php，如果其他＆URL vars

Question

URL

x.php?foo=100

x.php

$x = $_GET['foo'];

if ($x = 100) {
   echo "yeah";    
}else{
   echo "no";
}

My code doesnt work, where is error?

Answer 1

if ($x == "100")

要么

if (intval($x) === 100)

Answer 2

You are using a single equal mark ( = ) instead of two in the IF statement ( if ($x = 100) { ).

A single equal mark will set the value 100 into $x, and then evaluate the IF statement with it - which evaluates to true in PHP.

Answer 3

if ($x = 100) {

sets $x to 100 and evaluates the result. You want:

if ($x == '100') {

Answer 4

这是因为您使用赋值语句而不是检查行中的相等性：if（$ x = 100）而是尝试if（$ x == 100）。

Answer 5

you are not comparing value of x, you are saying that x is equal to 100

try

if($x == 100) { echo "yeah"; }

Answer 6

首先使用isset($_REQUEST['foo'])或isset($_GET['foo'])