PHP If Statement if (!$_POST) - What does !$_POST mean?
For instance, in some legacy code I'm looking at, the code reads:
<form action="paypalorders.php" method="POST">
<input type="hidden" name="orderList" value="' . $orderList . '">
<input type="submit" value="Archive">
</form>';
if (!$_POST) {
file_put_contents('/orders.txt', $output_line1);
}
I've looked at a bunch of other threads and haven't seen this exact question asked, so I'm trying to find out what it is. I've seen threads that format it like this:
if(!empty($_POST)) {
But not quite the same as what I'm looking for. Is it the same thing, just shorthand? I'm not sure, which is why I'm asking. I've Googled around and looked at a handful of threads and I'm still not sure.
Thank you.
The ! (not) logical operator returns true if the argument on its right-hand side is not true. It forces the argument to be evaluated as a boolean. In this case, when the $_POST array is evaluated as a boolean it will evaluate as true if it is not empty and false if it is. (See converting to boolean.)
if (!$_POST) { should be a safe way to detect whether or not anything is in $_POST if you want to do that. empty isn't necessary in that case because superglobals are always set, and since you aren't referring to a specific key, you don't need to worry about an undefined index notice.
I think it's also worth mentioning that if the only point of the check is to see what type of request was sent, it is probably better to just check the request method directly, because !$_POST does not mean the request wasn't a post, since a post request can be empty.
Since $_POST is an array, if it's empty his value is null, so if(!$_POST) would look like this:
if(!null){
//code
}
The following code returns true or false, but the objective of both is the same.
if(!empty($_POST)){
//code
}
Hope it helps you!
(bool)$array evaluates to true if $array contains elements, and false if it is empty.
Since $_POST is an array, !$_POST returns true if $_POST is empty.
Another way to interpret this, you are performing conditional tasks for the case where this page was not reached through a HTTP POST method.
when you cast an array to bool, it will be cast to bool(false) if the array is empty (eg if count($arr)===0), or bool(true) otherwise. ! casts whatever it's checking to bool. because $_POST always exists, if (!$_POST) { and if(empty($_POST)) { both do the exact same thing, they check if $_POST is empty. it's not even about being legacy code, this is still perfectly valid for 7.3.0. the difference between the 2 approaches will only become apparent when you're checking variables that may not exist, !$arr will trow an Notice: Undefined variable-error if $arr doesn't exist, empty($arr) will not.
Related
I am maintaining some old PHP code and I find many places in the code a test for a variable being non-empty of the following form:
if (!(isset($field_name) && strlen($field_name) > 0))
To my way of thinking, the following much simpler form would do exactly the same thing:
if ($field_name)
Am I missing some subltety whereby the more complex form is more precise? I have tended to use the simpler form any place where I need to add new code.
You can use empty() to replace your first line:
if (!empty($field_name))
The problem with your second example is that it will generate a warning if the variable is not set. Both empty() and isset() will not generate a warning for non-existing variables.
Note that you always have to account for possible values so if your value can be 0 or '0', this will not work as after $var = 0;, empty($var) will evaluate to true.
if ($field) will fail if $field is null, 0, 0.00, false or anything that can be casted false
if(!isset($field))
will fail if $field has not been declared at all
They do not do the same thing. The first code sample:
Checks if $field_name exists (I figure the ! is unintentional as it doesn't make sense the way it is written)
Checks if $field_name has a string length greater than zero
The second code sample:
Checks if the variable has a Boolean true value
Will throw an error if $field_name is not set.
The first snippet is clear and precise in its intent and performs a specific task.
The second snippet is very basic and only verifies the variable has a Boolean true value (and there are better ways to do that). If $field_name contains a string "0" this snippet will cause a hard to spot bug as it will fail as "0" evaluates to false. The first snippet would catch this.
You are right that the code you found is odd. It should probably be either:
if(!(isset($field_name)) { ... do something }
or
if(isset($field_name) && strlen($field_name) > 0 ) { ... do something }
...As you can appreciate, there is no need to test the length of a variable that is not defined.
However, if($field_name) is not the same as if(!(isset($field_name)), and the difference is not subtle. Indeed, the former will earn you a Undefined variable: fieldname if by some stroke of bad luck $field_name is not defined.
