How to add values to an array inside an array obtained through AJAX

Question

In my PHP file, I'm receiving a total of 4 variables $data, $date, $shift and $val1. $data is an array and the other 3 are date and 2 strings obtained through AJAX with no problem. What I'm trying to do is to insert these 3 values inside my $data variable.

I tried using array merge, and a For each loop with multiple instances but no luck so far.

I obtained my variables like this:

if (isset($_POST['date'])){
    $date = $_POST['date'];
    $date = json_encode($date);
    $date = json_decode($date);
}

if (isset($_POST['shift'])){
    $shift = $_POST['shift'];
    $shift = json_encode($shift);
    $shift = json_decode($shift);
}

if (isset($_POST['val1'])){
    $val1 = $_POST['val1'];
    $val1 = json_encode($val1);
    $val1 = json_decode($val1);
}

if (isset($_POST['data'])){
    $dat = $_POST['data'];
    $data = json_decode($dat, true);
}

$values = array($date,$shift,$val1);

$r = (array_merge($data, $values));

My data array looks something like this:

Array (
    [0] => Array ( 
        [data] => Array ( 
            [0] => Array (
                [0] => 1
                [1] => 2
                [2] => 3 
                [3] => 0 
                [4] => Mat1 
                [5] => Box1 
                [6] => 100 
                [7] => 100 
                [8] => Piece1
                [9] => Loc1 
                [10] => Mach1 
                [11] => 1000
                [12] => Accepted 
                )  
            ) 
        ) 
    [1] => 2019-04-09 
    [2] => First
    [3] =>  Value1
)

But what I want to achieve is this:

Array (
    [0] => Array ( 
        [data] => Array ( 
            [0] => Array (
                [0] => 1
                [1] => 2
                [2] => 3 
                [3] => 0 
                [4] => Mat1 
                [5] => Box1 
                [6] => 100 
                [7] => 100 
                [8] => Piece 1
                [9] => Suc1 
                [10] => Mach1 
                [11] => 1000
                [12] => Accepted
                [13] => 2019-04-09 
                [14] => First
                [15] => Value1
                )  
            ) 
        ) 
    )

What am I doing wrong? Or How can I achieve what I'm trying to do?

Edit: Since I can get more than one array at my array, something like this

Array (
    [0] => Array ( 
        [data] => Array ( 
            [0] => Array (...)
            [1] => Array (...)
            [2] => Array (...)
            [3] => Array (...)  
            ) 
        ) 
    )

I just added this code to @HelgeB answer, I'm leaving it here in case someone might need it in the future.

$count = count($data[0]['data']);

for ($i=0; $i < $count ; $i++) { 
  $data[0]['data'][$i][] = $date;
  $data[0]['data'][$i][] = $shift;
  $data[0]['data'][$i][] = $val1;
}

Would something like $_POST['data'][0][] = $_POST['date'] before assigning $r do what you're looking for? — Daniel G
– Daniel G, Commented Apr 17, 2019 at 21:23
I've tried that and it just gives me an offset warning or it just adds another parentheses to the result — carlos.p
– carlos.p, Commented Apr 18, 2019 at 2:31

HelgeB · Accepted Answer · 2019-04-18 07:15:12Z

2

As far as I can see from your merged output, your $data array structure is $data[0]['data'][0] = [1,2,3,...,'Accepted']. So in my opinion you need to insert the values exactly on the level $data[0]['data'][0] to obtain your result.

The simplest way to achieve this would be:

$data[0]['data'][0][] = $date;
$data[0]['data'][0][] = $shift;
$data[0]['data'][0][] = $val1;

If you want to use your merge approach you need to merge on the correct level like this:

$r = [0 => ['data' => [0 => (array_merge($data[0]['data'][0], $values))]]];

answered Apr 18, 2019 at 7:15

HelgeB

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1 Comment

carlos.p Over a year ago

Well this worked like a charm! Thanks! As a side note, I can also get more than 1 variable at $data[0]['data'][0], so I just added a count and a for loop to add it to every other array. I will edit my question to add that in case someone needs it in the future.

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