Unable to update table inside of while loop - php

I am querying table in a while loop and has a button to update data for each row when clicked. But the update isn't happening .
The table column "Enroll" is to be updated based on clicking the submit button for that row.
<?php
$connection = mysql_connect('localhost', 'root', '');
mysql_select_db('mydata');
$query = "SELECT * FROM internship";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
if($row['Enroll']=="unenroll")
{
echo '<div class="intern">';
echo '<h1>'.$row['course'] . "</h1>" ;
echo '<h2>'.$row['compay'] . "</h2>";
echo "<h3>Stipend:Rs." . $row['stipend'] ."<h3>";
echo "<h3>Duration:".$row['duration']."month</h3>";
echo "<h3>Start-date:".$row['start date']."</h3>";
echo '</div>';
echo '<form method="post" action="">';
echo '<button name="add_to_cart" type="submit" ><h2>Enroll</h2></button>';
echo '</form>';
if(isset($_POST["add_to_cart"]))
{
$selectedProduct = $row["ID"];
$sql='UPDATE internship SET Enroll="enrolled" WHERE ID="$selectedProduct"';
mysql_query($connection,$sql);
header("location:enrolled.php");
}
}
}
mysql_close($connection);
?>
I am not able to update. Please help.

There are few issues with your code, such as:
You haven't included any ID value in the button of each row.
echo '<button name="add_to_cart" type="submit" ><h2>Enroll</h2></button>';
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ --> no value="..."
You are not using the selected row i.e. the row for which the button has been clicked to update the table row with enrolled.
$selectedProduct = $row["ID"];
^^^^^^^^^^
So your code should be like this:
<?php
$connection = mysql_connect('localhost', 'root', '');
mysql_select_db('mydata');
if(isset($_POST["add_to_cart"])){
$selectedProduct = $_POST['add_to_cart'];
$sql ="UPDATE internship SET Enroll = 'enrolled' WHERE ID = '$selectedProduct'";
mysql_query($connection,$sql);
header("location:enrolled.php");
exit();
}
$query = "SELECT * FROM internship";
$result = mysql_query($query);
while($row = mysql_fetch_array($result)){
if($row['Enroll']=="unenroll"){
?>
<div class="intern">
<h1><?php echo $row['course']; ?></h1>
<h2><?php echo $row['compay']; ?></h2>
<h3>Stipend:Rs <?php echo $row['stipend']; ?><h3>
<h3>Duration: <?php echo $row['duration']; ?> month</h3>
<h3>Start-date: <?php echo $row['start date']; ?></h3>
</div>
<form method="post" action="">
<button name="add_to_cart" value="<?php echo $row['ID']; ?>" type="submit"><h2>Enroll</h2></button>
</form>
<?php
}
}
mysql_close($connection);
?>
Sidenote: Don't use mysql_* functions, they are deprecated as of PHP 5.5 and are removed altogether in PHP 7.0. Use mysqli or pdo instead. And this is why you shouldn't use mysql_* functions.

Related

How to submit just one button from a set of buttons that are displayed using a php while-loop

