></td>";
echo "<td>".$row['Product_Type']."</td>";
echo "<td>".$row['Console_Type']."</td>";
echo "</tr>";
}
//end of loop
echo "</table>";
echo "<p>No Reviews available for this product.<br> To post a review of this product, fill up the below form.</p>";
//end of else
}
else
{
?>
<table class="table-style-one">
<tr>
<th>Product Image</th>
<th>Product Name</th>
<th>Description</th>
<th>Product Type</th>
<th>Console Type</th>
</tr>
<?php
while($row = mysqli_fetch_array($result)) { //loops through records
echo "<tr>";
echo "<td><img src='".$row['picture']."'/>";
echo "<td>".$row['Product_Name']."</td>";
echo "<td>".$row['Description']." <center><b><br>".$row['Trailer']."<br></b></center></td>";
echo "<td>".$row['Product_Type']."</td>";
echo "<td>".$row['Console_Type']."</td>";
echo "</tr>";
echo "<br>";
?>
<?php
while($row = mysqli_fetch_array($result)) {
echo "<table class=table-style-one align=center>";
// echo "<tr><th>Review ID</th><td>".$row['Review_ID']."</td></tr>";
echo "<tr><th>Review By:</th><td>".$_SESSION['client_name']."</td></tr>";
echo "<tr><th>Review Title</th><td>".$row['Review_Title']."</td></tr>";
echo "<tr><th>Rate:</th><td>".$row['Review_Rate']."/5</td></tr>";
echo "<tr><th>Review</th><td colspan=2>".$row['Review']."</td></tr>";
echo "<tr><th>Submitted On</th><td>".$row['Review_Date']."</td></tr>";
echo "<br>";
?>
<?php
}
//end of loop
echo "</table>";
//end of else
}
}
?>
<table class="table-style-one" align="center">
<tr>
<form method="POST" action="submitreview.php">
<!--<th>Product Name:</th><td> <input type="text" size="30" id="ReviewTitle" name="ReviewTitle" pattern=".{5,}" required title="5 characters minimum" placeholder="Review Title"></td> -->
<th>Review Title:</th><td> <input type="text" required size="30" id="ReviewTitle" name="ReviewTitle" pattern=".{5,}" required title="5 characters minimum" placeholder="Review Title"> </td>
<th>Rate: </th>
<td>
<select name="Review_Rate" required>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>
</td>
<tr>
<td colspan="4">
<textarea name="WriteReview" id="WriteReview" required rows="10" cols="50" wrap="physical" placeholder="Write your Review here" style="margin: 0px; width: 437px; height: 150px;"></textarea>
</td>
</tr>
<td align="center" colspan="2"><input type="submit" value="submit"></td>
<td colspan="2"><input type="Reset"></td>
</form> </tr>
</table>
</h2>
<br>
<br>
<br>
</main>
</body>
<footer>
<p>Made by Humaid Al Ali - H00233671</p>
<div id="google_translate_element"></div><script type="text/javascript">
function googleTranslateElementInit() {
new google.translate.TranslateElement({pageLanguage: 'en', layout: google.translate.TranslateElement.InlineLayout.HORIZONTAL, multilanguagePage: true}, 'google_translate_element');
}
</script><script type="text/javascript" src="//translate.google.com/translate_a/element.js?cb=googleTranslateElementInit"></script>
</footer>
</html>
and here the php file where I insert the review into the table
<?php
if (!isset($_SESSION)) {session_start();} //start session
if (!isset($_SESSION['client_ID'])) {
//echo "<script>alert('not logged in');</script>";
header("Location: index.html" );
}
?>
<?php
//new connection
$con = new mysqli("localhost", "student", "student", "cib4003_h00233671_at");
if ($con->connect_errno) { //failed
echo "Failed to connect to MySQL: (" . $con->connect_errno . ") " . $con->connect_error;
}
//success
//if ($_SERVER['REQUEST_METHOD'] === 'POST') {
// run sql
$sql ="INSERT INTO `cib4003_h00233671_at`.`reviews`(`Review_ID`, `Product_Name`, `client_ID`, `Review_Title`, `Review_Rate`, `Review`) VALUES (NULL, '".$_SESSION['product_name_RID']."', '".$_SESSION['client_ID']."', '".$_POST["ReviewTitle"]."', '".$_POST['Review_Rate']."', '".$_POST['WriteReview']."');";
if ($con->query($sql) === TRUE) {echo "<h3> New record created successfully</h3>";
header('Location: '. $_SERVER['HTTP_REFERER'] );
} else {
echo "Error : " . $sql . "<br>" . $con->error;
}
$con->close();
?>
Problem:
As the images display, the first review will never show up, only starting from the second and above
That's because you're fetching the first row which includes the first review but you don't display it, rather you starting fetching next reviews and display them. Look at the two conditions in while block inside the else block.
