Solution Checker

Question

Is there a difference between isset and !empty. If I do this double boolean check, is it correct this way or redundant? and is there a shorter way to do the same thing?

isset($vars[1]) AND !empty($vars[1])

Answer 1

Solution 1

This is completely redundant. empty is more or less shorthand for !isset($foo) || !$foo, and !empty is analogous to isset($foo) && $foo. I.e. empty does the reverse thing of isset plus an additional check for the truthiness of a value.

Or in other words, empty is the same as !$foo, but doesn't throw warnings if the variable doesn't exist. That's the main point of this function: do a boolean comparison without worrying about the variable being set.

The manual puts it like this:

empty() is the opposite of (boolean) var, except that no warning is generated when the variable is not set.

You can simply use !empty($vars[1]) here.

Answer 2

Solution 2

isset() tests if a variable is set and not null:

http://us.php.net/manual/en/function.isset.php

empty() can return true when the variable is set to certain values:

http://us.php.net/manual/en/function.empty.php

To demonstrate this, try the following code with $the_var unassigned, set to 0, and set to 1.

<?php

#$the_var = 0;

if (isset($the_var)) {
  echo "set";
} else {
  echo "not set";
}

echo "\n";

if (empty($the_var)) {
  echo "empty";
} else {
  echo "not empty";
}
?>

Answer 3

Solution 3

The accepted answer is not correct.

isset() is NOT equivalent to !empty().

You will create some rather unpleasant and hard to debug bugs if you go down this route. e.g. try running this code:

<?php

$s = '';

print "isset: '" . isset($s) . "'. ";
print "!empty: '" . !empty($s) . "'";

?>

https://3v4l.org/J4nBb

Answer 4

Solution 4

isset($vars[1]) AND !empty($vars[1]) is equivalent to !empty($vars[1]).

I prepared simple code to show it empirically.

Last row is undefined variable.

+-----------+---------+---------+----------+---------------------+
| Var value | empty() | isset() | !empty() | isset() && !empty() |
+-----------+---------+---------+----------+---------------------+
| ''        | true    | true    | false    | false               |
| ' '       | false   | true    | true     | true                |
| false     | true    | true    | false    | false               |
| true      | false   | true    | true     | true                |
| array ()  | true    | true    | false    | false               |
| NULL      | true    | false   | false    | false               |
| '0'       | true    | true    | false    | false               |
| 0         | true    | true    | false    | false               |
| 0.0       | true    | true    | false    | false               |
| undefined | true    | false   | false    | false               |
+-----------+---------+---------+----------+---------------------+

And code

$var1 = "";
$var2 = " ";
$var3 = FALSE;
$var4 = TRUE;
$var5 = array();
$var6 = null;
$var7 = "0";
$var8 = 0;
$var9 = 0.0;

function compare($var)
{
    print(var_export($var, true) . "|" .
        var_export(empty($var), true) . "|" .
        var_export(isset($var), true) . "|" .
        var_export(!empty($var), true) . "|" .
        var_export(isset($var) && !empty($var), true) . "\n");
}

for ($i = 1; $i <= 9; $i++) {
    $var = 'var' . $i;
    compare($$var);
}

@print(var_export($var10, true) . "|" .
    var_export(empty($var10), true) . "|" .
    var_export(isset($var10), true) . "|" .
    var_export(!empty($var10), true) . "|" .
    var_export(isset($var10) && !empty($var10), true) . "\n");

Undefined variable must be evaluated outside function, because function itself create temporary variable in the scope itself.

Answer 5

Solution 5

$a = 0;
if (isset($a)) { //$a is set because it has some value ,eg:0
    echo '$a has value';
}
if (!empty($a)) { //$a is empty because it has value 0
    echo '$a is not empty';
} else {
    echo '$a is empty';
}

Answer 6

Solution 6

Empty just check is the refered variable/array has an value if you check the php doc(empty) you'll see this things are considered emtpy

* "" (an empty string)
* 0 (0 as an integer)
* "0" (0 as a string)
* NULL
* FALSE
* array() (an empty array)
* var $var; (a variable declared, but without a value in a class)

while isset check if the variable isset and not null which can also be found in the php doc(isset)

Answer 7

Solution 7

It is not necessary.

No warning is generated if the variable does not exist. That means empty() is essentially the concise equivalent to !isset($var) || $var == false.

php.net

Answer 8

Solution 8

From the PHP Web site, referring to the empty() function:

Returns FALSE if var has a non-empty and non-zero value.

Thats a good thing to know. In other words, everything from NULL, to 0 to will return TRUE when using the empty() function.

Here is the description of what the isset() function returns:

Returns TRUE if var exists; FALSE otherwise.

In other words, only variables that dont exist (or, variables with strictly NULL values) will return FALSE on the isset() function. All variables that have any type of value, whether it is 0, a blank text string, etc. will return TRUE.

Answer 9

Solution 9

"Empty": only works on variables. Empty can mean different things for different variable types (check manual: http://php.net/manual/en/function.empty.php).

"isset": checks if the variable exists and checks for a true NULL or false value. Can be unset by calling "unset". Once again, check the manual.

Use of either one depends of the variable type you are using.

I would say, it's safer to check for both, because you are checking first of all if the variable exists, and if it isn't really NULL or empty.

Answer 10

Solution 10

if we use same page to add/edit via submit button like below

<input type="hidden" value="<?echo $_GET['edit_id'];?>" name="edit_id">

then we should not use

isset($_POST['edit_id'])

bcoz edit_id is set all the time whether it is add or edit page , instead we should use check below condition

!empty($_POST['edit_id'])

Why check both isset() and !empty()