I am coding a website for an online university portal where I have a programs/courses page in which I am displaying the programs/courses on the page using data from the database in a PHP while-loop I have the enroll buttons also being displayed in that same while loop. but I'm having a bit of difficulty submitting the enroll buttons as when I click one of them all of them get submitted.
can anyone please let me know what I'm doing wrong here or if I have to use any javascript in this case!
<?php
session_start();
$con = mysqli_connect('localhost', 'root', '');
mysqli_select_db($con, 'htdatabase');
if ($con->connect_error) {
die("Connection failed: " . $con->connect_error);
}
$id = $_SESSION['userID'];
$sql = "SELECT * FROM programs";
$result = $con->query($sql);
if ($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
$i = '';
$progID = $row["progID"];
$name = $row["progName"];
$halfTime = $row["halfTDuration"];
$fullTime = $row["fullTDuration"];
$fee = $row["fee"];
$descrip = $row["description"];
$stringname = strval($name);
$spaceRemoved = str_replace(' ', '', $stringname);
?>
<div class="card-header" id="headingOne">
<h5 class="mb-0">
<?php echo "<button class='btn btn-link' type='button' data-toggle='collapse' data-target='#$spaceRemoved' aria-expanded='false' aria-controls='$spaceRemoved'> $name </button>"; ?>
</h5>
</div>
<?php echo "<div id='$spaceRemoved' class='collapse' aria-labelledby='headingOne' data-parent='#accordionExample'>"; ?>
<div>
<div class="ccard-body col-md-9">
<h6><?php echo $descrip; ?></h6>
<hr>
<h5>Duration:</h5>
<h6>Full time: <?php echo $fullTime; ?></h6>
<h6>Half time: <?php echo $halfTime; echo $i; ?></h6>
<hr>
<h5 style="display: inline-block;">Estimated fees: $</h5><h5 style="display: inline-block;"><?php echo $fee ?></h5>
</div>
<form action="programs.php" method="post">
<div id="enroll" class="col-md-3">
<?php
$sql1 = "SELECT * FROM userprograms WHERE userID = '$id' AND progID = '$progID'";
$result1 = $con->query($sql1);
if ($result1->num_rows > 0) {
echo '<div id="enrolled" name="enrolled">ENROLLED</div>';
} else {
if (isset($_POST["enroll"])) {
$enrollqry = "insert into userprograms (userID, progID) values ('$id' , '$progID')";
mysqli_query($con, $enrollqry);
}
echo "<button name='enroll'type='submit'>ENROLL</button>";
}
?>
</div>
</form>
</div>
</div>
<?php
}
} ?>
You can specify a value for the button. like
<button name='enroll' value="<?php echo $program_id?>" type='submit'>ENROLL</button>
Then when checking for $_POST['enroll'] check the value and also validate it before entry to db.
After clicking the submit button a browser will send a POST request to programs.php with a form data, that includes values of input & button tags.
<input type="submit" name="course1" value="42">Subscribe</input>
<input type="text" name="first_name" placeholder="Your name"/>
Will send
course1=42
first_name=...
So you should either give a unique name to each submit button to be able to distinguish them on the server-side, or set up distinct values, as #mohamed-jailam mentioned above.

Delete from database depending on button clicked

I have a script that prints in the screen all the data of a table. Associated to each row of data, I have a delete button, and I would like that, when a button of any row is cliked, the row is deleted. To do so, I have got the following code:
$con = mysqli_connect("","","","");
$result = mysqli_query($con,"SELECT * FROM `clientes_pmt`");
while($row = mysqli_fetch_array($result)){
?> <button name="delete" value="<?php echo $row['id']; ?>" type="submit"><img src="paginas/borrar.jpg" /></button>
<a href="page<?php echo $row['nombre']; ?>">
<div><p><?php echo $row['nombre']; ?></p></div>
<div><p><?php echo $row['pais']; ?></p></div>
</a>
<section class="clearboth"></section><br><?php
}
if(isset($_POST['delete'])){
$id = $_POST['delete'];
$result = mysqli_query($con,"DELETE FROM `clientes_pmt` WHERE id = '$id'");
}
mysqli_close($con);
I receive no errors but the row is not being deleted.
Enclose the button inside a form element and set appropriate action and method attributes, something like:
<form action="/delete.php" method="POST">
<button ...> [your button]
</form>
You have to modify these line
$con = mysqli_connect("localhost","root","","YOUR DATABASE NAME");
To see the result you can put a header() to the end of this part of code:
if(isset($_POST['delete'])){
$id = $_POST['delete'];
$result = mysqli_query($con,"DELETE FROM `clientes_pmt` WHERE id = '$id'");
header('location:your_page_where_the_rows_are.php');
}
I think it may be because the $id was not getting added into the query.
Try bellow?
if(isset($_POST['delete'])){ $id = $_POST['delete']; $result = mysqli_query($con,"DELETE FROM clientes_pmt WHERE id = '".$id."'); }
Use Form tag and Header for redirect your page on same page.
Try This -
<?php
$con = mysqli_connect("","","","");
$result = mysqli_query($con,"SELECT * FROM `clientes_pmt`");
while($row = mysqli_fetch_array($result)){
?>
<form action="" method="post" >
<button name="delete" value="<?php echo $row['id']; ?>" type="submit"><img src="paginas/borrar.jpg" /></button>
</form>
<a href="page<?php echo $row['nombre']; ?>">
<div><p><?php echo $row['nombre']; ?></p></div>
<div><p><?php echo $row['pais']; ?></p></div>
</a>
<section class="clearboth"></section><br><?php
}
if(isset($_POST['delete'])){
$id = $_POST['delete'];
$result = mysqli_query($con,"DELETE FROM `clientes_pmt` WHERE id = '$id'");
header('location:'.$_SERVER['PHP_SELF']);
}
mysqli_close($con);
?>

How to get ID of a query asked long ago and post it to another table?

