Question

我有两个表，第一个表有一些数据，第二个表第二个，所有数据现在都保留在联接查询中，并显示在json响应中null值如何替换空值我正在使用邮递员，在json字段中更新鲜。

如果有数据，则数据将像这些字段一样显示

“ athlete_attendance_id”：“ 48”，“ coach_id”：“ 302”，“ athlete_id”：“ 380”，“ athlete_attendance”：“ 1”

{"responseCode":200,"responseMessage":"Athlete details Successfully display","data":[{"user_id":"380","athlete_attendance_id":"48","coach_id":"302","athlete_id":"380","athlete_attendance":"1"}]}

如果表1st中没有数据然后将空值替换为“ 0”，以显示邮递员的响应

“ athlete_attendance_id”：“ null”，“ coach_id”：“ null”，“ athlete_id”：“ null”，“ athlete_attendance”：“ null”

{
    "responseCode": 200,
    "responseMessage": "Athlete details Successfully display",
    "data": [
        {
            "user_id": "377",
            "athlete_attendance_id": null,
            "coach_id": null,
            "athlete_id": null,
            "athlete_attendance": null
        }
]
}

我想要这样的响应

{
        "responseCode": 200,
        "responseMessage": "Athlete details Successfully display",
        "data": [
            {
                "user_id": "377",
                "athlete_attendance_id": 0,
                "coach_id": 0,
                "athlete_id": 0,
                "athlete_attendance": 0
            }
    ]
    }

这是模特

            public function showAthleteData($team_id2,$coach_id2){
 $this->db->select('user.*,team.team_id,teams_athlete.team_id,dev_athlete_attendance.*');
          $table = array('user');
          $this->db->from($table);
                 $this->db->join('teams_athlete', 'user.user_id=teams_athlete.user_id');
                $this->db->join('dev_athlete_attendance' ,'dev_athlete_attendance.athlete_id = dev_teams_athlete.user_id','left' );
              $this->db->join('team','team.team_id = teams_athlete.team_id');
                $this->db->where('team.user_id',$coach_id2);

                $result = $this->db->get();
               if($result->num_rows() > 0 ){
                      return $result->result_array();
                                 }else{
                                     return 0;
                                }
 }

控制器

$team_id2= $this->input->post('team_id');
         $coach_id2= $this->input->post('coach_id'); //coach_id

         $userCount['result'] = $userCount1 = $this->Querydata->showAthleteData($team_id2,$coach_id2);
                    if($userCount['result']>0){


                         $data_arr1 = array(
                        "responseCode" =>  $this->res = 200,
                         "responseMessage" =>  $this->login = 'Athlete details Successfully display',
                        "data" =>$userCount['result']);
                       echo json_encode($data_arr1);

Answer 1

使用IFNULL检查您的列是否为空，然后将默认值设置为0

例如：

   select IFNULL(coach_id, '0') AS coach_id,     
    IFNULL(athlete_id, '0') AS athlete_id,     
    IFNULL(athlete_attendance, '0') AS athlete_attendance ....

编辑：

SELECT dev_user.*, IFNULL(team.team_id, 0) as team_id1,
 IFNULL(teams_athlete.team_id, 0) as team_id2,
IFNULL(dev_athlete_attendance.coach_id,0) as coach_id,
IFNULL(dev_athlete_attendance.athlete_id,0) as athlete_id,
IFNULL(dev_athlete_attendance.athlete_attendance,0) as athlete_attendance 
FROM dev_user JOIN dev_teams_athlete ON dev_user.user_id=dev_teams_athlete.user_id 
LEFT JOIN dev_athlete_attendance ON dev_athlete_attendance.athlete_id = dev_teams_athlete.user_id 
JOIN dev_team ON dev_team.team_id = dev_teams_athlete.team_id 
WHERE dev_team.user_id = '301'

我有两个表第一个表有一些数据另一个表第二个有所有数据现在联接查询空值显示在json响应中

1 个答案: