php: '0' as a string with empty()

Question

I want a 0 to be considered as an integer and a \'0\' to be considered as a string but empty() considers the \'0\' as a string in the example below,

$var = \         


        

Answer 1

If you want skip empty $filter and dont skip $filter = '0' and others values

$filter = ''; //or $filter = '0'; or $filter = '1';

//trim the $filter

if( isset($filter) and ( $filter != '' or $filter == '0') ) {

//$filter data 

};

Answer 2

You can't with only empty(). See the manual. You can do this though:

if ($var !== '0' && empty($var)) {
   echo "$var is empty and is not string '0'";
}

Basically, empty() does the same as:

if (!$var) ...

But doesn't trigger a PHP Notice when the variable is not set.

Answer 3

$var = '0';

// Evaluates to true because $var is empty
if (empty($var) && $var !== '0') {
    echo '$var is empty or the string "0"';
}

Answer 4

You can not, because integer,string,float,null does not matter for PHP.

Because it is cool :)

You must check characteristic features your variable is_numeric,isset,===,strlen, etc.

For example:

if (strlen(@$var)==0) {
    echo @$var . ' is empty';
}

or

if (@$var==="" || !isset($var)) {
    echo @$var . ' is empty';
}

or other examples :)

thanks for reading.

Answer 5

if ( (is_array($var) && empty($var)) || strlen($var) === 0 ) { echo $var . ' is empty'; }

Answer 6

In this case don't use empty, use isset() in place of it. This will also allow 0 as an integer.

$var = '0';
if (!isset($var)) {
    print '$var is not set';
}

$var = 0;
if (!isset($var)) {
    print '$var is not set';
}

Neither should print anything.