延长给定的长MySQL查询并获得完成和未完成的计数

时间：2016-08-06 15:35:51

标签： mysql

当我在MySQL查询下面写的时候，它给了我正确的输出，就像之后我想要完成并且没有完成计数一样。我已经尝试但失败了。

SQL小提琴：

http://sqlfiddle.com/#!9/c8dab/1

问题：

如何通过扩展以下查询来完成和未完成计数？

查询：

SELECT keyworddefs.name as sprint, 
   SUM(CASE WHEN bugs.bug_status IN ('CLOSED', 'VERIFIED')
             THEN bugs.cf1 + bugs.cf2
             ELSE 0 END) completed,
   SUM(bugs.cf1 + bugs.cf2) total,
   (CASE WHEN SUM(CASE WHEN bugs.bug_status IN ('CLOSED', 'VERIFIED')
             THEN bugs.cf1 + bugs.cf2
             ELSE 0 END)=SUM(bugs.cf1 + bugs.cf2)
             THEN 'Completed'
             ELSE 'Not Completed' END) status
FROM bugs
JOIN keywords ON bugs.bug_id = keywords.bug_id
JOIN keyworddefs ON keyworddefs.id = keywords.keywordid
GROUP BY keywords.keywordid
ORDER BY keyworddefs.name DESC;

输出：

+--------+------------------------+--------------------+---------------+
| name   |      completed         |        total       |    status     |
+--------+------------------------+--------------------+---------------+
| K2     |                    14  |                14  | Completed     |
| J2     |                    16  |                24  | Not Completed |
| J1     |                    0   |                5   | Not Completed |
+--------+------------------------+--------------------+---------------+

扩展查询后，预期输出只是：

+---------------+--------+
|   status      | count  |
+------------------------+
| Completed     | 1      |
| Not Completed | 2      |
+------------------------+

1 个答案:

答案 0 :(得分：1)

正如Strawberry所说，给它一个派生表名。

架构：

-- bugs table1 (master table) :

CREATE TABLE `bugs` 
(   `bug_id` int(11) NOT NULL, 
    `bug_date` date NOT NULL, 
    `cf1` int(11) NOT NULL, 
    `cf2` int(11) NOT NULL, 
    `bug_status` varchar(200) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `bugs` (`bug_id`, `bug_date`, `cf1`, `cf2`, `bug_status`) VALUES 
(101, '2016-07-19', 3, 2, 'RESOLVED'), 
(102, '2016-07-19', 2, 1, 'VERIFIED'), 
(103, '2016-07-19', 2, 1, 'VERIFIED'), 
(103, '2016-07-19', 2, 1, 'VERIFIED'), 
(1363, '2016-07-19', 2, 1, 'VERIFIED'), 
(1352, '2016-07-19', 2, 1, 'VERIFIED'), 
(102, '2016-07-19', 2, 1, 'VERIFIED'), 
(102, '2016-07-22', 2, 2, 'CLOSED'), 
(103, '2016-07-22', 2, 2, 'CLOSED'), 
(103, '2016-07-22', 2, 2, 'CLOSED'), 
(102, '2016-07-19', 3, 2, 'NEW'), 
(102, '2016-07-19', 2, 1, 'REOPENED'), 
(102, '2016-07-19', 2, 1, 'CLOSED'), 
(102, '2016-07-19', 2, 1, 'VERIFIED'), 
(1363, '2016-07-19', 2, 1, 'VERIFIED'), 
(1352, '2016-07-19', 2, 1, 'VERIFIED'), 
(565, '2016-07-19', 2, 1, 'VERIFIED'), 
(398, '2016-07-22', 2, 2, 'CLOSED'), 
(565, '2016-07-22', 2, 2, 'CLOSED'), 
(9872, '2016-07-22', 2, 2, 'CLOSED');

-- keywords table2 (having keyword ids):

CREATE TABLE `keywords` 
(   `bug_id` int(11) NOT NULL, 
    `keywordid` varchar(11) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `keywords` (`bug_id`, `keywordid`) VALUES 
(101, '1'), (102, '2'), (103, '3'), (104, '4'), (105, '1'), (106, '1'), (107, '2'), (108, '3'), (109, '4');

-- keyworddefs table3 (having keyword names according to keywordid):

CREATE TABLE `keyworddefs` 
(   `id` int(11) NOT NULL, 
    `name` varchar(200) NOT NULL,
    `description` varchar(200) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

INSERT INTO `keyworddefs` (`id`, `name`, `description`) VALUES 
(1, 'J1', 'My J1 item'), (2, 'J2', 'My J2 item'), (3, 'K2', 'My K2 item'), (4, 'K2', 'My K2 item');

您的查询：

SELECT keyworddefs.name as sprint,  
       SUM(CASE WHEN bugs.bug_status IN ('CLOSED', 'VERIFIED') 
                 THEN bugs.cf1 + bugs.cf2 
                 ELSE 0 END) completed_story_points, 
       SUM(bugs.cf1 + bugs.cf2) total_story_points, 
       (CASE WHEN SUM(CASE WHEN bugs.bug_status IN ('CLOSED', 'VERIFIED') 
                 THEN bugs.cf1 + bugs.cf2 
                 ELSE 0 END)=SUM(bugs.cf1 + bugs.cf2) 
                 THEN 'Completed' 
                 ELSE 'Not Completed' END) sprint_status 
  FROM bugs 
  JOIN keywords ON bugs.bug_id = keywords.bug_id 
  JOIN keyworddefs ON keyworddefs.id = keywords.keywordid 
 GROUP BY keywords.keywordid 
 ORDER BY keyworddefs.name DESC;

您的输出：

+--------+------------------------+--------------------+---------------+
| sprint | completed_story_points | total_story_points | sprint_status |
+--------+------------------------+--------------------+---------------+
| K2     |                     14 |                 14 | Completed     |
| J2     |                     16 |                 24 | Not Completed |
| J1     |                      0 |                  5 | Not Completed |
+--------+------------------------+--------------------+---------------+

通缉：

+---------------+--------+
|   status      | count  |
+------------------------+
| Completed     | 1      |
| Not Completed | 2      |
+------------------------+

懒惰的方法（意思是，没有多想）：

SELECT sprint_status,count(*) AS count 
FROM 
( 
 SELECT keyworddefs.name as sprint,   
       SUM(CASE WHEN bugs.bug_status IN ('CLOSED', 'VERIFIED') 
                 THEN bugs.cf1 + bugs.cf2 
                 ELSE 0 END) completed_story_points, 
       SUM(bugs.cf1 + bugs.cf2) total_story_points, 
       (CASE WHEN SUM(CASE WHEN bugs.bug_status IN ('CLOSED', 'VERIFIED') 
                 THEN bugs.cf1 + bugs.cf2 
                 ELSE 0 END)=SUM(bugs.cf1 + bugs.cf2) 
                 THEN 'Completed' 
                 ELSE 'Not Completed' END) sprint_status 
  FROM bugs 
  JOIN keywords ON bugs.bug_id = keywords.bug_id 
  JOIN keyworddefs ON keyworddefs.id = keywords.keywordid 
 GROUP BY keywords.keywordid 
 ) xDerived  
 GROUP BY sprint_status 
 ORDER BY sprint_status;

输出：

+---------------+-------+
| sprint_status | count |
+---------------+-------+
| Completed     |     1 |
| Not Completed |     2 |
+---------------+-------+

按什么排序

每个派生表都需要一个名称。上面包裹的块是一个。 所以我们选择了名字xDerived。之后它没有被名称使用（虽然它本来可以）。但是它仍然需要一个名字或者会出现错误。

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