How to define condition based on get url value - php

Question

When a user clicks on "List View" link then I want to show them the "List View" HTML and when they click on "Grid View" I want to show "Grid View" HTML.
I've defined the following links to click-
List View <br>
Grid View
Then I defined the following condition with PHP get method to show user the desired output-
<?php
if( isset( $_GET['view'] ) == 'list' ){
echo "This is List view";
}else if( isset($_GET['view'] ) == 'grid' ){
echo "This is Grid view";
}
This condition is not working. If I change "view" to "view_1" and "view_2" from URL then my condition is working as well.
<?php
if( isset( $_GET['view_1'] ) == 'list' ){
echo "This is List view";
}else if( isset($_GET['view_2'] ) == 'grid' ){
echo "This is Grid view";
}
?>
<br>
List View <br>
Grid View
But I don't want to change the "view" key. I just want to keep the same key and different value for both URL to do the conditional statement.
Is it possible?

Answer 1

isset() only checks if a variable exists and isn't null and returns a boolean (true/false).
To check the value, you first need to check if it exists (isset) and then check the value, like this::
if (isset($_GET['view']) && $_GET['view'] == 'list') {
// your code here.
}
You can read more here: http://php.net/manual/en/function.isset.php
Alternative
If you want to make your code and conditions slightly more readable, you could to this:
// Get the value of $_GET['view'] once, if it exists (pre PHP 7)
$view = isset($_GET['view']) ? $_GET['view'] : null;
// PHP 7.x (new shorter syntax for the above)
$view = $_GET['view'] ?? null;
if ($view == 'list') {
// your code
}

Answer 2

Did you try like this..
if(isset($_GET['view']))
{
if($_GET['view'] == 'list')
{
//code here
}
elseif ($_GET['view'] == 'grid')
{
//code here
}
}