I have virtually no programming experience and trying this first project, I am a bit stuck on how to update the database, so I click on edit and the correct record gets loaded into the edit screen update.php
When I click update, I get the message from updated.php saying that the database has been updated, but the database does not get updated, when I display the records they are the same as before the update, thanks in advance for all your help.
the following code:
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
"http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Form Edit Data</title>
</head>
<body>
<table border=1>
<tr>
<td align=center>Form Edit Employees Data</td>
</tr>
<tr>
<td>
<table>
<?
$user_name = "";
$password = "";
$database = "";
$server = "localhost";
mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database);
$id = $_GET['id'];
$order = "SELECT * FROM MY_ID where ID = ' " .$id . " ' ";
$result = mysql_query($order);
$row = mysql_fetch_array($result);
?>
<form method="post" action="edit_data.php"?id=<?= $id ?>>
<input type="text" name="id" value="<? echo "$row[ID]"?>">
<tr>
<td>First Name</td>
<td>
<input type="text" name="FirsName" size="20" value="<? echo "$row[FirstName]"?>">
</td>
</tr>
<tr>
<td>Sur Name</td>
<td>
<input type="text" name="SurName" size="40" value="<? echo "$row[SurName]"?>">
</td>
</tr>
<tr>
<td>Address</td>
<td>
<input type="text" name="Address" size="40" value="<? echo "$row[Address]"?>">
</td>
</tr>
<tr>
<td align="right">
<input type="submit" name="submit" value="submit">
</td>
</tr>
</form>
</table>
</td>
</tr>
</table>
</body>
</html>
and here is the other file
<?php
$user_name = "";
$password = "";
$database = "";
$server = "";
mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database);
$id = $_REQUEST['ID'];
$FirstName = trim(mysql_real_escape_string($_POST["FirstName"]));
$SurName = trim(mysql_real_escape_string($_POST["SurName"]));
$Address = trim(mysql_real_escape_string($_POST["Address"]));
$sql = "UPDATE MY_ID SET FirstName='$FirstName',SurName='$SurName',Address='$Address' WHERE ID='$id'";
$result=mysql_query($sql);
if ($result){
echo "Successful";
echo "<BR>";
echo "<a href='edit.php'>View result</a>";
}
else {
echo "ERROR";
}
?>
Looks like you forget the double quotation mark and the full stop. You should write it as: '".$example."'
$sql = "UPDATE MY_ID SET FirstName='".$FirstName."',SurName='".$SurName."',Address='".$Address.:' WHERE ID='".$id."'";
It is because your form method is POST, and you are trying to GET ID.
Probably ID returns null.
My suggestion is to put a hidden input in your form as with name="ID", then read it in your posted page as $_POST["ID"];
Yes, the answer is as Mansours said. You should not use single quota to your variable.
So, it's bad practice writing code something like this:
<input type="text" value="<?php echo "$row[name]"; ?>">
it should be
<input type="text" value="<?php echo $row['name']; ?>">
it would be clear, and also, when inserting or updating the record you should write as follow:
$sql = "UPDATE MY_ID SET FirstName='" . $FirstName . "',
SurName='" . $SurName . "',
Address='" . $Address . "'
WHERE ID='" . $id . "'";
mysql_query($sql);
Related
I have a form and successfully connect them to the database.
Now I'm trying to update the data. Unfortunately, nothing happened when I click the submit button. I'm sure I miss something. Please help me, thank you.
