i cant format json array to desired format

Question

i have json array that i generate from php. However the format is not what i want. i have related questions but it is not the format i want Please see below what i did and tried vs what i want

    //////////////
$data variable array
///////////////

Array
    (
        [0] => Array
            (
                [id] => 2
                [start_date] => 2018-05-17 08:40
                [end_date] => 2018-05-17 09:00
            )

        [1] => Array
            (
                [id] => 3
                [start_date] => 2018-05-17 08:40
                [end_date] => 2018-05-17 09:00
            )
      )

This is what i do

$array[] = $data;
echo json_encode($array);

The output

[
 {"id":2,"start_date":"2018-05-17 08:40","end_date":"2018-05-17 09:00"},
 {"id":3,"start_date":"2018-05-17 08:40","end_date":"2018-05-17 09:00"}
]

This is what i tried

$array["data"] = $data;
echo json_encode($array);

The output

{"data":
 {"id":2,"start_date":"2018-05-17 08:40","end_date":"2018-05-17 09:00"},
 {"id":3,"start_date":"2018-05-17 08:40","end_date":"2018-05-17 09:00"}
}

The output i want

{
  "data": [
             {"id":2,"start_date":"2018-05-17 08:40","end_date":"2018-05-17 09:00"},
             {"id":3,"start_date":"2018-05-17 08:40","end_date":"2018-05-17 09:00"}
          ]
}

i am not following what you are saying. May you provide an example? — user7498826, Aug 16 '18 at 10:29
this is $data: $data['id']=2;$data['start_date']=2018-05-17; and so on @AlexShesterov — user7498826, Aug 16 '18 at 10:30
The 2nd output snippet is wrong then - it won't contain two objects without array (which is invalid json). — Alex Shesterov, Aug 16 '18 at 10:33
$data variable is a php array populated in a loop like so : $data['id']='2'; data['start_date']='2018-05-17 08:40'; $ata['end_date']='2018-05-17 08:40'; @AlexShesterov — user7498826, Aug 16 '18 at 10:33

JamesBond · Answer 1 · 2018-08-16T15:56:22.423

1

instead of using $array["data"] i put "data" into an array an then another array for $data.

$array = array("data" => $data);
$enData = json_encode($array); //encoding array to json
$deData = json_decode($enData, true); //decoding array

print_r($deData);

The above would output something like this:

Array
(
    [data] => Array
        (
            [id] => 3
            [start_date] => 2018-05-17 08:40
            [end_date] => 2018-05-17 09:00
        )

)

and if you do this:

print_r($deData['data']);

//Output

Array
(
    [id] => 3
    [start_date] => 2018-05-17 08:40
    [end_date] => 2018-05-17 09:00
)

edited Aug 16 '18 at 15:56

answered Aug 16 '18 at 11:26

JamesBond

312
2
17

1

Can you share some more context? – Nico Haase Aug 16 '18 at 12:23
@JamesBond: Please edit your answer, include some information about what your code does and how it solves the problem/answers the question. Otherwise, it may be deleted as *Very Low Quality* or attract downvotes – Our Man in Bananas Aug 16 '18 at 15:20
@OurManinBananas Updated – JamesBond Aug 16 '18 at 15:26

score 0 · Answer 2 · answered Aug 16 '18 at 10:29

You need to make $data an array first.

For example:

$data[] = ["id" => 2, "start_date" => "2018-05-17 08:40", "end_date" => "2018-05-17 09:00"];

$array["data"] = $data;
echo json_encode($array);

Which returns:

{
  "data":[
           {"id":2,"start_date":"2018-05-17 08:40","end_date":"2018-05-17 09:00"}
         ]
}

score 0 · Accepted Answer · answered Aug 16 '18 at 10:37

0

you have to build an array first, then construct an object. See below

$array = [
    [
        "id" => 2,
        "start_date" => "2018-05-17 08:40",
        "end_date" => "2018-05-17 09:00"
    ],
    [
        "id" => 3,
        "start_date" => "2018-05-17 08:40",
        "end_date" => "2018-05-17 09:00"
    ]
];
$data["data"] = $array;
echo json_encode($data);

answered Aug 16 '18 at 10:37

NicossB

408
2
13

i have an array, please edited code. i did exactly that but the output is as indicated on the third ouptut – user7498826 Aug 16 '18 at 10:41
This is strange. It looks like your first data variable array is not interpreted as a try array. Can you please var_dump it ? Point is : check this answer, looks similar https://stackoverflow.com/questions/11195692/json-encode-sparse-php-array-as-json-array-not-json-object – NicossB Aug 16 '18 at 10:54

score 0 · Answer 4 · answered Aug 16 '18 at 10:41

It seems that you are getting the $data from loop over sql output:

$array = ["data"=>[]];// initialize $array outside your loop

while($data = mysqli_fetch_assoc($result)){
    $array['data'][] = $data;  // push $data into $array['data']
}

echo json_encode($array);

i cant format json array to desired format

4 Answers4