I have a table in data base with columns like ID,question,date. and an php page (1.php)which outputs question and date from db table and takes answer in textarea. when clicked submit, it redirects to another php page (2.php) where i m supposed to store the typed answer into another table in db.
here, my problem is how can i add question_ID to the second table in db along with answer.
1.PHP script as follows::
<div class="name">
<?php
$sql = "SELECT * FROM input";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$index = 0;
while($row = $result->fetch_assoc()) {
$index++;
?>
<div id="q">
<?php echo $row["question"]; ?> </B>
<?php
echo '<button class="add" id="add_'.$index.'"><B>Add Answer</B></button>';
echo '<form style="display:none;" name="answer_'.$index.'" method="post" action="output.php">'; // I dont think openning form from row to row would be nice!
echo '<textarea type="text" class="addtext" name="addtext" required id="addtext_'.$index.'" placeholder="Please type your answer here.." ></textarea>';
echo '<button onClick="addsubmit('.$index.');" type="submit" id="addsubmit_'.$index.'" class="addsubmit"><B>Submit</B></button>';
echo '</form>';
?>
<small><p><?php echo $row["date"]; ?></p></small>
2.PHP script as follows::
<?php include('1.php'); ?>
<?php
$servername = "localhost";
$dbusername = "root";
$dbpassword = "******";
$dbname = "the_database";
$addtext = $_POST['addtext'];
$date = date_default_timezone_set('Asia/Kolkata');
$date = date('d/m/Y H:i:s');
$conn = new mysqli ($servername, $dbusername, $dbpassword, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO output (answer, date)
VALUES ('$addtext', '$date')";
if ($conn->query($sql) === TRUE) {
echo '<script language="javascript">';
echo 'alert("Your Answer has been Succesfully posted")';
echo '</script>';
echo '';
}
else {
echo "ERROR" . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Answer and date saved in db table successfully but got stuck in getting question_ID...
Any help is greatly Appreciated.
Add the hidden field:
on 1.php
echo '<form style="display:none;" name="answer_'.$index.'" method="post" action="output.php">'; // I dont think openning form from row to row would be nice!
echo '<input type="hidden" name="questionid" value="<?php echo $row[ID]?>"/>'
echo '<textarea type="text" class="addtext" name="addtext" required id="addtext_'.$index.'" placeholder="Please type your answer here.." ></textarea>';
echo '<button onClick="addsubmit('.$index.');" type="submit" id="addsubmit_'.$index.'" class="addsubmit"><B>Submit</B></button>';
echo '</form>';
on page 2.php
$qid = $_POST['questionid']; //get the question id and insert in the table

How to delete multiple rows from mysql database with checkbox using PHP?

