Check if a variable is undefined in PHP

Question

Consider this JavaScript statement:

isTouch = document.createTouch !== undefined 

I would like to know if we have a similar statement in PHP, not being isset(), but literally checking for an undefined value. Something like:

$isTouch != ""

Is there something similar as the above in PHP?

Answer 1

You can use -

$isTouch = isset($variable); 

It will return true if the $variable is defined. If the variable is not defined it will return false.

Note: It returns TRUE if the variable exists and has a value other than NULL, FALSE otherwise.

If you want to check for false, 0, etc., you can then use empty() -

$isTouch = empty($variable); 

empty() works for -

• "" (an empty string)
• 0 (0 as an integer)
• 0.0 (0 as a float)
• "0" (0 as a string)
• NULL
• FALSE
• array() (an empty array)
• $var; (a variable declared, but without a value)

Answer 2

Another way is simply:

if($test){     echo "Yes 1"; } if(!is_null($test)){     echo "Yes 2"; }  $test = "hello";  if($test){     echo "Yes 3"; } 

Will return:

"Yes 3" 

The best way is to use isset(). Otherwise you can have an error like "undefined $test".

You can do it like this:

if(isset($test) && ($test!==null)) 

You'll not have any error, because the first condition isn't accepted.