config.php :
<?php
$conn=odbc_connect("dsn", "", "");
if (!$conn)
{
exit("Connection Failed : " . $conn);
}
?>
This is my code :
<?php
include "config.php";
ini_set('error_reporting', E_ALL);
error_reporting(-1);
$sql = odbc_exec( $conn, "SELECT
UserId,
UserName,
UserEmail
FROM DBA.tblUser
WHERE UserId='".$_GET['UserId']."'");
if(isset($_POST['submit']))
{
$UserId=$_POST["UserId"];
$UserName=$_POST["UserName"];
$UserEmail=$_POST["UserEmail"];
//UPDATE
$stmt = odbc_exec( $conn,
"UPDATE DBA.tableUsers SET
UserName = '$UserName',
UserEmail ='$UserEmail'
WHERE UserId=$UserId");
if ($stmt) {
echo "Update Success";
echo $UserName;
echo $UserEmail;
} else {
"Error : " . odbc_errormsg();
}
}
?>
Form :
<form class="form" method="post">
<tr>
<td class = "userid">User ID</td>
<td><?php echo $UserId = odbc_result($sql,'UserId'); ?></td>
</tr>
<tr>
<td class = "name">User Name<span class="required"> * </span></td>
<td><input type="text" name="UserName" value="<?php echo $UserName = odbc_result($sql,'UserName'); ?>"></td>
</tr>
<tr>
<td class = "email">Email<span class="required"> * </span></td>
<td><input type="text" name="UserEmail" value="<?php echo $UserEmail = odbc_result($sql,'UserEmail'); ?>"></td>
</tr>
<button name="submit" type="submit" value ="submit" >Update</button>
</form>
Replace
$UserName=$_POST["UserName"];
With
$UserName=$_POST["UserId"];
Why? Simply because in your form, you have name as UserId and not UserName
<input type="text" name="UserId" value="<?php echo $UserName = odbc_result($sql,'UserName'); ?>"></td>
I managed to get the answer with the same concept and variable.
config.php :
<?php
$conn=odbc_connect("dsn", "", "");
if ($conn) {
echo "Connected";
}
if (!$conn) {
exit("Connection Failed : " . $conn);
}
?>
code :
<?php
include "config.php";
$sql = odbc_exec( $conn, "SELECT
UserId,
UserName,
UserEmail
FROM DBA.tblUser
WHERE UserId='".$_GET['UserId']."'");
if(isset($_POST['submit'])) {
$UserName=$_POST["UserName"];
$UserEmail=$_POST["UserEmail"];
$stmt = odbc_exec($conn,
"UPDATE DBA.tblUser SET
UserName = '$UserName',
UserEmail ='$UserEmail'
WHERE UserId='".$_GET['UserId']."'");
if ($stmt) {
echo "Update Successfull";
echo $UserName;
echo $UserEmail;
}
else {
"Error : " . odbc_errormsg();
}
}
?>
html :
<!DOCTYPE html>
<html>
<head>
<title>Administration</title>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="css/userpage.css" media="screen"/>
<body>
<content>
<div class="user-form">
<div class="user-form-heading">Update User</div><br>
<form class="form" method="post">
<h><table class ="user">
<!-- .................................. Updatable ....................................... -->
<tr>
<td class = "name">User Name<span class="required"> * </span></td>
<td><input type="text" name="UserName" value="<?php echo $UserName = odbc_result($sql,'UserName'); ?>"></td>
</tr>
<tr>
<td class = "userid">User Id</td>
<td><?php echo $UserId = odbc_result($sql,'UserId'); ?></td>
</tr>
<tr>
<td class = "email">Email<span class="required"> * </span></td>
<td><input type="text" name="UserEmail" value="<?php echo $UserEmail = odbc_result($sql,'UserEmail'); ?>"></td>
</tr>
<tr>
<td>
<f><button name="submit" type="submit" value ="submit" >UPDATE</button></f>
</table>
</form>
</div>
</content>
</body>
</head>
</html>
This is my table html code. I tried sending the data using the normal insert but it only sends the last row data. I don't know how to send the full data . Can someone please help me with this.
<form action="admin_schedule_employee.php" id="schedule_employee" method="post" >
<input type="date" class="input-sm" name="scheduledate" style="margin:10px;">
<table class="table-responsive table table striped table-bordered">
<thead>
<tr>
<th style="width:20%">Employee First Name</th>
<th style="width:20%">Employee ID</th>
<th style="width:20%">Start Time</th>
<th style="width:20%">End Time</th>
</tr>
<?php while($row = mysqli_fetch_array($search_result)): ?>
<tr>
<td><input disabled name="employeename" type="text" value="<?php echo $row['fname']; ?>"></input></td>
<td><input disabled name="employeeid" type="number" value="<?php echo $row['employee_id']; ?>"></input></td>
<td><input name="starttime" type="time"></td>
<td><input name="endtime" type="time"></td>
</tr>
<?php endwhile; ?>
</thead>
<tbody>
</tbody>
</table>
<input type="submit" name="Schedule" value="Schedule">
</form>[This is how my table look like i want to send the whole data to sql database using php][1]
To start with, you will need to create multiple pages:
form.php
process.php
done.php
Creating your user form is simple, place the table in form tags like you have done above, here is an example. Save this page as form.php
<form id="new record" action="process.php" method="POST">
<table width="500px">
<tr>
<td width="50%">
<input type="text" name="fname" id="fname">
</td>
<td width="50%">
<input type="text" name="lname" id="lname">
</td>
</tr>
<tr>
<td width="50%">
</td>
<td width="50%">
<input type="submit" value="Add Record">
</td>
</tr>
</table>
</form>
Next, you will need to create a page which can process this data, and add it to your mysql database. For the following example, I have omitted my database details and substituted them, but you should add your own.
