Question

鉴于以下数据：

CREATE TABLE Bugs
    (`ID` int, `BugDate` DATE, `resolved` int)
;

INSERT INTO Bugs
    (`ID`, `BugDate`, `Resolved`)
VALUES
    (1, '2014-01-01', 1),
    (2, '2014-01-01', 0),
    (3, '2014-01-10', 1),
    (4, '2014-01-10', 1),
    (5, '2014-01-31', 0),
    (6, '2014-01-31', 0)
;

我需要输出如下：

weeks old  Resolved  Unresolved  total  Rate
---------  --------  ----------  -----  ----
    1        1            1        2     50%
    2        2            0        2    100%
    3        0            2        0    100%

我正在查看问题列表，我想查看每周的总数和每周解决的数字。我已经接受了，我可能需要计算前端的总列数和速率列，而不是让MySQL做到这一点。

我尝试了以下内容，但它并没有完全实现：

SELECT
  WEEKOFYEAR(NOW())-WEEKOFYEAR(BugDate) as WeeksOld,
  COUNT(Resolved) as NumberResolved,
  ( /* get the total number of issues for this time period */
    SELECT COUNT(Resolved)
    FROM Bugs
    WHERE WEEKOFYEAR(NOW())-WEEKOFYEAR(BugDate) = WeeksOld
  ) - COUNT(Resolved) as Unresolved
FROM Bugs
GROUP BY WEEKOFYEAR(NOW())-WEEKOFYEAR(BugDate), Resolved
ORDER BY WeeksOld DESC;

我是如此接近，但问题是，我已经连续几周获得双栏，其中既有已解决的问题也有未解决的问题。

我错过了什么？有更好的方法吗？

这是一个小提琴：http://sqlfiddle.com/#!2/eeb34/12

Answer 1

请勿按已解决的分组 - 最后两行应该是

GROUP BY WEEKOFYEAR(NOW())-WEEKOFYEAR(BugDate)
ORDER BY WeeksOld DESC;

Answer 2

您只想要条件聚合：

SELECT WEEKOFYEAR(NOW())-WEEKOFYEAR(BugDate) as WeeksOld,
       SUM(Resolved) as NumberResolved,
       SUM(1 - Resolved) as Unresolved,
       concat(format(AVG(Resolved)*100, 1), '%') as Rate
FROM Bugs
GROUP BY WEEKOFYEAR(NOW())-WEEKOFYEAR(BugDate)
ORDER BY WeeksOld DESC;

请注意，这会从Resolved中删除group by。它会将resolved的计算更改为使用sum()而不是count()。它添加了Rate列。它改变了Unresolved的计算。你可以把后者写成：

count(*) - sum(Resolved)

MySQL - 通过导致重复行分组

2 个答案: