如何将数据库表的一条记录中增加一个单独的字段来进行json输出

表user有两个字段，id,name
表user_friend有三个字段，id,userid,friendid

        $query = $this->db->query('SELECT * FROM user where id=46' );
        $user=$query->row();
        $queryfriend= $this->db->query('SELECT * FROM user_friend where userid='.$user->id );
        $userfriend=$queryfriend->result();

各位亲，请问下我要输出如下的json格式，应该怎样操作?
{
    "user": {
        "id": "46",
"name": "john",
         "userfriend": [  {
                "id": "13",
                "userId": "46",
                "friendId": "43"
                           },
            {
                "id": "15",
                "userId": "46",
                "friendId": "44"
                         }
]
    }
}

回复讨论(解决方案)

print_r($userfriend);
看看都是什么

print_r($userfriend)结果如下;
Array ( [0] => stdClass Object ( [id] => 13 [userId] => 46 [friendId] => 43 ) [1] => stdClass Object ( [id] => 15 [userId] => 46 [friendId] => 44  ) [2] => stdClass Object ( [id] => 16 [userId] => 46 [friendId] => 45  ) )

userfriend输出数据库查出来对应的[]格式就行了

这样试试

$user['userfriend']=$userfriend;
$data['user']=$user;
echo json_encode($data);

$user['userfriend']=$userfriend;
$data['user']=$user;
echo json_encode($data);
输出如下，不是我想要的格式
{
    "user": {
        "0": {
            "id": "45",
         
            "nickName": "Jerry",
            "gender": "1",
            "birthday": "2000年10月01日",
            "avatar": "upload/user/45/avatar/0434541421051694278.jpg",
        },
        "userfriend": [
            {
                "id": "5",
                "userId": "45",
                "friendId": "3",
                "endTime": "1426745161"
            },
            {
                "id": "7",
                "userId": "45",
                "friendId": "5",
                "endTime": "1426745161"
            },
            {
                "id": "9",
                "userId": "45",
                "friendId": "44",
                "endTime": "1426745161"
            },
            {
                "id": "10",
                "userId": "45",
                "friendId": "46",
                "endTime": "1426745161"
            }
        ]
    }
}

我的想法：
1.拼接json
    1）.循环$user，嵌套循环$userfriend
2.重组数组
    1）.两表联查
    2）.循环构造新数组

额外：
json_encode()注意编码UTF-8

这样就可以了。

$query = $this->db->query('SELECT * FROM user where id=46' );$user=$query->row();$queryfriend= $this->db->query('SELECT * FROM user_friend where userid='.$user->id );$userfriend=$queryfriend->result();$data = array();$data['user']['id'] = $user[0]['id'];$data['user']['name'] = $user[0]['name'];$data['user']['userfriend'] = $userfriend;echo json_encode($data);

先取出所有用户，foreach循环取出每个用户的朋友，在保存到一个数值里面

     正确的代码如下，感谢各位

       $user = $query->result_array();
        $data = $user[0];
        $data['userfriend'] = $userfriend;

       // $result = array('result' => array('status' => '1', 'message' => '登陆成功', 'user' => $data));
        $result = array( 'user' => $data);
        echo json_encode($result);