For this example, imagine my database has a table with only an fname and an lname column.
<meta http-equiv="refresh" content="0; url=/done.php" />
<?php
$servername = "your_server_name";
$username = "mysql_username";
$password = 'mysql_password';
$dbname = "database_name";
$fname = $_GET['fname'];
$lname = $_GET['lname'];
try {
$conn = new PDO("mysql:host=$servername; dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = "INSERT INTO online (fname, lname)
VALUES ('$fname', '$lname')";
// use exec() because no results are returned
$conn->exec($sql);
echo "New record inserted";
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
?>
Hopefully, that will work to insert the record. Now we need a table on the done.php page which can display all the records in the database. Use the following code:
<html lang="en">
<head>
<meta http-equiv="refresh" content="5; url=/done.php" />
<meta charset="utf-8" />
<title></title>
</head>
<body>
<?php
$servername = "your_server_name";
$username = "mysql_username";
$password = 'mysql_password';
$dbname = "database_name";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT * from table_name";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
echo $row["fname"]. ": ";
echo $row["lname"]. "<br /><br />";
}
} else {
echo "No messages";
}
mysqli_close($conn);
?>
</body>
</html>
Hopefully this will work for you.
I have created a php sticky form so data will not disappear when the submit button is clicked. A url link is being used to pass values to a form so they can be edited. However, the values from the url are not being passed into the form fields. Why are the values from the url not being passed into the form fields? Thank you so much for your time.
This is the code:
index.php
<?php
require_once('authorize.php');
?>
<!DOCTYPE html>
<html>
<head>
</head>
<body>
<?php
require_once('appvars.php');
require_once('connectvars.php');
$conn = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
$data = mysqli_query($conn, $query);
echo '<table>';
echo '<tr><th>Name</th><th>Caption</th><th>Action</th></tr>';
while ($row = mysqli_fetch_array($data)) {
//link
echo '<td><a href="link.php?id=' . $row['id'] . '&image=' . $row['image1'] . '&name=' . $row['name'] .
'&caption=' . $row['caption'] .
'&video=' . $row['video'] . '">Edit </a>';
echo '</td></tr>';
}
echo '</table>';
echo "<br><br>";
mysqli_close($conn);
?>
</body>
</html>
sticky_form.php
<!DOCTYPE html>
<html>
<head>
<title>Edit Conent</title>
</head>
<body>
<h3>Edit Conent</h3>
<?php
require_once('appvars.php');
require_once('connectvars.php');
$vid="";
$vname="";
$vcaption="";
$vvideo="";
$id ="";
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if(isset($_POST["button_edit"])){
$id = $_POST["id"];
$name = $_POST['name'];
$caption = $_POST['caption'];
$video = $_POST['video'];
$qry = mysqli_query($dbc,"Update table1 Set name='$name', caption='$caption', video='$video' Where id='$id'");
else if(isset($_GET["id"])){
$qry = mysqli_query($dbc,"Select * From table1 Where id='".$_GET["id"]."'");
while($row=mysqli_fetch_array($qry,MYSQLI_ASSOC)){
$vid=$row["id"];
$vname=$row["name"];
$vcaption=$row["caption"];
$vvideo=$row["video"];
}
}
?>
<body>
<form action='' method="post" enctype="multipart/form-data" >
