Checking if a constant is empty

Why is this not possible?

if(!empty( _MY_CONST)){
  ...

But yet this is:

$my_const = _MY_CONST;
if(!empty($my_const)){
  ...

---

define( 'QUOTA_MSG' , '' ); // There is currently no message to show

$message = QUOTA_MSG;
if(!empty($message)){
  echo $message;
}

I just wanted to make it a lit开发者_StackOverflowtle cleaner by just referencing the constant itself.

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See the manual: empty() is a language construct, not a function.

empty() only checks variables as anything else will result in a parse error. In other words, the following will not work: empty(trim($name)).

So you'll have to use a variable - empty() is really what you want in the first place? It would return true when the constant's value is "0" for example.

Maybe you need to test for the existence of the constant using defined() instead?

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Just letting you know that you can do

if(!empty(MY_CONST))

since PHP 5.5.

---

You can get along with this if for some reason you are still not using PHP 5.5.

if (defined('MY_CONST') && MY_CONST) {
    echo 'OK';
}

---

if (!empty(constant('MY_CONST')) {
    ...
}

mixed constant ( string $name )

Return the value of the constant indicated by $name, or NULL if the constant is not defined

http://php.net/manual/en/function.constant.php

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what about strlen?

if(strlen(MY_CONST) == 0) 

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If constant defined and need check is empty or not, for example you use it in config file, you can use !MY_CONST:

define('MY_CONST', '');
if (!MY_CONST) {
    echo 'MY_CONST is empty';
}