Solution:
To display reviews, you should do something like this:
<?php
// your code
if(mysqli_num_rows($result)==0){
// your code
}else{
echo "<table class='table-style-one'>";
echo "<tr>";
echo "<th>Product Image</th>";
echo "<th>Product Name</th>";
echo "<th>Description</th>";
echo "<th>Product Type</th>";
echo "<th>Console Type</th>";
echo "</tr>";
// fetch first row, which also contains first review
$row = mysqli_fetch_array($result);
echo "<tr>";
echo "<td><img src='" . $row['picture'] . "'/>";
echo "<td>" . $row['Product_Name'] . "</td>";
echo "<td>" . $row['Description'] . "<center><b><br>" . $row['Trailer'] . "<br></b></center></td>";
echo "<td>" . $row['Product_Type'] . "</td>";
echo "<td>" . $row['Console_Type'] . "</td>";
echo "</tr>";
echo "</table>";
echo "<table class=table-style-one align=center>";
echo "<tr>";
echo "<th>Review ID</th>";
echo "<th>Review By:</th>";
echo "<th>Review Title</th>";
echo "<th>Rate:</th>";
echo "<th>Review</th>";
echo "<th>Submitted On</th>";
echo "</tr>";
do{
// display first review and then fetch rest of the reviews
echo "<tr>";
echo "<td>" . $row['Review_ID'] . "</td>";
echo "<td>" . $_SESSION['client_name'] . "</td>";
echo "<td>" . $row['Review_Title'] . "</td>";
echo "<td>" . $row['Review_Rate'] . "/5</td>";
echo "<td colspan=2>" . $row['Review'] . "</td>";
echo "<td>" . $row['Review_Date'] . "</td>";
echo "</tr>";
}while($row = mysqli_fetch_array($result));
echo "</table>";
?>
</table>
<?php
}
// your code
?>
The problem looks to me to be in your SQL query. It looks as though your SQL query should actually be doing a join on the related column to bring together data across two different tables
$sql="SELECT *
FROM products AS p,
JOIN reviews AS r
ON p.Product_Name = r.Product_Name
WHERE products.Product_Name = '$product'";
But I think the easiest solution would be to run another query within the table output loop where you get the reviews for the product name you are currently outputting and loop over those.
I have an HTML table being displayed from php data and I'm trying to move data from one table to another based on what is clicked. I cannot get it to move from table to another but I"m not getting any code error.
<html>
<head>
<title>View Requests</title>
</head>
<body>
<div class="breadcrumbs">
<center>
Home · Requests
</center>
</div>
<?php
require_once '../../scripts/app_config.php';
require_once '../../scripts/database_connection.php';
// get results from database
$result = mysql_query("SELECT * FROM temp")
or die(mysql_error());
echo "<table border='1' cellpadding='10'>";
echo "<tr> <th>ID</th> <th>First Name</th> <th>Last Name</th> <th>Username</th> <th></th> <th></th></tr>";
// loop through results of database query, displaying them in the table
while($row = mysql_fetch_array( $result )) {
// echo out the contents of each row into a table
echo "<tr>";
echo '<td>' . $row['User_id'] . '</td>';
echo '<td>' . $row['First_name'] . '</td>';
echo '<td>' . $row['Last_name'] . '</td>';
echo '<td>' . $row['Username'] . '</td>';
echo '<td>Approve</td>';
echo '<td>Delete</td>';
echo "</tr>";
}
// close table>
echo "</table>";
?>
</body>
</html>
PHP EDIT SCRIPT:
<html>
<head>
<title>View Requests</title>
</head>
<body>
<div class="breadcrumbs">
<center>
Home · Requests · All Users
</center>
<?php
require_once '../../scripts/app_config.php';
require_once '../../scripts/database_connection.php';
$id = $_GET['id'];
$sql = ("INSERT INTO users
SELECT * FROM temp WHERE User_id = $id ");
mysql_query($sql)
or die(mysql_error());
?>
</body>
</html>
I needed to use
$id = $_get['id];
and in the insert statement i had to replace id with User_id
<html>
<head>
<title>View Requests</title>
</head>
<body>
<div class="breadcrumbs">
<center>
Home · Requests · All Users
</center>
<?php
require_once '../../scripts/app_config.php';
require_once '../../scripts/database_connection.php';
$id = $_GET['id'];
$sql = ("INSERT INTO users
SELECT * FROM temp WHERE User_id = $id ");
mysql_query($sql)
or die(mysql_error());
?>
</body>
</html>
I need to delete a specific row in php.. so how could I get the ID or other way to delete a specific record
dbconnect.php just a simple database connection
<?php
$con = mysqli_connect("localhost","root","","phpractice");
if(mysqli_connect_errno()){
echo "Database connection failed";
}
?>
index.php the page where the user can see
<html>
<head>
<link rel="stylesheet" type="text/css" href="../styles/index.css">
</head>
<title>Home Page</title>
<?php include'../script/dbconnect.php';?>
<body>
<div id="container">
<div id="table">
<?php
$result = mysqli_query($con,"SELECT * FROM users");
echo "<table border='1'>
<tr>
<th>First Name</th>
<th>Last Name</th>
<th>Username</th>
<th>Password</th>
<th colspan=2>Controls</th>
</tr>
";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>".$row['firstname']."</td>";
echo "<td>".$row['lastname']."</td>";
echo "<td>".$row['username']."</td>";
echo "<td>".$row['password']."</td>";
echo "<td>"."<a href='#'>EDIT</a>"."</td>";
echo "<td><a href='../script/delete.php?id=".$row['user_id']."'>DELETE</a></td>";
echo "</tr>";
}
echo "</table>";
?>
Add User
</div><!--table-->
</div><!--container-->
</body>
</html>
delete.php the delete script
<?php
include '../script/dbconnect.php';
$id = $_GET['user_id'];
$query = "DELETE FROM users WHERE user_id = $id";
mysqli_query($con, $query) or die (mysqli_error($con));
echo "DELETE USER SUCCESSFUL!";
echo "</br>";
echo "<a href='../main/index.php'>RETURN TO DISPLAY</a>";
?>
thanks in advance
in index.php use:
echo "<td><a href='../script/delete.php?user_id=".$row['user_id']."'>DELETE</a></td>";
then use $_GET['user_id']; in delete.php