<table>
<tr>
<td>ID</td>
<td><input type="text" name="id" value="<?php echo $vid;?>"></td></tr>
<tr>
<td>Name</td>
<td><input type="text" class="bigger_textbox" name="name" value="<?php if (isset($_POST['name'])) {echo htmlentities($_POST['name']);}?>"></td></tr>
<tr><td>Caption</td>
<td><input type="text" class="bigger_textbox" name="caption" value="<?php if (isset($_POST['caption'])) {echo htmlentities($_POST['caption']);}?>"></td></tr>
<tr><td>Video</td>
<td><input type="text" class="bigger_textbox" name="video" value="<?php if (isset($_POST['video'])) {echo htmlentities($_POST['video']);}?>"></td></tr>
<tr><td colspan="2">
<input type="submit" name="button_edit" value="Edit Content"></td></tr> </table>
</form>
<table border=1>
<tr><th>Name</th><th>Caption</th>
<th>Video</th> <th>Action</th></tr>
<?php
if (isset($_GET["id"])) {
$qry =mysqli_query($dbc, "Select * From table1 Where id='".$_GET["id"]."'");
while($row=mysqli_fetch_array($qry,MYSQLI_ASSOC)) {
echo '<tr><td>'.$row["name"].'</td>';
echo '<td>'.$row["caption"].'</td>';
echo '<td>'.$row["video"].'</td>';
echo '<td>Edit </td></tr>';
}
}
?>
</table>
</body>
</html>
Apparently you already have the values you need in stick_form.php:
else if(isset($_GET["id"])){
$qry = mysqli_query($dbc,"Select * From table1 Where id='".$_GET["id"]."'");
while($row=mysqli_fetch_array($qry,MYSQLI_ASSOC)){
$vid=$row["id"];
$vname=$row["name"];
$vcaption=$row["caption"];
$vvideo=$row["video"];
}
Try replacing this part of the code of stick_form.php:
<td><input type="text" class="bigger_textbox" name="name" value="<?php if (isset($_POST['name'])) {echo htmlentities($_POST['name']);}?>"></td></tr>
<tr><td>Caption</td>
<td><input type="text" class="bigger_textbox" name="caption" value="<?php if (isset($_POST['caption'])) {echo htmlentities($_POST['caption']);}?>"></td></tr>
<tr><td>Video</td>
<td><input type="text" class="bigger_textbox" name="video" value="<?php if (isset($_POST['video'])) {echo htmlentities($_POST['video']);}?>" </td></tr>
With:
<td><input type="text" class="bigger_textbox" name="name" value="<?php echo $vname; ?>"></td></tr>
<tr><td>Caption</td>
<td><input type="text" class="bigger_textbox" name="caption" value="<?php echo $vcaption; ?>"></td></tr>
<tr><td>Video</td>
<td><input type="text" class="bigger_textbox" name="video" value="<?php echo $vvideo; ?>"></td></tr>
Update
As you commented, after clicking the edit button, your form fields get empty. That's because you're not setting the correct variables in this part of your code:
if(isset($_POST["button_edit"])){
$id = $_POST["id"];
$name = $_POST['name'];
$caption = $_POST['caption'];
$video = $_POST['video'];
$qry = mysqli_query($dbc,"Update table1 Set name='$name', caption='$caption', video='$video' Where id='$id'");
Change it to:
if(isset($_POST["button_edit"])){
$vid = $_POST["id"];
$vname = $_POST['name'];
$vcaption = $_POST['caption'];
$vvideo = $_POST['video'];
$qry = mysqli_query($dbc,"Update table1 Set name='$vname', caption='$vcaption', video='$vvideo' Where id='$vid'");
Hope it helps.
my problem it's I'm trying to develop a back-end where I need to update but my problem is, when I update one field my script update all fields and all data of my mysqli database.
