In php how can I access an array's values without using square brackets around the key? My particular problem is that I want to access the elements of an array returned by a function. Say function(args) returns an array. Why is $var = function(args)[0]; yelling at me about the square brackets? Can I do something like $var = function(args).value(0); or am I missing something very basic?
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php refers to this as function array referencing. It has been included in php since version 5.4 php.net/manual/en/migration54.new-features.phpSteve– Steve2015-11-13 15:56:24 +00:00Commented Nov 13, 2015 at 15:56
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5 Answers
As the others have said, you pretty much have to use a temporary variable:
$temp = myFunction();
$value = $temp[0];
But, if know the structure of the array being returned it is possible to avoid the temporary variable.
If you just want the first member:
$value = reset(myFunction());
If you want the last member:
$value = end(myFunction());
If you want any one in between:
// second member
list(, $value) = myFunction();
// third
list(, , $value) = myFunction();
// or if you want more than one:
list(, , $thirdVar, , $fifth) = myFunction();
3 Comments
Pekka
Hey, clever!
list() never ceases to amaze me. +1.
goat
reset() and end() require the arguments be references. You get an E_STRICT notice in recent versions of php.
amb
yes, i have been doing this with temporary variables, but was wondering whether i really needed to. now i'm just wondering why i have to. but in any case your list() usage is pretty clever. thanks!
In PHP, when getting an array as a function result, you unfortunately have to do an extra step:
$temp_array = function($args);
$var = $temp_array[0];
For objects, this has been relaxed in PHP 5. You can do:
$echo function($args)->property;
(provided function returns an object of course.)
3 Comments
Luke Magill
It sucks but this is the only way to do this. One of the many reasons to hate PHP.
Pekka
@Luke true, but not that bad imo. Who knows, it may get fixed in PHP 7 :)
nickf
@Luke - that's a bit strong isn't it?
function getKey($array, $key){
return $array[$key];
}
$var = getKey(myFunc(args), $key);
There is no way to do this without adding a user function unfortunately. It is just not part of the syntax.
You could always just do it the old fashion way
$array = myFunc();
$value = $array[0];
Comments
What exactly matches your expecting is:
echo pos(array_slice($a=myFunc(), pos(array_keys(array_keys($a), 'NameOfKey'));
answered Kinetix Kin, Taipei
Comments
if you want this, its probably best to be returning an object (unfortunately, its totally lame php doesnt support this). Heres a crazy way i was able to figure out though, out of novelty (please dont do this!):
function returnsArray(){
return array("foo" => "bar");
}
echo json_decode(json_encode((object)returnsArray()))->foo;
//prints 'bar'
So yeah..until they add support for array dereferencing in php, i think you should probably just cast the return array as an object:
return (object)array("foo" => "bar");
and then you can do returnsArray()->foo, since php relaxes dereferencing for objects but not arrays.. or of course write a wrapper function like others have suggested.
1 Comment
mutexkid
also, looks like support for this has been recently added to php? wiki.php.net/rfc/functionarraydereferencing