What's the difference?
if($field_name) tests if the existing variable $field_name evaluates to TRUE. For instance, it's value might be "my dog", and that evaluates to TRUE
if(!(isset($field_name)) tests if the variable $field_name exists at all.
i used this way to testing an array $arr if empty
if(!empty($arr)){ //do something }
recently, i saw someone use an another like:
if($arr){ //do something }
just want to know is't the second way a simply way to testing an array or is there some potential risk here ?
An empty array like array() is regarded as equal to false. So a simple if ($arr) works perfectly fine.
empty does the same kind of comparison, but does not trigger a NOTICE about missing variables, should the variable $arr not exist at all. You should not use empty if you are sure the variable exists, since it suppresses valuable error reporting. Only use empty if you really don't know whether a variable exists or not and have no control over it.
For more information about empty see The Definitive Guide To PHP's isset And empty.
The second casts the array to a boolean. Empty arrays are cast to false, anything else to true. So if($arr) and if(!empty($arr)) are functionally identical.
Both methods are functionally equivalent. The documentation for empty() mentions the following things are considered to be empty
"", 0, 0.0, "0", NULL, FALSE, array(), var $var;
Taking a look at casting to a bool, we can see that the list matches the list, which means both methods handle different types in the same way.
The second way is not ideal. If you use the second method and determine you're dealing with an array (and you aren't) and pass it to a foreach statement, you'll end up with an error. It is also more instructive for what you're checking to do more than test with if($arr).
My preference is:
if (is_array($arr) && count($arr) > 0) {
//work with array
}
Edit: I think my underlying point here is that the ability to test an array's existence is only part of the problem. If $arr turns out to be a string, a more robust check is needed.
I noticed this weird evaluation yesterday after searching for a few hours in my code for an error. i am passing scores into php, sometimes the score=0 which causes an issue.
send php ?blah=blah&score=0
if(!empty($_REQUEST['score']){
//do database update stuff
}else{
// show entire webpage
}
It works great unless the score=0 the if() will evaluate to false and return the entire webpage to my ajax handler and error. I have temporarily changed !empty to isset but this will cause problems in the future because isset evaluates to true even if the score key is in the url string without a value.
ex: (?blah=blah&score=&something=else)
my question is: what is the best way to recode this to work correctly now and in the future?
edit: there are a few working answers here, i appreciate everyones time. it was difficult to choose an answer
As the manual says, a variable is considered empty() if it has an empty or zero value.
So it will treat your variable wrongly as empty even though 0 is a perfectly acceptable value in your case.
If you need score to be a number, you could use isset() in combination with a is_numeric() check instead:
if((isset($_REQUEST['score']) and (is_numeric($_REQUEST['score'])){
Check out the manual page to see the kinds of values is_numeric() accepts. If score is always an integer, you can also use is_int((int)$_REQUEST['score']) but that will convert invalid input values to 0.
Additionally, as #sightofnick says, it's better to use explicit $_GET or $_POST instead of $_REQUEST.
Re your update:
In that case I would
Do check whether the variable is "0" (string "zero")
If it is "0", make it 0 (integer "zero")
If it is not 0, convert it to an integer (int)$_REQUEST["score"])
If the conversion resulted in 0, it was invalid input - exit
You have a valid integer variable.
empty() will return false if a value is zero. Use isset() or array_key_exists() instead, if you want to check if a variable in an array is set:
if (array_key_exists('score', $_REQUEST)) {...}
Try doing
if (isset($_REQUEST['score']) && ($_REQUEST['score'] !== '')) {
...
}
The isset will handle the presence/absence of the query parameter, and the strict string (!==) comparison will handle the case where the 'score' query is present but has no value. PHP treats all data coming from _GET/_POST/_REQUEST as strings, so this test is 100% reliable.
if(isset($_REQUEST['score']) && $_REQUEST['score'] != ''){
//do database update stuff
}else{
// show entire webpage
}
You may be able to solve that with
if (isset($_REQUEST['score']) && is_numeric($_REQUEST['score'])) {}
That of course if scrore can only contain numeric value
I am new to the concept of empty and null. Whilst I have endeavoured to understand the difference between them, I am more confused. I came across an article at http://www.tutorialarena.com/blog/php-isset-vs-empty.php however I still don't see when you would use isset and empty when validating forms. Seeing that I don't grasp the difference, I don't want to be using the incorrect functions as well as not be able to use the functions in other areas. Can someone give examples that will help me understand? I am very new to coding so would appreciate if someone could give me real world examples and at the same time keep it simply enough for noob to follow.