My code for now is:
<html>
<body>
<?php
ini_set('display_errors', 1);
error_reporting(~0);
$serverName = "localhost";
$userName = "root";
$userPassword = "";
$dbName = "hotel_vaniet";
$strCustomerID = null;
if(isset($_GET["cod"]))
{
$cod = $_GET["cod"];
}
$serverName = "localhost";
$userName = "root";
$userPassword = "";
$dbName = "hotel_vaniet";
$conn = mysqli_connect($serverName,$userName,$userPassword,$dbName);
$sql = "SELECT * FROM quartos WHERE cod=$cod";
$query = mysqli_query($conn,$sql);
$result=mysqli_fetch_array($query,MYSQLI_ASSOC);
?>
<div id="main">
<form action="editar_quartos_final.php" name="frmAdd" method="post">
<br><h1>Página de Edição</h1>
<br><hr/>
<div id="login2">
<table width="284" border="1">
<tr>
<th width="120">Tipo</th>
<td width="238"><input type="text" name="tipo" size="50" value="<?php echo $result["tipo"];?>"></td>
</tr>
<tr>
<th width="120">Capacidade</th>
<td><input type="text" name="capacidade" size="50" value="<?php echo $result["capacidade"];?>"></td>
</tr>
<tr>
<th width="120">Preço p/ Noite</th>
<td><input type="text" name="preco" size="50" value="<?php echo $result["preco"];?>"></td>
</tr>
<tr>
<th width="120">Reservado</th>
<td><input type="text" name="reservado" size="50" value="<?php echo $result["reservado"];?>"></td>
</tr>
</table>
<br><input id="submitbuttoneditar" type="submit" value=" Editar " name="submit"/><br />
</div>
</form>
<?php
mysqli_close($conn);
?>
</body>
</html>
This the first page ,this page send me to another where makes all changes. The second page :
<html>
<head>
<title>Página de Edição do Cliente</title>
</head>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "hotel_vaniet";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE quartos SET
tipo = '".$_POST["tipo"]."' ,
capacidade = '".$_POST["capacidade"]."' ,
preco = '".$_POST["preco"]."' ,
reservado = '".$_POST["reservado"]."'
WHERE cod=cod";
if ($conn->query($sql) === TRUE) {
echo "Dados actualizados com sucesso!";
header("Location: quartos.php");
} else {
echo "Erro na edição dos dados! " . $conn->error;
header("Location: quartos.php");
}
$conn->close();
?>
</body>
</html>
On your first page you have $cod variable which equals $_GET["cod"].
On your second page $cod variable is not defined. So your try to update
WHERE cod=cod
means - update where value of field cod is the same as value of field cod. And as it is true for all records - all your records are updated.
So, the solution is to pass your $cod value to your second script.
For example, you can do it with a hidden field from your first form:
<form action="editar_quartos_final.php" name="frmAdd" method="post">
<br><h1>Página de Edição</h1>
<br><hr/>
<input type="hidden" name="cod" value="<?php echo $cod?>" />
<div id="login2">
<table width="284" border="1">
<tr>
<th width="120">Tipo</th>
<td width="238"><input type="text" name="tipo" size="50" value="<?php echo $result["tipo"];?>"></td>
</tr>
<tr>
See this field with hidden type?
And in your second script use $_POST['cod']:
$sql = "UPDATE quartos SET
tipo = '".$_POST["tipo"]."' ,
capacidade = '".$_POST["capacidade"]."' ,
preco = '".$_POST["preco"]."' ,
reservado = '".$_POST["reservado"]."'
WHERE cod=" . $POST['cod'];
And of course, your code is vulnerable to sql injections.
So you should start using prepared statements asap.
<form action="book.php" method="post">
<table>
<thead>
<tr>
<td>FlightID</td>
<td>From</td>
<td>Destination</td>
</tr>
</thead>
<tbody>
<tr>
<td name="flightID" value="1">1</td>
<td name="From" value="Sydney">Sydney</td>
<td name="Destination" value="Bali">Bali</td>
<td class="tdBook"><button class="btnBook" type=submit name="booking"> Book </button>
</tr>
<tr>
<td name="flightID" value="2">2</td>
<td name="From" value="London">London</td>
<td name="Destination" value="HongKong">Hong Kong</td>
<td class="tdBook"><button class="btnBook" type=submit name="booking"> Book </button>
</tr>
</tbody>
</table>
</form>
I created a table like this. At the end of each row, it has a book button.
What I am trying to do is when the user clicked the button, the selected row data(ID,From,Des) will pass to the 'book.php', then the PHP file will do the rest of the job.
But I tried to catch the value using $_POST['name'] in 'book.php', like this
<?php
if(isset($_POST['booking'])){
$ID = $_POST['flightID'];
$From = $_POST['From'];
$To = $_POST['Destination'];
}
?>
It shows all of those values are undefined. Any help would be appreciated.
The problem is that the values in <td> cannot be passed from the form to your PHP file by themselves. You could use hidden inputs for this. Additionally, each row in the table should be its own form to assure that all data is not submitted at the same time.
