If null use other variable in one line in PHP

Question

Is there in PHP something similar to JavaScript's:

alert(test || 'Hello'); 

So, when test is undefined or null we'll see Hello, otherwise - we'll see the value of test.

I tried similar syntax in PHP but it doesn't seem to be working right... Also I've got no idea how to google this problem..

thanks

Edit

I should probably add that I wanted to use it inside an array:

$arr = array($one || 'one?', $two || 'two?'); //This is wrong 

But indeed, I can use the inline '? :' if statement here as well, thanks.

$arr = array(is_null($one) ? "one?" : $one, is_null($two) ? "two ?" : $two); //OK 

Answer 1

you can do echo $test ?: 'hello';

This will echo $test if it is true and 'hello' otherwise.

Note it will throw a notice or strict error if $test is not set but...

This shouldn't be a problem since most servers are set to ignore these errors. Most frameworks have code that triggers these errors.

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Edit: This is a classic Ternary Operator, but with the middle part left out. Available since PHP 5.3.

echo $test ? $test : 'hello'; // this is the same echo $test ?: 'hello';        // as this one 

This only checks for the truthiness of the first variable and not if it is undefined, in which case it triggers the E_NOTICE error. For the latter, check the PHP7 answer below (soon hopefully above).