i recently had to do a "test" for a job, and i got feed back saying that this statement was incorrect:
$images = $flickr->get_images(5, !empty($_GET['pg']) ? $_GET['pg'] : 1);
The "supposed" error was generated via the ternary operator on the first time the page was loaded, as there was no "?pg=1" (or whatever) passed via the query string.
The feed back said i should have used isset instead. I have looked at various posts both here (question 1960509) and blogs, but cannot find any definitive answer.
Is this really an error? How can i replicate this issue? do i need to put on E_STRICT or something in my php.ini file? Or might this be due to an older version of php?
Note: please don't tell me about how i should validate things.. i know this... it was a test to just see if i could use the flickr api calls.
This is perfectly fine. empty is not an actual function, it's a language construct. It does not issue a warning if a variable is not set (in that case the variable is considered empty, thus the 'function' returns TRUE just as you want), and additionally it checks for empty or zero values.
You could see empty as a normal isset check with an additional loose comparison to FALSE:
empty($var) === (!isset($var) || $var == FALSE)
$images = $flickr->get_images(5, (isset($_GET['pg']&&($_GET['pg']))) ? $_GET['pg'] : 1);
without isset you'll get error so combine them
I'd use
$images = $flickr->get_images(5, array_key_exists('pg', $_GET) ? $_GET['pg'] : 1);
Combine with !empty($_GET['pg']) if needed (i.e. array_key_exists('pg', $_GET) && !empty($_GET['pg'])), but array_key_exists is the intended function for this job.
I think in a situation like this isset is the correct function to use as it is checking the existence of the array element rather than checking if the value of the element has been set. As Martin notes, the best thing to do here is combine them as this will only check the value if the element exists, meaning that the error will not occur on the first page load.
Also, I think this will only give a warning if E_NOTICE is on (or perhaps E_WARNING as well)
The reason you would get an error is because the empty function is designed to check the value of an existing variable, whearas isset() is designed to tell you whether a variable has been instantiated, however because empty() is a language construct technically it doesn't throw an error or create a warning so most people don't see the difference.
From the docs:
empty() is the opposite of (boolean) var, except that no warning is generated when the variable is not set.
isset — Determine if a variable is set and is not NULL. So "isset" is the correct function to use for checking for value is set or not.
More details :http://php.net/manual/en/function.isset.php
Related
Knowing the differences of array_key_exists() and isset() in PHP, I see a lot of advocates on the web suggesting for replacing array_key_exists() with isset(), but I am thinking is it safe to do so?
In one project, I have the value of $var submitted by users. The value can be anything, even NULL or not set at all. I want to use !empty($var) to check if $var holds a non-empty value, but I understand that using !empty($var) alone is dangerous since if $var isn't pre-defined, PHP will throw an error.
So I have isset($var) && !empty($var) for checking whether $var holds a non-empty value.
However, things get complicated when I have a value stored in an assoc array. Compare the followings, assuming array $arr always exists but the key foo may or may not exist in $arr.
// code snipplet 1
$arr = array();
echo isset($arr['foo']);
// code snipplet 2
$arr = array();
echo array_key_exists('foo', $arr) && !is_null($arr['foo']);
Code snipplet 2 will always work but it seems clumsy and harder to read. As for code snipplet 1, I had bad experience... I wrote something like that before in the past on a development machine. It ran fine, but when I deployed the code to the production machine, it threw errors simply because key didn't exist in array. After some debugging, I found that the PHP configs were different between the development and the production machines and their PHP versions are slightly different.
So, I am thinking is it really that safe to just replace array_key_exists() with isset()? If not, what can be of better alternatives of code snipplet 2?
In one project, I have the value of $var submitted by users.
How does that work? Users shouldn't be able to set variables. They should be able to, e.g., submit form values which end up in $_POST. I repeat, they should not be able to directly create variables in your scope.
If "users" here means some sort of plugin system where people write and include PHP code… then you may want to think about defining a more stable interface than setting variables.
The value can be anything, even NULL…
Not if it's a value submitted through HTTP. HTTP has no concept of null. It's either an empty string or doesn't exist at all.
using !empty($var) alone is dangerous since if $var isn't pre-defined, PHP will throw an error
That is wrong. empty specifically exists to test a variable against false without throwing an error. empty($var) is the same as !$var without triggering an error if the variable doesn't exist.
So I have isset($var) && !empty($var) for checking whether $var holds a non-empty value.
