Php MySQLi Undefined values(variables)

Question

Okay so I fixed my mistake but I am still getting an error. I am not use to MySQLi but I would like to help my friend get something working for his site he is making.

    <?php

if(isset($_POST['submit']))
{
$firstname = filter($_POST['firstname']);
$lastname = filter($_POST['lastname']);
$age = filter($_POST['age']);
}

$db = new MySQLi('localhost', 'root', '', 'register');
    if ($db->connect_error) {
            $message = $db->connect_error;
            die($message);
    }


    $sql = 'SELECT * FROM users';
    $result = $db->query($sql);
    if ($db->error) {
            $message = $db->error;
            die($message);
    }

    $db->query("INSERT INTO users (firstname,lastname,age)
    VALUES ('".$firstname."', '".$lastname."', '".$age."')");

    mysqli_close($db);

?>

<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<form action="confirmation.php" method="POST">
First Name:<br />
<input type="text" name="firstname" placeholder="First name" />
Last Name:<br />
<input type="text" name="lastname" />
Age:<br />
<input type="text" name="age"  />
Submit:<br />
<input type="submit" name="submit" />
</form>
</body>
</html>

This is the error we are getting ->

( ! ) Notice: Undefined variable: firstname in C:\wamp\www\website\register.php on line 25
Call Stack
# Time Memory Function Location
1 0.0008 244952 {main}( ) ..\register.php:0

( ! ) Notice: Undefined variable: lastname in C:\wamp\www\website\register.php on line 25
Call Stack
# Time Memory Function Location
1 0.0008 244952 {main}( ) ..\register.php:0

( ! ) Notice: Undefined variable: age in C:\wamp\www\website\register.php on line 25
Call Stack
# Time Memory Function Location
1 0.0008 244952 {main}( ) ..\register.php:0

When I do this in the old MySQL I don't get no such error like this so I am very confused since it is working for me when I do it but with the SQLi I get the error.

You're running your query before you assign the values to your variables. This is both obvious and something that doesn't work no matter what API you use. — John Conde
– John Conde, Commented Feb 4, 2015 at 1:02
Oh wow I can't believe I missed that yes I see why it wouldn't work at all, thank you John for pointing that out. — GrimmRP
– GrimmRP, Commented Feb 4, 2015 at 1:05

Sphinxxx · Accepted Answer · 2015-02-04 01:22:17Z

1

This code block needs to be first, before any MySQL query attempt.

if(isset($_POST['submit']))
{
    $firstname = filter($_POST['firstname']);
    $lastname = filter($_POST['lastname']);
    $age = filter($_POST['age']);
}

This code block defines the variables used. In your current example, the variables aren't set until after the query attempt - that's why you're getting the error.

Thus:

<?php

if(isset($_POST['submit']))
{
    $firstname = filter($_POST['firstname']);
    $lastname = filter($_POST['lastname']);
    $age = filter($_POST['age']);
    $db = new MySQLi('localhost', 'root', '', 'register');
    if ($db->connect_error) {
            $message = $db->connect_error;
            die($message);
    }


    $sql = 'SELECT * FROM users';
    $result = $db->query($sql);
    if ($db->error) {
            $message = $db->error;
            die($message);
    }

    $db->query("INSERT INTO users (firstname,lastname,age)
     VALUES ('".$firstname."', '".$lastname."', '".$age."')");

    mysqli_close($db);
}

?>

edited Feb 4, 2015 at 1:22

Sphinxxx

13.1k5 gold badges59 silver badges89 bronze badges

answered Feb 4, 2015 at 1:04

Patrick Moore

13.4k5 gold badges41 silver badges63 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

John Conde Over a year ago

That's still an error as you're always running the query whether the values are set or not

GrimmRP Over a year ago

Yeah I am still getting the error I updated my post or question I hope it helps on explain my error I am getting, its the Value() part

Patrick Moore Over a year ago

Please see the updated code block from @Sphinxxx - his if() statement properly encapsulates the rest of the query code, so the query is not attempted unless this is a $_POST form submission.

Darren · Accepted Answer · 2015-02-04 01:34:05Z

1

You should make sure that the fields actually have data in them.

$required = array(
    'firstname',
    'lastname',
    'age'
);

$missing = array();
foreach($required as $item) {
    if(!isset($_POST[$item}) || empty($_POST[$item])) {
        $missing[] = $item;
    }
}

if(!empty($missing)) {
    die("You need to supply the required fields. (" .implode(", ", $missing). ")");
} else {
    // run the query here.
}

As you can see, we run the $required array to ensure the values are set in the $_POST superglobal and are not empty. As we require them to run this.

I still, however think your filter() function has something to do. Could you supply that?

answered Feb 4, 2015 at 1:34

Darren

13.1k4 gold badges44 silver badges81 bronze badges

Comments

Scott Powell · Accepted Answer · 2015-02-04 01:25:00Z

0

Give this a shot:

<?php

if(isset($_POST['submit']))
{
$firstname = filter($_POST['firstname']);
$lastname = filter($_POST['lastname']);
$age = filter($_POST['age']);


$db = new MySQLi('localhost', 'root', '', 'register');
    if ($db->connect_error) {
            $message = $db->connect_error;
            die($message);
    }


    $sql = 'SELECT * FROM users';
    $result = $db->query($sql);
    if ($db->error) {
            $message = $db->error;
            die($message);
    }

    $db->query("INSERT INTO users (firstname,lastname,age)
     VALUES ('".$firstname."', '".$lastname."', '".$age."')");

    mysqli_close($db);
}
else { die("No post data was submitted");  }
?>

answered Feb 4, 2015 at 1:25

Scott Powell

2152 silver badges6 bronze badges

Collectives™ on Stack Overflow

Php MySQLi Undefined values(variables)

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related