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@NogDogSep 14.2016 — #I took the liberty of formatting your code and embedding it in this forum's "php" tags.
I suspect those warnings are showing up in the code at the end that displays the form, as the variables used there won't be defined if the form has not been submitted, and thus the initial if() block will not be entered.
how to fetch data from database in php and display in fields
Can anyone please tell me what am i doing wrong in this code , i am trying to fill in the textboxes from the mysql database before updating it but it is giving me error : Notice: Undefined variable: password , Notice: Undefined variable: email , Notice: Undefined variable: phone
[code=php]
<?php
include “sessioncheck.php”;
?>
<?php
if(isset($_POST[“update”])){
//echo'<pre>’;print_r($_POST);exit();
//$user_name =$_POST[“name”];
$email = $_POST[“email”];
$password = $_POST[“password”];
$phone = $_POST[“tel”];
$id =$_POST[“id”];
//***** DB Connection *****//
$hostname = ‘localhost’;
$dbname = ‘shopping’;
$username = ‘root’;
mysql_connect(‘localhost’, ‘root’, ”) or DIE(‘Connection Not FOUND..!’);
mysql_select_db(‘shopping’) or DIE(‘Database name is not available!’);
//***** DB Connection END*****//
$query = “SELECT id, email , password, status FROM users WHERE id=’$id’ , email=’$email’ , password=’$password’ , status=’$phone'”;
//echo $query;exit();
$result = mysql_query($query);
$rows = mysql_num_rows($result);
// //echo’USER EXIST’;
if ($rows==1){
while($rs = mysql_fetch_array($result)){
$id = $rs[“id”];
$email = $rs[“email”];
$phone =$rs[“status”];
$password=$rs[“password”];
//echo $id.’ Email: ‘.$uemail;exit();
}
}
$query = “UPDATE users SET id = ‘$id’ , email=’$email’ , password = ‘$password’ , status= ‘$phone'”;
//echo $query;exit();
$result = mysql_query($query);
if($result==false){
die(mysql_error());
}
header(“Location: Grid.php”);
return mysql_affected_rows();
exit();
}
?>
<!DOCTYPE html>
<html>
<head>
<title>Edit Data</title>
</head>
<body>
<div class=”form”>
<h2>Edit Data</h2>
<form method=”post” action=”update1.php”>
<input type=”hidden” value=”<?php echo $id;?>” name=”id”>
<!– <input type=”text” name=”name” placeholder=”Username”/> –>
<input type=”password” name=”password” value=”<?php echo $password;?>” placeholder=”Password”/>
<input type=”email” name=”email” value=”<?php echo $email;?>” placeholder=”Email Address”/>
<input type=”tel” name=”tel” value=”<?php echo $phone;?>” placeholder=”Phone Number”/>
<input type=”submit” name = “update” value=”Update”>
</form>
</div>
</body>
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@NogDogSep 14.2016 — #I suspect those warnings are showing up in the code at the end that displays the form, as the variables used there won't be defined if the form has not been submitted, and thus the initial if() block will not be entered.