Why does trim() make a non existent variable exist?

Question

I have the following code:

$a=$_POST['dsaghgjjkhfdsfdsfsdfdsjhkhkhgj'];

if(isset($a))
{
    echo"exists";
}
else
{
    echo"does not exist";
}

echos the value "does not exist", HOWEVER when i apply trim to the $_POST variable,

$a=trim($_POST['dsaghgjjkhfdsfdsfsdfdsjhkhkhgj']);

if(isset($a))
{
    echo"exists";
}
else
{
    echo"does not exist";
}

the code will echo "exists". Why does passing in a non-existing $_POST variable to trim() magically makes it exist?

You have no check in your code at all that checks whether `$_POST['dsaghgjjkhfdsfdsfsdfdsjhkhkhgj']` exists or not. So how can you say so that `$_POST['dsaghgjjkhfdsfdsfsdfdsjhkhkhgj']` is magically made into existance when used with `trim()`? You have neither showed that it did not exist nor that it does exist. You're obviously mixing things here, perhaps you're irritated by the fact that `trim()` returns a string, always? Perhaps you're running to a conclusion too fast while not checking the basics? — hakre, Jan 05 '14 at 11:15
Recommended reading: [How to get useful error messages in PHP?](http://stackoverflow.com/q/845021/367456) — hakre, Jan 05 '14 at 11:17

gd1 · Accepted Answer · 2014-01-05T10:55:39.370

This is exactly what happens, step by step.

When you refer to $_POST['...'] in the second code snippet this notice is issued:

Notice: Undefined index: ... on line ...

You don't get the notice as your error_reporting level does not include E_NOTICE. The intermediate $_POST['...'] is evaluated NULL and trim(NULL) returns an empty string. So $a is assigned an empty string.

If you prepend error_reporting(E_ALL) and ini_set("display_errors", "on") to your script, you will see the actual errors/warnings/notices issued.

score 3 · Answer 2 · answered Jan 05 '14 at 10:43

3

Because trim returns a string, regardless of its inputs, see http://ch2.php.net/manual/en/function.trim.php.

php > var_dump(NULL);
NULL
php > var_dump(trim(NULL));
string(0) ""

A string, even if it is empty, is declared as "defined". This is why isset returns true then.

answered Jan 05 '14 at 10:43

pdu

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it doesn't explain what happens – zerkms Jan 05 '14 at 10:43
Yes it does. trim creates a string from its inputs. – pdu Jan 05 '14 at 10:44
It *did not* explain what happens. Originally (when I posted my comment) your answer didn't have anything but code. Also it makes sense to mention implicit casting, since it's caused not by returning a string, but by casting a `null` – zerkms Jan 05 '14 at 10:44
This does not explain how $_POST['...'] results in NULL. – gd1 Jan 05 '14 at 10:55
It does not "result" in NULL. Accessing a var which is not defined automatically is "NULL" when accessing it. – pdu Jan 05 '14 at 12:41

score 2 · Answer 3 · answered Jan 05 '14 at 10:57

This is what isset does:

Returns TRUE if var exists and has value other than NULL, FALSE otherwise.

In your first example, the $_POST variable doesn’t exist. Non-existing variables have the value null:

A variable is considered to be null if:

it has been assigned the constant NULL.

it has not been set to any value yet.

it has been unset().

So $a does also equal null. Although $a exists, it has the value null, so isset returns false.

In your second example, null is passed to trim, which returns an empty string:

var_dump(trim(null)); // string(0) ""

So $a does also equal an empty string. And since $a exists and has value other than null, isset returns true.

I would also remark that a E_NOTICE is issued when referring to non existing variables and array indices. — gd1, Jan 05 '14 at 11:00

score 1 · Answer 4 · answered Jan 05 '14 at 10:47

You could try empty() function. trim() is returning an empty string != NULL:

$a=trim($_POST['dsaghgjjkhfdsfdsfsdfdsjhkhkhgj']);

if(empty($a))
{
    echo"does not exist";
}
else
{
    echo"exists";
}

From php.net:

The following things are considered to be empty:

"" (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)

Why does trim() make a non existent variable exist?

4 Answers4