PHP - How to merge not-specified number of arrays?

Question

Is it possible to merge non-specified number of array elements in PHP?

I received data through JSON string (n elements), then I used them to do n loops - each loop saves the result to a different array cell ($result[1], $result[2], $result[3], ... ,$result[n]). I would like to merge the results into a single array (I have to send single array through JSON). Everything works if I write manually array_merge ($result[1], $result[2], $result[3]), but I never know how many variables I will receive.

Thanks in advance

Why do you save the result in different arrays if you can just throw them in the same array in the first place? — Charlotte Dunois
– Charlotte Dunois, Commented Sep 8, 2014 at 21:32
Hm this is a bit complicated ;) In main loop I execute 5 different SQL statements and save results into arrays: $result1, $result2, $result3, $result4, $result5. So after all n loops I got $result1[0], $result1[1], $result1[n], then $result2[0], $result2[1] etc... — belialek
– belialek, Commented Sep 8, 2014 at 21:52

user1467267 · Accepted Answer · 2014-09-08 21:22:33Z

1

Try this:

$yourListOfArrays = [$result[1], $result[2], ...];

call_user_func_array('array_merge',$yourListOfArrays);

answered Sep 8, 2014 at 21:22

user1467267

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7 Comments

user1467267 Over a year ago

I wonder who down-voted this answer. Can you comment what's the issue?

AbraCadaver Over a year ago

I didn't downvote, but how is that different? You still need to know how many arrays.

user1467267 Over a year ago

No you dont. You can loop all your arrays into the $yourListOfArrays. Then you don't need to know them upfront. So for the OP it would probably just be $result. But it can not if the array contains other stuff.

belialek Over a year ago

If i could specifed $yourListOfArrays then i would be able to use simple array_merge($result[1], $result[2], ...) :) Rocket Hazmat's answer solved my problem.

user1467267 Over a year ago

My answer is exactly, 1:1, the same as his. I'm just showing the added array-buildup in case your $result object would contain more than just arrays. You might for example wanted to shift those non-array keys out of it. That's the reason. Besides that my answer is no different.

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