Undefined index?

Question

Okay, I changed my code for a few reasons which are irrelevant, but now I need your help again, since I'm very close on finishing one of first ''bigger'' projects in php and mysql, and I've changed my code to this:

<?php

if (isset($_POST['submitted'])) {


}

include('connect-with-mysql.php'); 

$Test1 = $_POST['test1']; 
$Test2 = $_POST['test2']; 
$sqlinsert = "INSERT INTO vanille"



?>
<html>
<head>
<title>Datenbank</title>
</head>
<body>

<h1>Überschrift</h1>
<form method="post" action="action.php">
<input type="hidden" name="submitted" value="true" />
<fieldset>
 <legend>New Ppl</legend>
 <label>Test1:<input type="text" name="test1" /></br>
 <label>Test2:<input type="text" name="test2" /></br>
</fieldset>
<br />
<input type="submit" value="add new informations" />    

</form>

</body>
</html>

EDIT:

Here's the content of the connect-with-mysql.php file:

<?php

DEFINE ('DB_USER', 'root'); 
DEFINE ('DB_PSWD', ''); 
DEFINE ('DB_HOST', 'localhost'); 
DEFINE ('DB_NAME', 'vanille'); 

$dbcon = mysqli_connect(DB_HOST, DB_USER, DB_PSWD, DB_NAME); 

?>

And then I'll get this error line:

Notice: Undefined index: test1 on line 11 (same with line 10)

Possible duplicate of [PHP: "Notice: Undefined variable" and "Notice: Undefined index"](http://stackoverflow.com/questions/4261133/php-notice-undefined-variable-and-notice-undefined-index) — gpinkas, Oct 30 '15 at 10:29

Niranjan N Raju · Answer 1 · 2015-10-30T10:24:31.793

1

you have closed the if()

Your database query, is open for sql injection and more over you will get errors if you run this query.

<?php

 if (isset($_POST['submitted'])) {

 include('connect-with-mysql.php'); // 

     $Test1 = $_POST['test1']; 
     $Test2 = $_POST['test2']; 
     $sqlinsert = "INSERT INTO vanille"
 }

?>

Query should look like,

$sqlinsert = "INSERT INTO vanille (`test1`,`test2`) values ($Test1,$Test2)";

mysqli_query($dbcon,"INSERT INTO vanille (`test1`,`test2`) values ($Test1,$Test2)");

See this for syntax

I have shown the query for your reference, but make sure you use PDO

edited Oct 30 '15 at 10:24

answered Oct 30 '15 at 10:14

Niranjan N Raju

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and how do i fix that? – DarkYagami Oct 30 '15 at 10:17
for your reference i have changed the code of insert. – Niranjan N Raju Oct 30 '15 at 10:18
please share the content of `connect-with-mysql.php`. – Niranjan N Raju Oct 30 '15 at 10:18
i edited my question now you can see the content – DarkYagami Oct 30 '15 at 10:21
i have changed my answer for insert, please check... – Niranjan N Raju Oct 30 '15 at 10:24
yes im doing that now – DarkYagami Oct 30 '15 at 10:25
that shud come inside `if()` – Niranjan N Raju Oct 30 '15 at 10:25
so it should be like this: if ($sqlinsert = "INSERT INTO vanille (`test1`,`test2`) values ($Test1,$Test2)"; mysqli_query($dbcon,"INSERT INTO vanille (`test1`,`test2`) values ($Test1,$Test2)"); ? – DarkYagami Oct 30 '15 at 10:26
no... im telling this line shud be inside `if(isset()){ post values //here }` – Niranjan N Raju Oct 30 '15 at 10:28
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/93797/discussion-between-darkyagami-and-niranjan-n-raju). – DarkYagami Oct 30 '15 at 10:30

score 0 · Answer 2 · answered Oct 30 '15 at 10:18

Once you request the page at initially stage it is not a post request. So, It will be get request and you are accessing array of $_POST which is not exist. So, you got this error.Check that post array is present then go for accessing it's index.

<?php

 if (!empty($_POST)) {

     include('connect-with-mysql.php'); 

     $Test1 = $_POST['test1']; 
     $Test2 = $_POST['test2']; 
     $sqlinsert = "INSERT INTO vanille"
 }

?>

Undefined index?

2 Answers2