I try to delete my data in "admin" database, but the delete button does not function.
This is my top part
<?php
$host="localhost"; // Host name
$username="root"; // Mysql username
$password=""; // Mysql password
$db_name="admin"; // Database name
$tbl_name="admin"; // Table name
// Connect to server and select databse.
mysql_connect("$host", "$username", "$password")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
$count=mysql_num_rows($result);
?>
This is my checkbox code
<tbody>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td><?php echo $rows['course_code']; ?></td>
<td><?php echo $rows['course_name']; ?></td>
<td><?php echo $rows['lecture_id']; ?></td>
<td><input name="checkbox[]" type="checkbox"
id="checkbox[]" value="<?php echo $rows['course_code'];?>"></td>
<td><form>
</form>
</td>
</tr>
<?php
}
?>
</tbody>
and, this is my button code
<input type='button' id="delete" value='Delete' name='delete'>
This is my php function code
<?php
if(isset($_POST['delete'])){
for($i=0;$i<$count;$i++){
$del_id = $checkbox[$i];
$sql = "DELETE FROM $tbl_name WHERE course_code='$del_id'";
$result = mysql_query($sql);
}
if($result){
echo "<meta http-equiv=\"refresh\" content=\"0;URL=delete.php\">";
}
}
mysql_close();
?>
include all the input elements within your <form> tags: <form> all inputs are here </form>
update:
<input name = "checkbox[]" type="checkbox" id="checkbox[]" value="<?php echo $rows['course_code'];?>">
to (id doesn't matter here):
<input name="checkbox[]" type="checkbox" value="<?php echo $rows['course_code'];?>"/>
and your button code:
<input type='button' id="delete" value='Delete' name='delete'>
to
<input type="submit" value="Delete"/>
set opening <form> tag to <form action="delete.php" method="post">
Note:
I assume below codes are in delete.php file. if not replace "delete.php" with that name in above opening form tag.
your delete.php file:
<?php
$cheks = implode("','", $_POST['checkbox']);
$sql = "delete from $tbl_name where course_code in ('$cheks')";
$result = mysql_query($sql) or die(mysql_error());
mysql_close();
?>
Note:
Since mysql_ will deprecate on future, better is use mysqli extension. But before use that, you have to enable it on your server. mysqli is a part of php and newer version of php has it but not enabled. To enable this, view php info page and find the path of php.ini file in "Loaded Configuration File" row on that page.
You can see php info page by loading below php file in the browser:
<?php
phpinfo();
?>
open that php.ini file in a text editor and un-comment or add a line extension=php_mysqli.dll at the extensions list there.
also search for "extension_dir" and open the directory it says and make sure php_mysqli.dll file is there.
(you may have .so extension if you not use windows OS)
Then restart your server and you are done!
By Fred -ii-
Using mysqli_ with prepared statements is indeed a better and
safer method. However, some will even suggest PDO, but even PDO
doesn't have some of the functionalities that mysqli_ offers;
strangely that. Even PDO needs sanitization. Many think that using PDO will solve injection issues, which is false.
-Thanks Fred.
try this code. it is working well.
connection.php
<?php $hostname_conection = "localhost"; /* this is the server name(assigned to variable) which is localhost since it runs on local machine */
$database_conection = "company"; /* this is the database name( assigned to variable)*/
$username_conection = "root"; /* user name (assigned to variable)*/
$password_conection = ""; /*password (assigned to variable) */
$conection = mysql_connect($hostname_conection, $username_conection, $password_conection) or trigger_error(mysql_error(),E_USER_ERROR); /* Mysql_connect function is used to conncet with database it takes three parameters server/hostname, username,and password*/
mysql_select_db($database_conection,$conection) or die(mysql_error("could not connect to database!")); /* Mysql_select is used to select the database it takes two parameters databasename and connection variable in this case $conection */
?>
multiple_delete.php
<?php require_once('conection.php'); ?>
<?php
in
