Do you really need to check for both isset() and empty() at the same time?

Comments:

• hahahahacker2009, 13.06.24 08:10

  Hi!

  if(!empty($var)) would be exactly the same as just if($var) which is way more clean and readable

  But doing if($var) on variable that doesn't exist will result in a warning, so if I want to check, sayy, I will have to use !empty instead?

  Reply:

  If you want to check a $_GET or $_POST array key, then you must use isset() (or, strictly speaking - array_ke_exists())

  If you want to check a $_GET or $_POST array value, then your best bet is to validate it properly, instead of just checking for being "empty". 0 could be a legit value, but empty won't let it through. An array is will cause an error where a string is expected, but empty() won't warn you.

  So I wouldn't use empty() for input validation at all.

• Federico Bau, 09.01.22 14:48

  Are you somehow related to 'Kyle Simpson' the author of You-Dont-Know-JS?
  Sometimes reading here I catch my self thinking I'm reading that book but than I see the blue light color theme of the website and i remember that nope, I PHP.

  No really joke a part, you should write something similar, like You-Don't Know PHP :P

  Reply:

  Hello Federico!

  Sadly, I am nether acquainted with Kyle Simpson nor with You-Dont-Know-JS book. Actually my knowledge and interest in JS are quite limited. but I am glad there is something similar about this great language.

  Thank you for the suggestion, I will definitely think of creating one!

• Hayley Watson, 25.10.20 10:41

  A bit of a digression, because it has nothing to do with empty(), but...

  Really, how much of a use case is there for isset($plainvariable)? Is your function so wildly out of control that you can't keep track of whether your variables have even been initialized or not? That would be the real problem.

  isset($compositevariable['key']), sure: if you get passed an array, PHP doesn't give you any guarantee it has all the fields you might want to look at (and note that array_key_exists is more strict about what counts as "set"). isset($object->field) if there's duck-typing going on. Decoded JSON and such.

  Oh, maybe you want to see if the variable contains an actual null-type null? You should say so for the sake of self-documenting code: $plainvariable === null or is_null($plainvariable). Often, though, passing nulls around a lot is a bit of a code smell (it's short for "I really don't know what value to put in this variable, I shouldn't even be using it to be honest with myself"). If you have to put up with them coming in from outside, check for them as above and move on.

  But isset($plainvariable)? You need to step back a bit and ask yourself why.

  Reply:

  Hello Hayley!

  Thank you for the excellent addition to the article!

  You are absolutely correct, and sadly, not so many people are sharing your concerns. Actually some time ago I wrote a similar rant, but it would be a good idea to add the link here, as well as a link to your comment. So I did!

• morpheus, 04.09.20 17:03

  There are situations when "0"(or maybe 0) is valid. For example user input, i found a solution that includes the trim function: if (isset($_POST['fieldName']) && trim($_POST['fieldName']) !== "") { echo "There's something here!"; }

• bob, 30.05.19 16:54

  does this apply to $somevar=array(); Thanks for the effort you have extended to the rest of us.

  Reply:

  Hello Bob!

  An empty array considered a FALSE-like variable in PHP. You may find these tables useful: https://www.php.net/manual/en/types.comparisons.php

• Petko Yotov, 15.10.18 08:59

  Thanks for your articles and examples!

  You seem to have a typo in this line:

  if (isset($someVar)) && $someVar)

  There is one right parenthesis too much.

  Reply:

  Hello Petko!

  Thank you very much, indeed it was a typo. Fixed!

• Everton Muniz, 07.04.18 05:28

  And if the variable is not set, will not it generate a notice?

  Reply:

  Hello Everton!

  Nope, it will not.