PHP array_merge if not empty

Question

Trying to merge 4 arrays, but some may be empty at certain times.

$array_1 = array('something1', something2);
$array_2 = array('something3', something4);
$array_3 = array();
$array_4 = array('something1', something2);

$list = array_merge($array_1,$array_2,$array_3,$array_4);

print_r($list);

But if one of the arrays are empty, there will be an error. I've been googling forever, but I can't find a solid simple answer on how to check for empty arrays before merging.

Argument #2 is not an array

Or whichever array is empty is the argument number. How do I strip out empty arrays before merging?

if you make array of arrays of them, you can loop through it, perform multiple merges and do the empty() checks there — MightyPork
– MightyPork, Commented Nov 21, 2013 at 21:13
There is NO error with an empty array. There is only an error if the arg is NOT an array. — AbraCadaver
– AbraCadaver, Commented Nov 21, 2013 at 21:23

AbraCadaver · Accepted Answer · 2013-11-21 21:25:14Z

19

There is NO error with an empty array. There is only an error if the arg is NOT an array.

You could check is_array() or:

$list = array_merge(
(array)$array_1,
(array)$array_2,
(array)$array_3,
(array)$array_4
);

answered Nov 21, 2013 at 21:25

AbraCadaver

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Comments

MightyPork · Accepted Answer · 2013-11-21 21:33:47Z

15

Okay here you go, this should do the trick (if you make array of the initial arrays):

$arrs = array();

$arrs[] = array('something1', something2);
$arrs[] = array('something3', something4);
$arrs[] = array();
$arrs[] = array('something1', something2);

$list = array();

foreach($arrs as $arr) {
    if(is_array($arr)) {
        $list = array_merge($list, $arr);
    }
}

print_r($list);

edited Nov 21, 2013 at 21:33

answered Nov 21, 2013 at 21:20

MightyPork

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5 Comments

David Over a year ago

This worked perfect, never thought of making the variables arrays and looping through. Thanks!

Yada Over a year ago

isn't $list empty on the first iteration?

MightyPork Over a year ago

@Yada tbh I dont think array_merge has a problem with empty array, I suspect David tried to feed null or undefined variable into it.

jszobody Over a year ago

if(!empty($arr)) should be if(is_array($arr)). As others have pointed out, you need to ensure that it is an array, it doesn't matter if it is empty.

Abhijit Jagtap Over a year ago

Yes really good but you should avoid empty array as well, if (is_array($arr)&&!empty($arr)) { $list = array_merge($list, $arr); } , got problem in json_encode. :)

Community · Accepted Answer · 2023-11-17 20:33:26Z

5

Array merge supports empty array()

Doc:

Example #3 Simple array_merge() example https://www.php.net/array_merge

<?php
$array1 = array();
$array2 = array(1 => "data");
$result = array_merge($array1, $array2);
?>

Result
Array
(
    [0] => data
)

You are getting notice because something2, something4 should be quoted as string or $ as variable.

PHP Notice:  Use of undefined constant something2 - assumed 'something2'

edited Nov 17, 2023 at 20:33

CommunityBot

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answered Nov 21, 2013 at 21:25

Yada

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