I've been out of the PHP world for a couple years and I've recently inherited a PHP project. Something that should be fairly easy in PHP is eluding me though. In Python I can do the following:
value = some_dict.get(unknown_key, default_value)
My first guess to do the same in PHP was:
$value = $some_array[$unknown_key] || $default_value;
But $value becomes a boolean because PHP doesn't support value short-circuiting. I also get a Notice: Undefined index: the_key but I know I can suppress that by prefixing #.
Is there I way to accomplish similar behavior in PHP to Python's dict.get(key, default)? I checked PHP's list of array functions but nothing stood out to me.
I guess you want something along the lines of the following:
$value = array_key_exists($unknown_key, $some_array) ? $some_array[$unknown_key] : $default_value;
This checks that the key exists and returns the value, else will assign the default value that you have assigned to $default_value.
$value = isset($some_array[$unknown_key]) ? $some_array[$unknown_key] : $default_value;
This way you won't need to suppress warnings
Try this:
$value = $some_array[$unknown_key] ?: $default_value;
This is equivalent to this line of code in Python:
value = some_dict.get(unknown_key, default_value)
Note that in the edge-case that the key exists but its value is falsey, you will get the $default_value, which may not be what you intend.
Related
I'm building my first application with Kohana, and using a basic templating system within that. In my templates I want to echo variables for the various contents of the page, but only if each variable is set, and I want to keep the code in the templates as short as possible, so something like this:
<?=$foo?>
works fine if the variable is set, but if it's not I get a notice. So I thought a ternary operator would do the trick nicely:
<?=$foo?:''?>
according to the PHP manual, from 5.3 it's ok to leave out the middle part and the above should output nothing if the variable isn't set, but I still get an error notice."Notice: Undefined variable: foo in /"
I can get the desired result with a slight alteration to suppress the notice:
<?=#$foo?:''?>
but I know that's generally not beset practice and would like a better solution if possible, while still keeping the code to a minimum in the template files.
the following works, but it's not as concise (mainly because my actual variables can be quite long):
<?=isset($foo)?$foo:'';?>
am I missing something or doing something wrong?
The ternary operation is not meant to replace checking with isset() - it needs it's variable defined or else you get a notice.
Template engines usually offer a way to use a default value instead, but they also do not use pure PHP code. You you are out of luck here: Either suppress the notice, or use the longer code variant. Or ensure that every variable is set, which enables you to consider any notice an error.
To avoid notices for undefined variables, you can create custom function that takes first parameter by reference
function tplvar(&$value, $default = '') {
return ($value !== null) ? $value : $default;
}
<?=tplvar($foo, 'bar');?>
Uninitialized variables passed by reference will be seen as nulls.
I'm wondering if there is a more elegant way to do this:
$foo = (isset($bar) and array_key_exists('meh', $bar)) ? $bar['meh'] : '';
If I remove the isset part, PHP issues a warning if $bar isn't an array, if I remove the array_key_exists part, PHP issues a warning if the meh key isn't in the array. Is there a more graceful, warning free, way of achieving the same end?
You take exactly the steps required to "secure" code against the warnings you mention.
However, the warnings are there for a reason. Typically it makes more sense to check, if you can prevent situations where the variables you are trying to access are not initialized.
You can reference the key directly. Inside isset it won't even throw an exception if $bar is undefined.
$foo = isset($bar['meh']) ? $bar['meh'] : '';
The difference between array_key_exists and isset is that isset will return FALSE if the key corresponds to a NULL value. In my code above, a NULL value will therefore result in $foo begin the empty string, not NULL. If that is a problem your current approach will be the best.
Sorry if the title isn't clear enough, but in Javascript you can do this :
var input = null;
var fruit = input || 'default';
Just wondering if PHP supports that kind of checking, currently I'm using :
$fruit = !empty($input)?$input : 'default' ;
Which is quite good, but of course the Javascript method is more elegant.
Thanks
|| returns a boolean in PHP, so this operator cannot be used for it. Same for or.
However, you can use the binary ?: operator if you are using PHP5.3+:
$fruit = $input ?: 'default';
However, it has one great disadvantage destroying its usefulness in the single case where it would be most useful (when importing request variables):
$fruit = $_REQUEST['fruit'] ?: 'strawberry';
This would throw an E_NOTICE in your face if $_REQUEST['fruit'] didn't exist. So in this case you still need the ternary version of it with an isset or !empty check.
PHP doesn't because it returns a Boolean, not the last evaluated expression like in JavaScript or Ruby.
You can use a ternary or the new binary form that ThiefMaster mentions.
In PHP, the || operator returns a boolean, thus you will always get either true or false from it. The reason it's possible in JS is because it will return a value.
So, no, not possible.
I just noticed PHP has an type casting to (unset), and I'm wondering what it could possibly be used for. It doesn't even really unset the variable, it just casts it to NULL, which means that (unset)$anything should be exactly the same as simply writing NULL.
# Really unsetting the variable results in a notice when accessing it
nadav#shesek:~$ php -r '$foo = 123; unset($foo); echo $foo;'
PHP Notice: Undefined variable: foo in Command line code on line 1
PHP Stack trace:
PHP 1. {main}() Command line code:0
# (unset) just set it to NULL, and it doesn't result in a notice
nadav#shesek:~$ php -r '$foo = 123; $foo=(unset)$foo; echo $foo;'
Anyone ever used it for anything? I can't think of any possible usage for it...
Added:
Main idea of question is:
What is reason to use (unset)$smth instead of just NULL?
As far as I can tell, there's really no point to using
$x = (unset)$y;
over
$x = NULL;
The (unset)$y always evaluates to null, and unlike calling unset($y), the cast doesn't affect $y at all.
The only difference is that using the cast will still generate an "undefined variable" notice if $y is not defined.
There's a PHP bug about a related issue. The bug is actually about a (in my mind) misleading passage elsewhere in the documentation which says:
Casting a variable to null will remove the variable and unset its value.
And that clearly isn't the case.
I’d guess (knowing PHP and it’s notaribly... interesting choices for different things, I may be completely wrong) that it is so that the value does not need setting to a var. For exact reason to use it for a code, I can’t think of an example, but something like this:
$foo = bar((unset) baz());
There you want or need to have null as argument for bar and still needs to call baz() too. Syntax of function has changed and someone did a duck tape fix, like what seems to be hot with PHP.
So I’d say: no reason to use it in well-thought architecture; might be used for solutions that are so obscure that I’d vote against them in first place.
As of PHP 8.0.X, (unset) casting is now removed and cannot be used.
For example it can be used like this
function fallback()
{
// some stuff here
return 'zoo';
}
var_dump(false ? 'foo' : fallback()); // zoo
var_dump(false ? 'foo' : (unset) fallback()); // null
Even if fallback() returns "zoo" (unset) will clear that value.
In a php page I have following code:
if($_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.