Code will explain more:
$var = 0;
if (!empty($var)){
echo "Its not empty";
} else {
echo "Its empty";
}
The result returns "Its empty". I thought empty() will check if I already set the variable and have value inside. Why it returns "Its empty"??
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 $var; (a variable declared, but without a value in a class)
Note that this is exactly the same list as for a coercion to Boolean false. empty is simply !isset($var) || !$var. Try isset instead.
I was wondering why nobody suggested the extremely handy Type comparison table. It answers every question about the common functions and compare operators.
A snippet:
Expression | empty($x)
----------------+--------
$x = ""; | true
$x = null | true
var $x; | true
$x is undefined | true
$x = array(); | true
$x = false; | true
$x = true; | false
$x = 1; | false
$x = 42; | false
$x = 0; | true
$x = -1; | false
$x = "1"; | false
$x = "0"; | true
$x = "-1"; | false
$x = "php"; | false
$x = "true"; | false
$x = "false"; | false
Along other cheatsheets, I always keep a hardcopy of this table on my desk in case I'm not sure
From a linguistic point of view empty has a meaning of without value. Like the others said you'll have to use isset() in order to check if a variable has been defined, which is what you do.
empty() returns true for everything that evaluates to FALSE, it is actually a 'not' (!) in disguise. I think you meant isset()
The following things are considered to be empty:
From PHP Manual
In your case $var is 0, so empty($var) will return true, you are negating the result before testing it, so the else block will run giving "Its empty" as output.
From manual: Returns FALSE if var has a non-empty and non-zero value.
The following things are considered to be empty:
empty should mean empty .. whatever deceze says.
When I do
$var = '';
$var = '0';
I expect that var_dump(empty($var)); will return false.
if you are checking things in an array you always have to do isset($var) first.
use only ($_POST['input_field_name'])!=0 instead of !($_POST['input_field_name'])==0 then 0 is not treated as empty.
Not sure if there are still people looking for an explanation and a solution. The comments above say it all on the differences between TRUE / FALSE / 1 / 0.
I would just like to bring my 2 cents for the way to display the actual value.
If you're working with a Boolean datatype, you're looking for a TRUE vs. FALSE result; if you store it in MySQL, it will be stored as 1 resp. 0 (if I'm not mistaking, this is the same in your server's memory).
So to display the the value in PHP, you need to check if it is true (1) or false (0) and display whatever you want: "TRUE" or "FALSE" or possibly "1" or "0".
Attention, everything bigger (or different) than 0 will also be considered as TRUE in PHP. E.g.: 2, "abc", etc. will all return TRUE.
If you're working with a number datatype, the way it is stored is the same.
To display the value, you need to tell PHP to handle it as a number. The easiest way I found is to multiply it by 1.
proper example. just create int type field( example mobile number) in the database and submit an blank value for the following database through a form or just insert using SQL. what it will be saved in database 0 because it is int type and cannot be saved as blank or null. therefore it is empty but it will be saved as 0. so when you fetch data through PHP and check for the empty values. it is very useful and logically correct.
0.00, 0.000, 0.0000 .... 0.0...0 is also empty and the above example can also be used for storing different type of values in database like float, double, decimal( decimal have different variants like 0.000 and so on.