php array_diff compare arrays with new array

Question

I have two arrays:

$userBuildingIds

print_r():

Array
(
    [0] => 4
 )

and $allRequiredIds

print_r:

Array
(
    [0] => 4
    [1] => 1
)

now I want check if one element of $userBuildingIds exists in the $allRequiredIds array. And if so, I want get a new array of all elements they are NOT in the first array like this:

Array
(
    [0] => 1
)

(because 1 isn't in $userBuildingIds) I try this with array_diff but it gives me this result (with the key of array 2):

Array
(
    [1] => 1
)

Is it possible to get an array in which are all the elements of array $allRequiredIds where are not in $userBuildingIds, but without copy the keys from $allRequiredIds?

If you don't care about the keys of the returned array then you can just use array_values() on it to get a new array having the keys starting with 0. So, the complete code will be array_values(array_diff($allRequiredIds, $userBuildingIds)). — axiac
– axiac, Commented Jan 29, 2015 at 14:39
thanks, this will work.. you can answer this, so I can accept your answer. — goldlife
– goldlife, Commented Jan 29, 2015 at 14:42
I am not sure what are you trying to achive and how different values could be in your array. but I am sure you know that for $arr1 = [1,1,1,1,1]; $arr2 = [1,2]; array_diff($arr1, $arr2); will return emty array? that is really your point? — Alex
– Alex, Commented Jan 29, 2015 at 14:50

axiac · Accepted Answer · 2015-01-29 14:46:29Z

3

If you don't care about the keys of the returned array then you can just use array_values() on it to get a new array having the keys starting with 0.

The code will be:

$diffIds = array_values(array_diff($allRequiredIds, $userBuildingIds));

It produces a list of values from $allRequiredIds that does not exist in $userBuildingIds. The returned list has numeric keys starting with 0 (no association with the original keys from $allRequiredIds is $userBuildingIds is kept, on purpose).

answered Jan 29, 2015 at 14:46

axiac

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