Undefined index: command [duplicate] - php

Question

This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
undefined index [closed]
(2 answers)
Closed 9 years ago.
These are my codes And i'm getting error
Notice: Undefined index: command in E:\xampp\htdocs\Shop_cart\products.php on line 17
Javascript:
<script language="javascript">
function addtocart(pid){
document.form1.productid.value=pid;
document.form1.command.value='add';
document.form1.submit();
}
</script>
HTML Code:
<form name="form1" action="">
<input type="hidden" name="productid" />
<input type="hidden" name="command" />
<input type="button" value="Add to Cart" onclick="addtocart(1)" />
</form>
PHP Code:
if($_REQUEST['command']=='add' && $_REQUEST['productid']>0){
$pid=$_REQUEST['productid'];
addtocart($pid,1);
header("location:shoppingcart.php");
exit();

Answer 1

In your PHP code add isset check to the condition:
if ( isset($_REQUEST['command']) && $_REQUEST['command'] == 'add' && $_REQUEST['productid']>0 ) {}

Answer 2

Change
if($_REQUEST['command']=='add' && $_REQUEST['productid']>0){
to
if(isset($_REQUEST['command']) && $_REQUEST['command']=='add' && $_REQUEST['productid']>0){

Answer 3

Same to other answer but I would literally proceed something like this for security purposes and better layout of my code.
//Check if request has been made
if (isset($_REQUEST['command'])){
//proceed for checking
if($_REQUEST['command'] == 'add' && $_REQUEST['productid']>0 )
{
//Your code here
}
}else{
//redirect or whatsoever
}