/* now to display the data from the database which we inserted in above form we */ /* we make the query to select data from the table EMP */
$display = "select * from test_mysql";
$result = mysql_query($display, $conection) or die(mysql_error()); /* the query is executed and result of the query is stored in variable $result */
if ($result == FALSE) {
die(mysql_error()); /* displays error */
} ?> <h1 align="center"> Displaying Recods in Table </h1>
<form method="get" action="" id="deleteform" >
<table width="245" border="1" align="center">
<tr>
<td width="51">
<input type="submit" name="delete" id="button" value="delete" onclick="document.getElementById('deleteform').action = 'delete.php';document.getElementById('deleteform').submit();"/> <!--- here on clicking the button the form is submitted and action is set to delete.php Here we have used javaScript document refers to this whole page and now we can access any tag that has its id with help of getElementById() method and after the we specify the operation we want to perform in this case action and submit. --->
</td>
<td width="50">id</td>
<td width="55">name</td>
<td width="47">lastname</td>
</tr>
<?php
while ($rows = mysql_fetch_array($result))
{ /* here we make use of the while loop which fetch the data from the $result int array form and stores in $row now we can display each field from the table with $row[‘field_name’] as below */
?>
<tr>
<td>
<input type="checkbox" name="empids[]" value="<?php echo $rows['id']; ?>" /> <!--here with each checkbox we send the id of the record in the empids[] array --->
</td>
<td>
<?php echo $rows['id'] ?>
</td>
<td>
<?php echo $rows['lastname'] ?>
</td>
<td><?php echo $rows['name'] ?></td>
<?php } ?>
</tr>
</table>
</form> ?>
</body>
</html>
delete.php
<?php
require_once('conection.php');
?>
<?php
if (isset($_GET['delete'])) /* checks weather $_GET['delete'] is set*/
{
if (isset($_GET['empids'])) /* checks weather $_GET['empids'] is set */
{
$checkbox = $_GET['empids']; /* value is stored in $checbox variable */
if (is_array($checkbox))
{
foreach ($checkbox as $key => $your_slected_id) /* for each loop is used to get id and that id is used to delete the record below */
{
$q="DELETE FROM test_mysql WHERE id=$your_slected_id "; /* Sql query to delete the records whose id is equal to $your_slected_id */
mysql_query($q,$conection) ; /* runs the query */
}
header("location:multiple_delete.php"); /* Goes back to index.php */
}
} else
{
echo" you have not selected reords .. to delete";
}
} ?>
$sql = "SELECT * FROM blacklist";
$result = $link->query($sql);
$count=mysqli_num_rows($result);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc())
{
echo "<table>";
echo "<th>";
echo "<td>" . "ID: " . $row["id"]."</td>";
echo "<td>" . " Dial Target: " . $row["dial_target"]."</td>";
echo "<td>" . " Destination: " . $row["pozn"]."</td>";
echo "<td>" . " Date: " . $row["block_date"] . "</td>";
echo "<td>" . "<div class='background' style='position: relative; top:8px;'>" . "<form>" . "<input action='index.php' method='post' type='checkbox' name='chechbox[]' value='".$row["id"]."'/>" ."</form>" . "</div>" . "</td>";
echo "</th>";
echo "</table>";
echo "</br>";
}
}
else
{
echo "0 results";
}
if(isset($_POST['Delete']))
{
for($i=0;$i<$count;$i++)
{
$del_id = $checkbox[$i];
$del = "DELETE FROM blacklist WHERE Delete='$del_id'";
$result = $link->query($del);
}
if($result)
{
echo "<meta http-equiv=\"refresh\" content=\"0;URL=index.php\">";
}
}
<!-- DELETE BUTTON -->
<form>
<input type='Submit' id="Delete" value='Delete' name='Delete'/>
</form>
<?php
$args1 = array(
'role' => 'Vendor',
'orderby' => 'user_nicename',
'exclude' => $user_id.',1',
'order' => 'ASC'
);
$subscribers = get_users($args1); foreach ($subscribers as $user) {
$fvendorck = $wpdb->get_row("select * from wp_vandor where parent_id = '".$user_id."' and child_id = '".$user->id."'");
$isfavvendor = $fvendorck->child_id;
if(!empty($isfavvendor)) {
?>
<li><input type="checkbox" id="listID" value='<?php echo $user->id; ?>' name="chk1[]" checked=""/><?php echo $user->headline; ?></li>
<?php }else{ ?>
<li><input type="checkbox" id="listID" value='<?php echo $user->id; ?>' name="chk1[]" /><?php echo $user->headline; ?></li>
<?php } }?>
</ul>

