[rakesh00718]

When I click on change password following error is shown but password changed successfully

Why This type of Problem is occured?

Quote

Notice: Undefined index: email in D:\xampp\htdocs\rakesh_work\rakesh_work_2\pract\change_password.php on line 5

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Notice: Undefined index: password in D:\xampp\htdocs\rakesh_work\rakesh_work_2\pract\change_password.php on line 6

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Notice: Undefined index: newpassword in D:\xampp\htdocs\rakesh_work\rakesh_work_2\pract\change_password.php on line 7

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Notice: Undefined index: confirmpassword in D:\xampp\htdocs\rakesh_work\rakesh_work_2\pract\change_password.php on line 8

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Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 5 in D:\xampp\htdocs\rakesh_work\rakesh_work_2\pract\change_password.php on line 16

New password and confirm password must be the same!

if($_POST['email']!="") { $email = $_POST['email']; } else die("No email was passed"); if($_POST['password']!="") { $password = $_POST['password']; } else die("No Password was passed"); if($_POST['newpassword']!="") { $newpassword = $_POST['newpassword']; } else die("No NewPassword was passed");

This post has been edited by Dormilich: 26 April 2012 - 05:14 AM
Reason for edit:: please use [code] [/code] tags when posting code

[Dormilich]

you have that problem because the data you want to process are not there. and if you try to get an array value with a non-existent key, that’s the type of error you get.

[codeprada]

The undefined index simply means that you're trying to access a an index in an array that doesn't exist. Before using variables you should use empty() or isset() to verify that they actually exist.

For a more comprehensive answer you should post your source code.

[Dormilich]

codeprada, on 26 April 2012 - 02:17 PM, said:

Before using variables you should use empty() or isset() to verify that they actually exist.

either that or filter functions.

[Cbeppe]

Also, this is gonna make your life 10x easier. Trust me!

<?php

if (isset($_POST['email'])) { 
    $email = $_POST['email']; 
} else {
    die("No email was passed"); 
}

if (isset($_POST['password'])) { 
    $password = $_POST['password']; 
} else {
    die("No password was passed"); 
}

if (isset($_POST['newpassword'])) { 
    $newpassword = $_POST['newpassword']; 
} else {
    die("No new password was passed");
}

?>

[Dormilich]

you can condense that even more:

$email = filter_input(INPUT_POST, "email", FILTER_VALIDATE_EMAIL);
if (!isset($email, $_POST['password'], $_POST['newpassword']) or !$email)
{
    throw new InvalidArgumentException("Invalid Mail or Password encountered.");
}

[Cbeppe]

^ ^ ^ -- Filter functions are great, but I was actually referring to the code readability . Having all that on one line is very impractical. But I forgot that isset() takes several arguments...

This post has been edited by Cbeppe: 26 April 2012 - 12:06 PM

[Dormilich]

btw. I wouldn’t die() in an error case, that usually breaks your HTML code.