CODEWITHSUNDEEP
php
jerrygarciuh
jerrygarciuh

Reputation: 22018

Testing for $_POST var affects variable even though $_POST should be undefined

I have used for years this method of testing for a $_POST var and assigning a value to $var simultaneously.

Today I ran into a server that seems to be different in its handling of this.

I'm loading the page http://site.com/foo.php?var=bar No form post made... and testing for $_POST var seems to unset my initial assignment 

$var= $_GET['var'];

echo "var is $var\n";  // correctly echos $_GET['var'] 

if ($var= $_POST['foo']) {
    echo "POST foo seen\n";  // does not fire, condition not met
}

echo "var is $var\n";  // does not echo $_GET['var']

When I switched to 

if (isset($_POST['foo'])) {
    $var = $_POST['foo'];
...

I got rid of the unsetting behavior.

So my question is whether this if ($var= $_POST['foo']) { is bad form?

Upvotes: 0

Views: 220

Answers (2)

Baba
Baba

Reputation: 95141

Try

$var = isset ( $_GET ['var'] ) ? $_GET ['var'] : null;
$foo = isset ( $_GET ['foo'] ) ? $_GET ['foo'] : null;

echo "var is $var\n"; // correctly echos $_GET['var']

if ($var === $foo) {
    echo "POST foo seen\n"; // does not fire, condition not met
}

echo "var is $var\n"; //

Upvotes: 0

gahooa
gahooa

Reputation: 137492

You are using the assignment operator = iinstead of the comparison operator ==. Also, note that PHP will throw an E_STRICT warning when accessing an undefined array element. Therefore, the best bet is typically:

$name = trim(isset($_POST['name']) ? $_POST['name'] : '');

or for integers:

$age = intval(isset($_POST['age']) ? $_POST['age'] : '');

Upvotes: 1

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