show single entry on a new page with sending id of row

i am very novice to php and mysqli and found a great tutorial but am needing some help.
i am wanting a row to be linkable and send it to another page named single.php?id=ROWID so it will show the single entry
this is what i got so far.
<html>
<head>
<title>MySQLi Tutorial</title>
</head>
<body>
<?php
//include database connection
include 'db_connect.php';
$action = isset($_GET['action']) ? $_GET['action'] : "";
if($action=='delete'){ //if the user clicked ok, run our delete query
$query = "DELETE FROM users WHERE id = ".$mysqli->real_escape_string($_GET['id'])."";
if( $mysqli->query($query) ){
echo "User was deleted.";
}else{
echo "Database Error: Unable to delete record.";
}
}
$query = "select * from users";
$result = $mysqli->query( $query );
$num_results = $result->num_rows;
echo "<div><a href='add.php'>Create New Record</a></div>";
if( $num_results ){
echo "<table border='1'>";//start table
//creating our table heading
echo "<tr>";
echo "<th><a href=\"single.php?id={$id}\">Firstname</></th>";
echo "<th>Lastname</th>";
echo "<th>Username</th>";
echo "<th>Action</th>";
echo "</tr>";
//loop to show each records
while( $row = $result->fetch_assoc() ){
//extract row
//this will make $row['firstname'] to
//just $firstname only
extract($row);
//creating new table row per record
echo "<tr>";
echo "<td>{$firstname}</td>";
echo "<td>{$lastname}</td>";
echo "<td>{$username}</td>";
echo "<td>";
echo "<a href='edit.php?id={$id}'>Edit</a>";
echo " / ";
echo "<a href='#' onclick='delete_user( {$id} );'>Delete</a>";
echo "</td>";
echo "</tr>";
}
echo "</table>";//end table
}else{
//if table is empty
echo "No records found.";
}
//disconnect from database
$result->free();
$mysqli->close();
?>
<script type='text/javascript'>
function delete_user( id ){
//this script helps us to
var answer = confirm('Are you sure?');
if ( answer ){ //if user clicked ok
//redirect to url with action as delete and id to the record to be deleted
window.location = 'index.php?action=delete&id=' + id;
}
}
</script>
</body>
</html>
i am right in thinking i would be sending the rows id in the url ?
echo "<th><a href=\"single.php?id={$id}\">Firstname</></th>";
but i am having issues with single.php what code would i have to put to show the single entry?
i have been on this a while and got no were near so i deleted the code and swallowed my pride to seek some help :/
thanks in advance
Thank you for the interesting question.
First, let me inform you that, although you are using a moder-looking database access library, the way you are using it is as ancient as a mammoth fossil.
Several things to consider
Never use mysqli as is, but only in the form of some higher level abstraction library.
Never use real_escape_string in the application code but use prepared statements only.
Never mix your database code with HTML output. Get your data first, then start for output.
Never use GET method to modify the data.
Here goes the example based on the above principles. It does ALL basic CRUD operations:
<?
include 'safemysql.class.php'; // a library
$db = new SafeMysql();
$table = "test";
if($_SERVER['REQUEST_METHOD']=='POST') {
if (isset($_POST['delete'])) {
$db->query("DELETE FROM ?n WHERE id=?i",$table,$_POST['delete']);
} elseif ($_POST['id']) {
$db->query("UPDATE ?n SET name=?s WHERE id=?i",$table,$_POST['name'],$_POST['id']);
} else {
$db->query("INSERT INTO ?n SET name=?s",$table,$_POST['name']);
}
header("Location: http://".$_SERVER['HTTP_HOST'].$_SERVER['PHP_SELF']);
exit;
}
if (!isset($_GET['id'])) {
$LIST = $db->getAll("SELECT * FROM ?n",$table);
include 'list.php';
} else {
if ($_GET['id']) {
$row = $db->getRow("SELECT * FROM ?n WHERE id=?i", $table, $_GET['id']);
foreach ($row as $k => $v) $row[$k]=htmlspecialchars($v);
} else {
$row['name']='';
$row['id']=0;
}
include 'form.php';
}
It is using templates to display the data:
list.php
Add item
<? foreach ($LIST as $row): ?>
<li><?=$row['name']?>
<? endforeach ?>
and form.php
<form method="POST">
<input type="text" name="name" value="<?=$row['name']?>"><br>
<input type="hidden" name="id" value="<?=$row['id']?>">
<input type="submit"><br>
Return to the list
</form>
<? if ($row['id']):?>
<div align=right>
<form method="POST">
<input type="hidden" name="delete" value="<?=$row['id']?>">
<input type="submit" value="Удалить"><br>
</form>
</div>
<?endif?>
here goes the part for display.
if ($_GET['id']) {
$row = $db->getRow("SELECT * FROM ?n WHERE id=?i", $table, $_GET['id']);
foreach ($row as $k => $v) $row[$k]=htmlspecialchars($v);
} else {
$row['name']='';
$row['id']=0;
}
include 'form.php';
if you don't want to show the form - create another template called single.php with whatever markup you wish
Single.php
I Use PDO if u want you can make it with MySQLi too.
<?php
include("db_connect.php"); // database configuration file
if(isset($_GET['id'])
{
$id = (int) $_GET['id'];
$sql = "SELECT * FROM `users` WHERE id=?";
$query = $conn->prepare($sql); // $conn is PDO object yours can be different
$query->bindValue(1,$id);
$query->execute();
if($query){
$row = $query->fetch(); //
}else{
echo "Error with Database";
}
}
else // Error for the Id selection
{
echo("ID is not selected");
}
?>
No while loop because you want just 1 record. $row variable is just for test because i don't know your fields in your DB
<table border="1">
<tr>
<td>ID</td>
<td>Firstname</td>
<td>Lastname</td>
</tr>
<tr>
<td><?php echo $row['id]; ?></td>
<td><?php echo $row['firstname']; ?></td>
<td><?php echo $row['lastname']; ?></td>
</tr>
</table>
in your single.php
$id=$_GET['id'];
$query="select * from users where id='".$id."'";

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