MySQL results to JSON with PHP

Question

I've seen lots of similar question on here, but I haven't been able to get my desired output. Could someone please help me out? How can generate the following JSON structure? My query is fine, I just can't figure out how to loop through and get this. Thanks

DESIRED OUTPUT

{
    "players": [
        {
            "id": 271,
            "fname": "Carlos",
            "lname": "Beltran",
            "position": "OF",
            "stats": [
                {
                    "year": "2010",
                    "hr": 32,
                    "rbi": 99,
                    "team": "NYM"
                },
                {
                    "year": "2011",
                    "hr": 35,
                    "rbi": 100,
                    "team": "STL"
                }, 
                {
                ............
                }
            ]
        },
        {
          ........
        }
    ]
}

CURRENT OUTPUT

{"0":{"cbs_id":"18817","fname":"Carlos","lname":"Beltran"},"stats":[{"year":"2007","hr":"33","rbi":"112"}]}

{"0":{"cbs_id":"174661","fname":"Willie","lname":"Bloomquist"},"stats":[{"year":"2007","hr":"2","rbi":"13"}]}

{"0":{"cbs_id":"1208693","fname":"Brennan","lname":"Boesch"},"stats":[{"year":"2010","hr":"14","rbi":"67"}]}

Which is generated with: (I know I'm way off)

if ($result = mysqli_query($link, $sql)) {

        $player = array();

        while ($row = $result->fetch_assoc())
        {

            $player[] = array (
            'cbs_id' => $row['cbs_id'],
            'fname' => $row['fname'],
            'lname' => $row['lname']
            );

            $player['stats'][] = array(
                'year' => $row['year'],
                'hr' => $row['hr'],
                'rbi' => $row['rbi']
            );

        }

        $json = json_encode($player);
        echo "<pre>$json</pre>";

        mysqli_free_result($result);
    }
}

NOTE: Each player can have more than one "stats" record (year, hr, rbi, etc)

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — Madara's Ghost
– Madara's Ghost, Commented Jul 21, 2012 at 20:33

sdhh · Accepted Answer · 2012-07-21 21:06:35Z

This may give what you want:

$players = array();

while ($row = $result->fetch_assoc())
{
  $id = (int)$row['cbs_id'];

  if ( ! isset($players[$id]))
  {
    // New player, add to $players array.
    // For the moment index players by ID so stats can be easily added
    // to an existing player. Without indexing (using $players[] = ...),
    // the same player would be added for each stats record related to
    // him.
    $players[$id] = array(
      'id'       => $id,
      'fname'    => $row['fname'],
      'lname'    => $row['lname'],
      'stats'    => array()
    );
  }

  // Add the stats
  $players[$id]['stats'][] = array(
    'year' => (int)$row['year'],
    'hr'   => (int)$row['hr'],
    'rbi'  => (int)$row['rbi']
  );
}

// Players are indexed by their ID in $players but need to be contained in
// a JSON array, so use array_values() to remove indices, e.g. convert
//
//     array(
//       271 => array('id' => 271, ...),
//       ...
//     )
//
// to
//
//     array(
//       array('id' => 271, ...),
//       ...
//     )
//
$data = array('players' => array_values($players));
$json = json_encode($data);

Eventually you might leave out the integer casts and let PHP automatically convert stringified numeric data (that's the default behaviour with MySQL query results) back to PHP numbers by using the MYSQLI_OPT_INT_AND_FLOAT_NATIVE mysqli connection option as described in example #5 of this page. Note that it requires the mysqlnd library to be used by PHP (you can check that with phpinfo).

alexn · Accepted Answer · 2012-07-21 18:48:00Z

2

You need an associative array. Something like this should get you started:

$data = array(
    'players' => array(
        array(
           'id' => 271,
           'fname': 'Carlos',
           'lname': 'Beltran',
           'position': 'OF',
           'stats' => array(
               array(
                   'year' => 2010,
                   'hr': 32,
                   'rbi': 99,
                   'team': 'NYM'
               ),
               ...
           )
        ),
        ...
    )
);

To encode it into json, use json_encode:

$json = json_encode($data);

answered Jul 21, 2012 at 18:48

alexn

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3 Comments

relyt Over a year ago

I guess the problem is that I'm having issues building the array correctly

relyt Over a year ago

I updated my question. I know I'm way off, but I've been staring at this for way too long.

Cole Tobin Over a year ago

WHy not take the JSON that you have (the 1 line thing), and just use that?

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