Convert PHP Array to JSON Array of object

Question

I'm using Google Orgchart in my project. In that I'm returning JSON OBJECT from PHP file.

Problem

My Problem is when I hardcode the value, It works fine. When I return data from PHP file. It did not work. I guess the data format which is returning from PHP file is not correct. File below.

$result = mysql_query("SELECT * FROM emp"); 
    while($row = mysql_fetch_array( $result )) {
        $arr1 = array(
            'v' => $row['name'],
            'f' => $row['name']+'<div style="color:red; font-style:italic">President</div>',
            '' => $row['rep'],
            '' => $row['des'],
        );
        array_push($dataarray, $arr1);
    }

echo json_encode($dataarray);

which returns object like below

How it Should be

My hardcorded JSON OBJECT below

   [
      [{v:'Prabhkar', f:'Prabhkar<div style="color:red; font-style:italic">President</div>'},
       '', 'The President'],
      [{v:'Raguram', f:'Raguram<div style="color:red; font-style:italic">GM</div>'},
       'Prabhkar', 'GM']
    ]

Console Screenshot below:

Do I need to create a one more array in PHP file. How I suppose to change the PHP array according to above screenshot. sorry for my english. Thank you.

Show the JSON output you get echo json_encode($dataarray);.. Also, what's stored in $dataarray initially? — Indrasis Datta
– Indrasis Datta, Commented Apr 18, 2017 at 6:28
@ObjectManipulator, I have parsed data from $dataarray and shown result in screenshot res. — Maria Jeysingh Anbu
– Maria Jeysingh Anbu, Commented Apr 18, 2017 at 6:30

RafH · Accepted Answer · 2017-04-18 06:31:21Z

2

You need to wrap 'v' and 'f' in a array, then push other values to parent array.

$result = mysql_query("SELECT * FROM emp"); 
    while($row = mysql_fetch_array( $result )) {
        $arr1 = array(
            array(
                'v' => $row['name'],
                'f' => $row['name'] . '<div style="color:red; font-style:italic">President</div>'
            ),
            $row['rep'],
            $row['des']
        );
        array_push($dataarray, $arr1);
    }

echo json_encode($dataarray);

answered Apr 18, 2017 at 6:31

RafH

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Comments

Zenel Rrushi · Accepted Answer · 2017-04-18 06:32:42Z

1

In your hard coded array the first key has an array inside so you must change your code like this

$result = mysql_query("SELECT * FROM emp"); 
$dataarray = [];
while($row = mysql_fetch_array( $result )) {
    $arr1 = array(
        array(
              'v'=> $row['name'], 
              'f' => $row['name'].'<div style="color:red; font-style:italic">President</div>',),
        $row['rep'],
        $row['des'],
    );
    array_push($dataarray, $arr1);
}

echo json_encode($dataarray);

answered Apr 18, 2017 at 6:32

Zenel Rrushi

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Comments

justderb · Accepted Answer · 2017-04-18 06:35:52Z

1

Your internal structure is wrong. Your internal structure is an array, with the first being a map, followed by two values. Your current implementation is an array, with only a map.

$result = mysql_query("SELECT * FROM emp"); 
    while($row = mysql_fetch_array( $result )) {
        $arr1 = array(
            array(
                'v' => $row['name'],
                'f' => $row['name'] . '<div style="color:red; font-style:italic">President</div>',
            ),
            $row['rep'],
            $row['des']);
        array_push($dataarray, $arr1);
    }

echo json_encode($dataarray);

edited Apr 18, 2017 at 6:35

answered Apr 18, 2017 at 6:30

justderb

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