A variable is NULL if it has no value, and points to nowhere in memory.
empty() is more a literal meaning of empty, e.g. the string "" is empty, but is not NULL.
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)
Source.
Example
$a is NULL.
$a = '' is empty, but not NULL.
Update
If $a='' is empty but not NULL, when do I use the empty() function and when do I use the isset() function.
isset() will return FALSE is the variable is pointing to NULL.
Use empty() when you understand what is empty (look at the list above).
Also when you say it points nowhere in memory, what does that mean exactly?
It means that $str = '' will be in memory as a string with length of 0.
If it were $str = NULL, it would not occupy any memory.
Null is a placeholder that generally means "no data about this is available".
The use of null for this is just a convention, but a rather widespread one, to the point where some programming languages support the convention directly. The reason this convention exists has IMHO historically to do with "pointers";
many times a procedure will be defined to return a pointer to an answer, and will return what is traditionally called a Null pointer if it could not produce an answer for some reason.
Empty means (if this is a set) that it has no members. That's an explicit answer, and it is very different than "no data about this is available".
In the PHP world, apparantly uninitialized variables have the Null value, and isset on such a variable returns FALSE.
For arrays and strings, PHP follows the convention that "empty" means "has no members" although arrays and strings are not technically sets.
PHP apparantly has this funny idea that 0 and 0.0 are also "empty", by PHP design. That's abusive of the concept of "empty" IMHO: Individual numbers are not sets, so 0 can't reasonably by "empty". THis just leads to obscure programming because it violates the principle of least surprise. I'm sure the PHP designers would are that "zero is the empty number" as some kind of vague analogy; but the if analogy is vague, why bother with it? But then PHP is full of silly ideas.
The table below is an easy reference for what these functions will return for different values. The blank spaces means the function returns bool(false).
refer this link for more https://www.virendrachandak.com/techtalk/php-isset-vs-empty-vs-is_null/
NULL is a special value which explicitly states that the variable has not been set to any value yet. Be careful with using the empty() function as you can't just determine that a variable is exactly NULL using it. For example the empty() function will return true if an int is set to 0. If you need to make sure a variable is exactly NULL use if($variable == NULL).
For more info on empty() see http://php.net/manual/en/function.empty.php
There are some good answers here, which I won't repeat. In the case of validating forms, though, when a form is submitted, the value of each form input element is sent to the server in the $_POST variable. You can check for the existence of a particular input by using isset().
isset($_POST['username'])
If this returns true, then this request to the server was the result of posting a form containing an input element named "username". Now that we know that we have a value for that form element, we can see if it has a valid value. empty() will tell us whether the user actually entered any data in the field, or whether they left it empty.
empty($_POST['username'])
If that returns true then the form submitted to the server had a field named "username" but the user didn't enter anything into before submitting the form.
Been awhile since i used PHP but if other languages are anything to go by empty will indicate an existing object/map/array that has no contents while null would indicate a variable that has no meaning/definition at all (uninitialised).
In database SQL, NULL means "no value".
The empty() is a nice fast way to see if the variable holds any useful info... that is for strings empty() returns true for a string of "" as well as a null string.
So you can write something like this:
if (! empty($name)) echo $name;
More info see here: PHP: empty()
isset() returns true if both these conditions are met:
The variable has been defined and has not yet been unset.
The variable has a non-null value in it.
A variable is automatically defined when it gets set to something (including null). This has a direct implication in arrays.
$a=array();
$a['randomKey']=true;
$a['nullKey']=null;
var_dump(isset($a['randomKey'])); // true
var_dump(isset($a['nullKey'])); // true, the key has been set, set to null!
var_dump(isset($a['unsetKey'])); // false !
unset($a['randomKey']);
var_dump(isset($a['randomKey'])); // false ! it's been unset!
From above, you can check if various $_POST fields have been set. For example, a page that has been posted to, stands to reason, has the submit button name in the $_POST field.
empty() on the other hand, tests if the variable holds a non zero value. This means that values that (int) cast to 0, return false too. You can use this to see if a specific $_POST field has data in it.
