我不能讓調用ID形成fuctions頁它顯示錯誤

Question

function.php

<a href = "details.php?pro_id='.$pro_id.'"><button type="button" class="btn btn-primary">View Detail</button> </a> 

====================== =====================

details.php

if(isset($_GET(['pro_id'])){ 

$product_id = $_GET['pro_id']; 
$get_pro = "select * from products where product_id='$product_id'"; 

$run_pro = mysqli_query($con, $get_pro); 

它不工作它會顯示以下錯誤：

Fatal error: Cannot use isset() on the result of a function call (you can use "null !== func()" instead) in C:\xampp\htdocs\Raj_Cassette\details.php on line 52

Answer 1

$ _GET不是函數。它是一個數組所以

不

$_GET(['pro_id'])

但是

$_GET['pro_id']