使用PHP的mysql数据库的JSON响应显示不正确的格式

Question

function getForum($class_divisionId) {
    $qry = "SELECT fm.uid,fm.class_divisionId,fm.username,fm.description,fm.date_time,si.name FROM forum as fm INNER JOIN studentinfo as si ON fm.username = si.username WHERE fm.class_divisionId='".mysql_real_escape_string($class_divisionId)."'";

    $exe = mysql_query($qry);

    if (mysql_num_rows($exe)> 0) {
        $response['error'] = 'false';
        while($result = mysql_fetch_array($exe)){
            $uid = $result['uid'];
            $response['forum'][$uid]['uid'] = $result['uid'];
            $response['forum'][$uid]['class_divisionId'] = $result['class_divisionId'];
            $response['forum'][$uid]['username'] = $result['username'];
            $response['forum'][$uid]['description'] = $result['description'];
            $response['forum'][$uid]['date_time'] = $result['date_time'];
            $response['forum'][$uid]['name'] = $result['name'];  
        }
        return $response;
    } else {
        return false;
    }
}

This is the getforum function in my php file. The response I am getting is:

{
    "error": "false",
    "forum": {
        "4": {
            "uid": "4",
            "class_divisionId": "ABC",
            "username": "XYZ_1",
            "description": "Forum description number 2",
            "date_time": "2015-08-15 10:17:18",
            "name": "XYZ"
        }
   }
}

I would like the response to be in the following format:

{
    "error": "false",
    "forum": [
        {
            "uid": "4",
            "class_divisionId": "ABC",
            "username": "XYZ_1",
            "description": "Forum description number 2",
            "date_time": "2015-08-15 10:17:18",
            "name": "XYZ"
        }
    ]
}

Can someone please advise what I am doing wrong?

Answer 1

use

$response['forum'][$uid]->'uid' = $result['uid'];
$response['forum'][$uid]->'class_divisionId' = $result['class_divisionId'];

This makes the forum value have an array denoted by uid values, for which you can set each entry's data.

Answer 2

It's not possible. Javascript arrays must have sequential key numering, as the [...] syntax doesn't allow you to specify keys. Since you're using ID numbers, which almost certainly won't be sequential or without gaps, the "array" must be encoded as an object.

eg

$foo = array(0 => 'a', 3 => 'b'); // missing keys 1 & 2
echo json_encode($foo);           // {"0":"a","3":"b"}
$bar = array('a', 'b');            // implicit keys 0,1
echo json_encode($bar);           // ["a","b"]
$baz = array(0 => 'a', 1 => 'b')  // explicit keys 0,1
echo json_encode($baz);           // ["a","b"]

As to why: since array [] -shortcut notation cannot specify keys, the json-encoding engine would be forced to specify EVERY SINGLE missing key directly, so consider a degenerate case:

$foo = array(0 => 'a', 999999999 => 'b');

The JSON would be absolutely MASSIVE, with 999,999,998 copies of x:null , just to "fill in the blanks".

Answer 3

Thats because you are not adding as array elements , but you are adding as assoc-array element with keys ,

this will have the output you posted

if (mysql_num_rows($exe)> 0) {
        $response['error'] = 'false';
        $response['forum'] = Array();
        while($result = mysql_fetch_array($exe)){
            $response['forum'][] = Array(
                                        'uid' => $result['uid'],
                                        'class_divisionId' => $result['class_divisionId'],
                                        'username' => $result['username'],
                                        'description' => $result['description'],
                                        'date_time' => $result['date_time'],
                                        'name' => $result['name']
                                        );
        }
        return $response;
    } else {
        return false;
    }