Notice: Undefined index $_POST and mysql - php

Question

If I set
$USER = (isset($_POST['username']));
with php have 0 errors BUT in db put '1' instead the it to user write on HTML input and if I put
`$USER = $_POST['username'];`
Notice: Undefined index: username in __FILE__ on line 5
this set in the db what user put in the HTML input, but get error
(Sorry for my bad English; I know it is so bad!)

Answer 1

Try below :-
if (isset($_POST['username'])) {
$USER = $_POST['username']);
}
As, isset() is a function which gives return values in boolean i.e; 1 or 0

Answer 2

When you do this :
$USER = (isset($_POST['username']));
You check if $_POST['username'] is set AND you set this result (true or false) into $USER
What you want to do is :
if (isset($_POST['username'])) // You check if username is set.
{
$USER = $_POST['username'];
// your query
}
else
{
// return error if isn't set.
}

Answer 3

First of all isset() is a checking operand type and returns 1 if true and 0 if false depending on something exist or not. I guess you got why this always output and saved 1 in the db.
So it should be $USER = $_POST['username'];
But the reason why you are getting Notice: Undefined index: username in __FILE__ on line 5 because when you first enter that page, it dont post anything. So as a matter of fact $_POST['username'] is unknown to php at that time. When you are submitting the form then php will know and recognize $_POST['username'].
So you need to put a check like below:
if(isset($_POST['username'])){
echo $_POST['username'];
}
Hope it will work and make you understand the problem.
Regards.