PHP json_encode return rows as arrays instead of objects

Question

From my experience, when I json_encode data from a mysql table, the output is an object containing each row as an object within. What I want to do is to get each row as an array, like the example below (don't need the column names in there). How can I do that?

{ "Data": [
    ["1","Internet Explorer 4.0","Win 95+","4","X","1"],
    ["2","Internet Explorer 5.0","Win 95+","5","C","2"],
    ["3","Internet Explorer 5.5","Win 95+","5.5","A","3"],
    ["4","Internet Explorer 6","Win 98+","6","A","4"],
] }

Felix Kling · Accepted Answer · 2011-07-10 22:26:34Z

4

Use mysql_fetch_row ^[docs] to get each row. This will get a row as a numerical array and those are also encoded as JSON arrays:

$data = array();

while(($row = mysql_fetch_row($result))) {
    $data[] = $row;
}

echo json_encode(array('Data' => $data));

answered Jul 10, 2011 at 22:26

Felix Kling

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3 Comments

ron tornambe Over a year ago

Where is $result defined?

FlipNovid Over a year ago

Where is $result defined? [2]

Felix Kling Over a year ago

@FlipNovid: $result is the result of running a query. If you look at the documentation I linked to you can see an example.

dobriy_santa · Accepted Answer · 2011-07-10 22:39:43Z

0

NOT CHECKED in interpreter: You can change the type of variable by simple types convertion:

$data = DB::get('SELECT * FROM `table` WHERE 1');
echo json_encode(array('Data' => (array)$data));

Also you can use next function: array mysql_fetch_array ( resource $result [, int $result_type ] )

edited Jul 10, 2011 at 22:39

answered Jul 10, 2011 at 22:34

dobriy_santa

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1 Comment

dobriy_santa Over a year ago

DB abstraction. You can use it as $data = mysql_fetch_array($mysql_result, MYSQL_NUM); Just convert object to array. It must help.

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