2017-02-25 148 views
1

我有一個表的類型是日期時間的列表。我只是想用兩個查詢,這是爲日期時間的日期時間的日期時間鑄造日期不反映在輸出

  1. select cast(attendance_student.created_date as date) from attendance_student
  2. select date(attendance_student.created_date) from attendance_student

它轉換爲日期,但我越來越輸出

7/16/2015 12:00:00 AM 
7/16/2015 12:00:00 AM 
7/16/2015 12:00:00 AM 
7/16/2015 12:00:00 AM 
7/16/2015 12:00:00 AM 
7/16/2015 12:00:00 AM 
7/16/2015 12:00:00 AM 
7/16/2015 12:00:00 AM 

爲什麼既能時代我沒有得到日期格式的輸出。

這裏是我的表腳本

CREATE TABLE `attendance_student` (
    `id` int(11) NOT NULL AUTO_INCREMENT, 
    `attendance_id` int(11) DEFAULT NULL, 
    `is_present` tinyint(1) DEFAULT '1', 
    `created_by` varchar(50) DEFAULT NULL, 
    `created_date` datetime DEFAULT NULL, 
    `modified_by` varchar(50) DEFAULT NULL, 
    `last_modified_date` datetime DEFAULT NULL, 
    `is_on_leave` tinyint(1) DEFAULT NULL, 
    `student_id` int(11) DEFAULT NULL, 
    `is_cocurricular_leave` tinyint(1) DEFAULT '0', 
    `is_activity_attendance` tinyint(1) DEFAULT '0', 
    `activity_id` int(11) DEFAULT NULL, 
    `is_sms_sent` tinyint(2) DEFAULT NULL, 
    PRIMARY KEY (`id`), 
    KEY `idx_attendance_attendance_student` (`attendance_id`) USING BTREE, 
    KEY `id_student_attendance_student` (`student_id`) USING BTREE, 
    KEY `fk_activity_student_attendance_id` (`activity_id`) 
) ENGINE=InnoDB DEFAULT CHARSET=latin1; 

回答

1

這是預期,因爲DATE函數返回日期相同,隨着時間定日期時間爲00:00或12小時制12:00 AM。

你可以使用DATE_FORMAT得到公正的日期部分的字符串:

select date_format(created_date, '%Y-%m-%d') as created_date 
from attendance_student