This concept can be better understood from mathematics. Have you ever tried dividing a number (not zero) by 0 using a calculator e.g 7/0? You will get a result that looks like something this: undefined, not a number, null etc. This means that the operation is impossible, for some reasons (let's leave those reasons to be discussed another day).
Now, perform this: 0/7. You will get the output, 0. This means that the operation is possible and can be executed, but you the answer is just 0 because nothing is left after the division. There is a valid output and that output is zero.
In the first example, not only was the output invalid, the operation was not possible to execute. This is akin to null. The second example is akin to empty.
Yesterday, I posted an answer to a question that included several (unknown to me at the time) very bad code examples. Since then, I've been looking at my fundamental knowledge of PHP that allowed me to think that such code is possible. This brings me to a question that I can't seem to find an answer to:
If I want to check for whether or not a variable has anything set, is it a valid practice to not use isset() or another helper function? here's a "for instance":
if($not_set){
//do something
} else {
//do something else
}
Rather than...
if(isset($not_set)){
//do something
} else {
//do something else
}
From the name of the variable, you can see that this variable is not set. Therefore the conditional would be false and the else portion would run. Up until now I have been using this practice, but after the posts yesterday, I now have an inkling that this is wrong.
Here's why I thought that it would be an ok practice to leave out the isset() function above. From PHP manual:
The if construct is one of the most
important features of many languages,
PHP included. It allows for
conditional execution of code
fragments. PHP features an if
structure that is similar to that of
C:
if (expr) statement
As described in the section about
expressions, expression is evaluated
to its Boolean value. If expression
evaluates to TRUE, PHP will execute
statement, and if it evaluates to
FALSE - it'll ignore it. More
information about what values evaluate
to FALSE can be found in the
'Converting to boolean' section.
And from the 'Converting to boolean section':
When converting to boolean
, the following values are considered
FALSE:
...
* the special type NULL (including unset variables)
Why would the manual go out of its way to state that unset variables are included if this is a bad practice? If it's unset, it gets converted to NULL and therefore is evaluated properly by the conditional. Using isset() will find the same result, but will take extra cycles to do so.
Have I been wrong this whole time, and if so, why? (And just how bad it is, maybe?)
If the variable is not set you get a Notice. If you use isset() you don't get a notice. So from an error reporting point of view, using isset() is better :)
Example:
error_reporting(E_ALL);
if($a) {
echo 'foo';
}
gives
Notice: Undefined variable: a in /Users/kling/test on line 5
whereas
error_reporting(E_ALL);
if(isset($a)) {
echo 'foo';
}
does not output anything.
The bottom line: If code quality is important to you, use isset().
It's okay but not good practice to use if to check for a set variable. Two reasons off the top of my head:
Using isset makes the intent clear - that you're checking whether the variable is set, and not instead checking whether a condition is true.
if ($not_set) will evaluate to false when $not_set is actually set but is equal to boolean false.
You will run in to problems if your variable is set, but evaluates to FALSE, like the following:
the boolean FALSE itself
the integer 0 (zero)
the float 0.0 (zero)
the empty string, and the
string "0"
an array with zero elements
an object with zero member
variables (PHP 4 only)
the special type NULL (including
unset variables)
SimpleXML objects created from empty
tags
Taken from the PHP manual.
Basically, using isset() is showing that you are explicitly checking if a variable exists and is not NULL, while the structure of your if statement only checks if the variable is true. It is more clear and less error-prone.
It is a common practise, but is not good -- you should always use isset!
If your $not_set is set, and is a bool with the value false, your "test" will fail!
isset works as a guard preventing you from using variables that do not actually exist.
if (isset($foo)) and if ($foo) do not mean the same thing. isset just tells you if the variable actually exists and if it's okay to use it, it does not evaluate the value of the variable itself*.
Hence, you should usually use one of these two patterns:
If the variable is sure to exist and you just want to check its value:
if ($foo == 'bar')
If the variable may or may not exist, and you want to check its value:
if (isset($foo) && $foo == 'bar')
If you're just interested that a variable is set and evaluates to true, i.e. if ($foo), you can use empty:
if (isset($foo) && $foo)
// is the same as
if (!empty($foo))
* it does check for null, where null is as good as not being set at all