See Why check both isset() and !empty(). (Spoiler: it's redundant.)
echo isset($arr['foo']);
echo array_key_exists('foo', $arr) && !is_null($arr['foo']);
These both do exactly the same thing. isset returns true if the value exists and its value is not null. The second line returns true if the array key exists and its value is not null. Same thing.
I had bad experience...
You'd need to be more detailed about that, since there should be no caveat to isset as you describe it.
See The Definitive Guide To PHP's isset And empty and Difference between isset and array_key_exists.
I was recently placed on a project with some PHP, and I don't know much about PHP. There are a couple instances in the site where upon clicking a button, the user is redirected to another page with some URL parameters. The next page then uses $_GET to get those parameters and move on.
Another issue in the code caused the page to reload the second page without the parameters, so using $_GET would return errors, but with the other issue fixed, I can't think of a reason why the parameters wouldn't be there.
While debugging, I came across advice to always check $_GET using isset(), but theoretically there should never be an instance when those parameters aren't there (otherwise something else is really wrong with the server or the code).
Is it still worth putting in the checks and working out a backup solution, even though there shouldn't be a need for it? I want to make sure I'm not ignoring some other potential issue that I may not be aware of.
It is always good practice+recommended to check your variables before applying any logic.
!empty() is recommended to use instead of isset() because it check both that variable is initialized and have some values too.
In case of array count($array)>0 can be used as a check.
Why to use !empty() check here:- !empty() Vs isset()
If you are expecting or requiring data to be sent via $_GET you should check if it's set. Especially since that data can be easily manipulated. Also like #Alive to Die said !empty() is better.
isset() checks if a variable has a value including ( False , 0 , or
empty string) , but not NULL. Returns TRUE if var exists; FALSE
otherwise. On the other hand the empty() function checks if the
variable has an empty value empty string , 0, NULL ,or False.
I have a simple form I created, and in it I have the following checkbox:
<input type="checkbox" name="test">
Note: this form is being submitted to itself.
Above the form, I have the following PHP:
if (empty($_POST['test'])) {
$thevalue = 0;
} else {
$thevalue = 1;
}
var_dump($thevalue);
When I process the form, I get what I would expect. If I check the box and submit, I get int(1) if I leave it unchecked I get int(0).
In the first line of my PHP code, I wanted to replace $_POST['test'] with some simple variable.
So I added the following line above my code:
$simplevar = $_POST['test']
I then replaced the condition in my if statement to be empty($simplevar)
But when I submit the form, I get a "Notice: Undefined index:" error message
Why is this happening?
Assuming it's possible to achieve what I was after (i.e. insert $_POST into $simplevar), how might I go about it?
Thanks in advance for your help!
PS: I may have a follow up to this question, but didn't want to clutter things by jamming it all in here.
Thanks again... oh, and Merry Christmas! ;-)
This happens because when you don't check the checkbox, the browser does not send any value to server for that control when the form is submitted. Because of this, $_POST['test'] is not defined, and you tried to use it without a check as to whether it existed, so you get a warning. One of the checks that empty() does is to see whether the value is set. So, when you use the $_POST keys directly in empty(), you don't get an error, but when you try and use it in an assignment without this check, you will get the error.
You can do roughly what you want to do, you just have to change the logic slightly. If you do:
$simplevar = !empty($_POST['test']);
// You could also do
// $simplevar = isset($_POST['test']);
if ($simplevar) {
// The box was checked
} else {
// The box was not checked
}
...it will do what you want without the error. Using this approach, $simplevar always holds a boolean indicating whether or not the box was checked.
When a checkbox is unchecked, it's not added to the $_POST array as a key, which is why $simplevar = $_POST['test'] returns the error you posted. Using empty() gets past this problem by empty() handling errors better (well, silently at any rate).
You haven't specified whether you get that error when the checkbox is checked or not, but the above explanation is the only one I can give. If you're unsure, try doing print_r($_POST) to see what $_POST actually contains.
A solution to your problem would be to use a ternary expression to handle the error a little better:
$simplevar = isset($_POST['test']) ? 0 : 1;
This will assign 0 to $simplevar if $_POST['test'] isn't set (checkbox isn't checked), or 1 otherwise.
Do make sure all your form processing code is put inside
if(!empty($_POST)) {
// Code
}
So that it's not executed every time the page loads, otherwise your error will show every time.
Checkbox values are only transmitted if the checkbox was checked. This means that unchecked checkboxes won't appear in the $_POST array.