Try this:
<table>
<thead>
<tr>
<td>FlightID</td>
<td>From</td>
<td>Destination</td>
</tr>
</thead>
<tbody>
<tr>
<form action="book.php" method="post">
<td><input type="hidden" name="flightID" value="1">1</td>
<td><input type="hidden" name="From" value="Sydney">Sydney</td>
<td><input type="hidden" name="Destination" value="Bali">Bali</td>
<td class="tdBook"><button class="btnBook" type=submit name="booking"> Book </button>
</form>
</tr>
<tr>
<form action="book.php" method="post">
<td><input type="hidden" name="flightID" value="2">2</td>
<td><input type="hidden" name="From" value="London">London</td>
<td><input type="hidden" name="Destination" value="HongKong">Hong Kong</td>
<td class="tdBook"><button class="btnBook" type=submit name="booking"> Book </button>
</form>
</tr>
</tbody>
i have the same problem as yours and tried to create an answer so i came up with this code to indicate each row in an HTML table with a special name using loops, i can now take the specified row and do as much PHP operations as i can with it without disturbing the table as a whole and it was well synchronized with my database, hope it helps!
and btw the whole "marking each row with a special name" code is in usersTable.php
users.sql
create table users(
id int,
username varchar(50),
password varchar(50)
);
users.php
<?php
$host = "localhost";
$username = "root";
$password = "";
$database = "wdl2hw4db";
$conn = mysqli_connect($host, $username, $password, $database);
if (mysqli_connect_errno()){
die("can't connect to the Database" . mysqli_connect_errno());
}else{
echo "Database is connected" . "<br>";
}
if (isset($_POST['insert'])){
$idN1= $_POST['id'];
$usernameN1 = $_POST['username'];
$passwordN1 = $_POST['password'];
$query = "insert into users(id, username, pass) values ('".$idN1."' , '".$usernameN1."' , '".$passwordN1."' )";
$result = mysqli_query($conn, $query);
}else if (isset($_POST['update'])){
$idN2 = $_POST['id'];
$usernameN2 = $_POST['username'];
$passwordN2 = $_POST['password'];
$query = "update users set pass = '". $passwordN2 ."'where id = " . $idN2;
$result = mysqli_query($conn, $query);
}else if (isset($_POST['Display'])){
header('Location: usersTable.php');
}
echo "<br>";
?>
<form method="post">
ID: <input type="text" name="id" ><br><br>
username: <input type="text" name="username" ><br><br>
password: <input type="password" name="password" ><br><br>
<input type="submit" name="insert" value="insert">
<input type="submit" name="Display" value="Display">
</form>
userTable.php
<?php
$host = "localhost";
$username = "root";
$password = "";
$database = "wdl2hw4db";
$conn = mysqli_connect($host, $username, $password, $database);
$query = "select * from users";
$result = mysqli_query($conn, $query);
echo "<table border=\"6px\"><thead><tr><th>ID</th><th>username</th><th>password</th><th>Delete</th><th>Update</th></tr></thead>";
$i = 1;
while ($row = mysqli_fetch_assoc($result)) {
echo "<tr><form method='post'><td>" . $row['id'] . "</td><td>" . $row['username'] . "</td><td>" . $row['pass'] . "</td><td><input type='submit' name='Delete" . $i . "' value='Delete'></td><td><input type='submit' name='Update" . $i . "' value='Update'><input type='text' name='UpdateText" . $i . "' placeholder='insert new password here'></td></form></tr>";
$i++;
}
echo "</table>";
$i = 1;
$result2 = mysqli_query($conn, $query);
while ($row2 = mysqli_fetch_assoc($result2)) {
if (isset($_POST['Delete' . $i])) {
$usernameN4 = $row2['username'];
$query2 = "delete from users where username ='" . $usernameN4 . "'";
$result2 = mysqli_query($conn, $query2);
header("Refresh:0");
break;
}
$i++;
};
$i = 1;
$result3 = mysqli_query($conn, $query);
while ($row3 = mysqli_fetch_assoc($result3)) {
if (isset($_POST['Update' . $i]) && $_POST['UpdateText' . $i] != null ) {
$id4 = $row3['id'];
$Utext = $_POST['UpdateText' . $i];
$query3 = "update users set pass ='" . $Utext . "' where id = " . $id4;
$result3 = mysqli_query($conn, $query3);
header("Refresh:0");
break;
}
$i++;
};
mysqli_free_result($result);