For example:
$array = [];
echo $array['bar']; // PHP NOTICE - trying to access not existing key`
$array['bar'][] = 'foo'; // Nothing
I understand that it creates array with that index 'bar', but how does PHP deals with that internally?
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That's just how it works.AbraCadaver– AbraCadaver2016-12-12 16:40:57 +00:00Commented Dec 12, 2016 at 16:40
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3@AbraCadaver That's how a religion starts, don't question just follow. OPs question is legitimate. Your answer is not.Andrei– Andrei2016-12-12 16:42:19 +00:00Commented Dec 12, 2016 at 16:42
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1@Andrew: That's funny, I didn't post an answer.AbraCadaver– AbraCadaver2016-12-12 16:44:33 +00:00Commented Dec 12, 2016 at 16:44
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Comment, whatever. Point stands.Andrei– Andrei2016-12-12 16:45:10 +00:00Commented Dec 12, 2016 at 16:45
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2This article will explain more about how PHP is internally handling arrays: nikic.github.io/2012/03/28/…Dragos– Dragos2016-12-12 17:04:11 +00:00Commented Dec 12, 2016 at 17:04
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1 Answer
$array['bar'][] = 'foo'; doesn't return a notice or error because there is no error. You are creating a new array index, and another index within that, and assigning a value to it. That's what the statement means. There's no error to return.
If you want to have behaviour for if a particular array index is not set, you can use array_key_exists (http://php.net/manual/en/function.array-key-exists.php):
if(array_key_exists('bar', $array)){
$array['bar'][] = 'foo';
} else {
// something else
}
That's if this question is functional (ie., you're trying to accomplish something specific). If the question is more conceptual - why doesn't PHP read the variable assignment as an error:
PHP is capable of initializing and assigning a variable in one line, ie $foo = 'bar'. This does not return an error even though $foo was not previously defined because PHP initializes the variable first. The same method holds true for array indexes. $array['foo'][] = 'bar' doesn't return an error or a notice because PHP is initializing the array index just as it would a variable.
6 Comments
Dragos
You still didn't answered the question, namely why (and how does PHP internally handle that) it DOES throw a notice when you're trying to access an inexisting key, but it does not throw any notice when you're trying to set a value for an inexisting key of an array
CGriffin
Because functionally speaking, you AREN'T trying to set a value for a nonexistent key. You're setting a key, and another key inside it, and assigning a value to THAT key. PHP is intelligent enough to cascade that key creation down. That's opposed to when you try to ACCESS a key that doesn't exist, in which case there's nothing PHP can do.
RWS
@Dragos The proceducre of creating a previously non-existing variable is called initialization. In PHP you can also initialize a variable by assigning a value to previously non-existing variable. There wouldn't be any notice in this case either. It is the same when you set a value for new key in an array.
povils
@PeregrineStudios thanks for answer and yes, my question is more conceptual. I can understand how PHP deals with that in abstract sense, but the part 'PHP is able to correct' is a bit mistery for me:)
CGriffin
Clarified my answer a little including Dragos' comment. I hope it clears things up a bit.
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