A way to suppress the notice from PHP is to use a reference instead of a variable:
$simplevar =& $_POST['test'];
if(empty($simplevar)) $thevalue = 1;
else $thevalue = 0;
That's expected behaviour. If you are assigning the variable like this:
$simplevar = $_POST['test'];
Then the $_POST variable might be absent. The Zend runtime then assigns the NULL value, but gives you a useful debug hint, should that not be what you wanted.
When you used empty() before, the check for variable existence was built in. empty() is a language construct. Like isset() it's often used to eschew such notices. The cumbersome syntax to emulate such language behaviour is:
$simplevar = empty($_POST['test']) ? NULL : $_POST['test'];
The language built-in for is:
$simplevar = #( $_POST['test'] );
Now, I will get roasted for mentioning it. (Using # is useful if you want to bring the debug notices back at some point, while the empty and isset constructs eternally suppress them.)
First, you should always check that variables in $_POST, $_REQUEST, and $_GET are set before attempting to use them. Always handle the condition where they are not set even if you simply output an error.
Because the error is an undefined index it seem the error is in test not being set in $_POST, though that doesn't make a lot of sense. I would add a check, maybe an echo or var dump to check $_POST. If it is set the other problem could be an issue with scope. $_POST is something called a super global which makes it available in any scope. Variables you set you may need to make global by defining them as such if you want to access them across scopes.
Yesterday, I posted an answer to a question that included several (unknown to me at the time) very bad code examples. Since then, I've been looking at my fundamental knowledge of PHP that allowed me to think that such code is possible. This brings me to a question that I can't seem to find an answer to:
If I want to check for whether or not a variable has anything set, is it a valid practice to not use isset() or another helper function? here's a "for instance":
if($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
In a php page I have following code:
if($_REQUEST['c']!="") // I get error on this line itself. Why?
{
$pidis=(int)($_REQUEST['c']);
}
I keep getting Undefined index error.
On Googling I manage to understand that if a page is access without parameters (in URL) which we are trying to access we can get this error/warning. I believe that if a parameter is not defined in the URL it should just return empty instead of giving error/warning message.
I know that it is possible to suppress errors and warning by adding
error_reporting(E_ALL ^ E_NOTICE);
But I do not want to do this.
This same page work just fine on our company's web server but does not work on our clients web server.
Why is this happening?
How to solve this problem?
You are getting that error because you are attempting to compare $_REQUEST['c'] to something when $_REQUEST['c'] does not exist.
The solution is to use isset() before comparing it. This will remove the warning, since the comparison won't happen if $_REQUEST['c'] doesn't exist.
if(isset($_REQUEST['c']) && $_REQUEST['c']!="")
{
$pidis=(int)($_REQUEST['c']);
}
It is an E_NOTICE level error, and your level of error reporting will affect whether the error shows up or not. Your client's server has E_NOTICE level error reporting turned on, which is why it shows up there.
It is a good idea to always develop using E_ALL so that you can catch this kind of error before moving your code to other servers.
Another solution is to use the following:
$pidis = isset($_REQUEST['c']) ? $_REQUEST['c'] : '';
You can also, if you prefer to return a value other than empty, by placing a default value within the final set of single quotes, e.g.
$pidis = isset($_REQUEST['c']) ? $_REQUEST['c'] : 'Default Value';
or return a different variable type, for instance an integer:
$pidis = isset($_REQUEST['c']) ? $_REQUEST['c'] : 34;
Instead of isset() you can also use: array_key_exists().
The difference between both methods is that isset() checks also whether the value of the variable is null. If it is null then isset returns false whereas array_key_exists() returns always true if the key exists (no mater which value). E.g.:
$array = array('c' => null);
var_dump(isset($array['c']))); // isset() returns FALSE here
var_dump(array_key_exists($array['c']); //array_key_exists() returns TRUE
Depending on the context, it is important to distinguish this. In your case I don't think it matters doesn't matter, as (I guess) a request parameter never will be null (except one overwrites it manually).
Use isset($_REQUEST['c']) to test if it exists first.
PHP is giving a notice (which is not an error : it's just a notice) when you are trying to use a variable that doesn't exists, or an array element that doesn't exist.
This is just to help you, and you should not mask those notices : they are here to help you -- for instance, to help you detect typos in variable names.
Before using that array index, if it's not always present, you should test if it's here, using isset :
if (isset($_REQUEST['c']) && $_REQUEST['c']!="") {
// ...
}
Clean way could be :
$pidis = $_REQUEST['c'] ?? null
this is same as checking isset request but shorter.