如何在帶有2個表的MySQL中編寫計算查詢？

Question

我有2湯匙如下。

請求tbl

id | request | created
1  | asdf    | 2013-07-04 14:39:03
2  | qwer    | 2013-07-10 12:06:37

歷史記錄

id | request_id | status       | date
1  |     1      | Pending      | 2013-07-04 14:39:03
2  |     1      | Reviewing    | 2013-07-05 01:10:14
3  |     1      | Implementing | 2013-07-06 11:25:54
4  |     1      | Completed    | 2013-07-07 12:36:32
5  |     2      | Pending      | 2013-07-10 15:05:56
6  |     2      | Reviewing    | 2013-07-11 03:08:04
7  |     2      | Implementing | 2013-07-13 11:45:48
8  |     2      | Completed    | 2013-07-17 14:28:15

我想在2 tbls以上顯示如下

Request | Reviewing Time | Implementing Time 
asdf    |      0         |      0            
qwer    |      1         |      2

具有request_id = 1的示例的理論是

審核到實施=（2013-08-06）-（2013-08-05）= 1天

待審核=（2013-08-05）-（2013-08-04）= 1天

審核時間 =（正在審核實施）-（正在審核）= 0天

審核到實施=（2013-08-06）-（2013-08-05）= 1天

實施到完成=（2013-08-07）-（2013-08-06）= 1天

實施時間 =（實施為強制實施）-（審查為實施）= 0

Answer 1

這是解決問題的長途之路...

 CREATE TABLE my_table 
 (id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
 ,request_id INT NOT NULL
 ,status       VARCHAR(20) NOT NULL
 ,date DATETIME NOT NULL
 ,UNIQUE(request_id,status)
 );

 INSERT INTO my_table VALUES
 (1  ,1      ,'Pending','2013-07-04 14:39:03'),
 (2  ,     1      ,'Reviewing','2013-07-05 01:10:14'),
 (3  ,     1      ,'Implementing','2013-07-06 11:25:54'),
 (4  ,     1     ,'Completed','2013-07-07 12:36:32'),
 (5  ,     2     ,'Pending','2013-07-10 15:05:56'),
 (6  ,     2     ,'Reviewing','2013-07-11 03:08:04'),
 (7  ,     2     ,'Implementing','2013-07-13 11:45:48'),
 (8  ,     2     ,'Completed','2013-07-17 14:28:15');

 SELECT request_id
      , DATEDIFF(implementing,reviewing) - DATEDIFF(reviewing,pending) rT
      , DATEDIFF(completed,implementing) - DATEDIFF(implementing,reviewing) iT
 FROM (
 SELECT x.request_id
      , MAX(CASE WHEN status = 'pending' THEN date END) pending 
      , MAX(CASE WHEN status = 'reviewing' THEN date END) reviewing
      , MAX(CASE WHEN status = 'implementing' THEN date END) implementing
      , MAX(CASE WHEN status = 'completed' THEN date END) completed
   FROM my_table x
  GROUP 
     BY request_id
     ) a;

 +------------+------+------+
 | request_id | rT   | iT   |
 +------------+------+------+
 |          1 |    0 |    0 |
 |          2 |    1 |    2 |
 +------------+------+------+

相同的sqlfiddle： http ：//www.sqlfiddle.com/#！2 / fc6db / 1

Answer 2

不確定我是否理解您的需求，但這是我從您的問題中讀出的內容的示例。 在示例中，我使用嵌套查詢來獲取每個請求的計算結果。 DATEDIFF以天為單位給出2個日期的差值。 我使用ABS，是因為我假設您不想要負數，而且我不確定這些日期是否始終為正數。 那些示例中的示例不帶任何絕對含義。

SELECT calc.request, ABS(calc.`reviewing to implementing`-calc.`pending to reviewing`) AS 'Reviewing Time', ABS(calc.`implementing to completed`-calc.`reviewing to implementing`) AS 'Implementing Time'
    FROM (
        SELECT t1.request,
        (
            SELECT ABS(DATEDIFF((SELECT `date` FROM history_tbl WHERE request_id = t1.id AND `status` = 'Reviewing'), (SELECT `date` FROM history_tbl WHERE request_id = t1.id AND `status` = 'Implementing')))
        ) AS 'reviewing to implementing',
        (
            SELECT ABS(DATEDIFF((SELECT `date` FROM history_tbl WHERE request_id = t1.id AND `status` = 'Pending'), (SELECT `date` FROM history_tbl WHERE request_id = t1.id AND `status` = 'Reviewing')))
        ) AS 'pending to reviewing',
        (
            SELECT ABS(DATEDIFF((SELECT `date` FROM history_tbl WHERE request_id = t1.id AND `status` = 'Implementing'), (SELECT `date` FROM history_tbl WHERE request_id = t1.id AND `status` = 'Completed')))
        ) AS 'implementing to completed'
        FROM request_tbl as t1
    ) AS calc

小提琴