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?
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2@Jleagle I assume because the question gets asked like ten times a day. I must say this is one of the better ones though.phihag– phihag2011-10-21 09:53:53 +00:00Commented Oct 21, 2011 at 9:53
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Ilya, for long term, you should read this: PHP OperatorsOvidiu– Ovidiu2011-10-21 11:54:24 +00:00Commented Oct 21, 2011 at 11:54
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6 Answers
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 $x, and then evaluate the IF statement with it - which evaluates to true in PHP.
Comments
if ($x == "100")
or
if (intval($x) === 100)
Comments
if ($x = 100) {
sets $x to 100 and evaluates the result. You want:
if ($x == '100') {
Comments
It is because you are using an assignment statement instead of checking for equality in the line: if ($x = 100) Try if ($x == 100) instead.
Comments
you are not comparing value of x, you are saying that x is equal to 100
try
if($x == 100) { echo "yeah"; }
Comments
first check with isset($_REQUEST['foo']) or isset($_GET['foo'])
