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when i run this code ,  i am getting undefined index for password , phonenumber and nickname.

 


form1.html:



<html>
<body>
 
<form action="welcome.php" method="post">
 
    E-mail: <input type="text" name="email"><br>
    Password: <input type="text" name="password"><br>
    Nickname: <input type="text" name="nickname"><br>
    Phonenumber:   <input type="text" name="phonenumber"><br>
    <input type="submit">
</form>
 
</body>
</html>



 

welcome.php:



<html>
<body>

Welcome <?php echo $_POST["name"]; ?><br>
Your email address is: <?php echo $_POST["email"]; ?>


<?php
//Step1
$db = mysqli_connect('localhost','root','','users')
or die('Error connecting to MySQL server.');
?>

<html>
<head>
</head>
<body>
<h1>PHP connect to MySQL</h1>

<?php


$email = $_POST['email'];
$password = $_POST['password'];
$phonenumber = $_POST['phonenumber'];
$nickname = $_POST['nickname'];



$sql ="INSERT INTO users (email,password,nickname,phonenumber) VALUES ('$email','$password','$nickname' , '$phonenumber' )";
mysqli_query($db,$sql);



//Step2
$query = "SELECT * FROM users";
mysqli_query($db, $query) or die('Error querying database.');

//Step3
$result = mysqli_query($db, $query);
//$row = mysqli_fetch_array($result);

while ($row = mysqli_fetch_array($result)) {
echo $row['email'] . ' ' . $row['password'] . ' '. $row['nickname'] . ' '. $row['phonenumber'] .'<br />';
}


//Step 4
mysqli_close($db);
?>





</body>
</html>


when i run this code ,  i am getting undefined index for password , phonenumber and nickname.

 

Edited by cyberRobot
separated code blocks

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Assuming that the error message is coming from the code at "step 3", why is your fetch commented out?

 

And - what is the point of the code at step 2???

Edited by ginerjm

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What are you looking to do? Is this a registration / sign-up form? Is it to log in?

 

If you're not doing so already, you need to look into password hashing. More information can be found here:

http://php.net/manual/en/faq.passwords.php

 

 

Assuming that the error message is coming from the code at "step 3", why is your fetch commented out?

There's another fetch in the while loop.

Edited by cyberRobot

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Make sure you are not going to welcome.php directly or through a link. Have to use the form in form1.html to get there. Which isn't good, but that's how you have it set up for now.

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