Question

我有两张桌子。一个包含应用程序列表。另一个包含每周与他们相关的计数。现在我想得到应用程序名称和本周以及之前的计数。让我解释一下。

应用程式：

+----+-------------+
| id | name        |
+----+-------------+
| 1  | Office 2007 |
+----+-------------+
| 2  | Office 2010 |
+----+-------------+
| 3  | Office 2013 |
+----+-------------+

数：

+----+--------+-------+------------+
| id | app_id | count | date       |
+----+--------+-------+------------+
| 1  | 1      | 200   | 2016-01-11 |
+----+--------+-------+------------+
| 2  | 2      | 500   | 2016-01-11 |
+----+--------+-------+------------+
| 3  | 3      | 750   | 2016-01-11 |
+----+--------+-------+------------+
| 4  | 1      | 180   | 2016-01-18 |
+----+--------+-------+------------+
| 5  | 2      | 378   | 2016-01-18 |
+----+--------+-------+------------+
| 6  | 3      | 1000  | 2016-01-18 |
+----+--------+-------+------------+

这是我需要的结果。我需要所有应用程序，包括本周和之前的计数：

+-------------+-----------------+-----------------+
| app         | count_this_week | count_prev_week |
+-------------+-----------------+-----------------+
| Office 2007 | 180             | 200             |
+-------------+-----------------+-----------------+
| Office 2010 | 378             | 500             |
+-------------+-----------------+-----------------+
| Office 2013 | 1000            | 750             |
+-------------+-----------------+-----------------+

每周运行一个填充count表的脚本。现在我需要每周收到一份报告。老实说我有点迷失，因为我不知道如何宣布列的条件。

Answer 1

您可以尝试先按DATEPART(WEEK,C.date),name分组，然后使用另一个GROUP BY将计数分成2列。像这样的东西

修改

如果每个应用每周只有1条记录，则只能使用一个group by这样的记录。

SELECT appname, SUM(CASE WHEN weekno = 0 THEN sumcount ELSE 0 END) as thisweek, SUM(CASE WHEN weekno = 1 THEN sumcount ELSE 0 END) as lastweek FROM ( SELECT DATEPART(WEEK,CURRENT_TIMESTAMP) - DATEPART(WEEK,C.date) as weekno, name as appname, count as sumcount FROM App A INNER JOIN CountTable C ON A.[id] = C.[app_id] WHERE DATEPART(WEEK,C.date) BETWEEN DATEPART(WEEK,CURRENT_TIMESTAMP) - 1 AND DATEPART(WEEK,CURRENT_TIMESTAMP) )T GROUP BY appname

<强>查询

SELECT appname, SUM(CASE WHEN weekno = 0 THEN sumcount ELSE 0 END) as thisweek, SUM(CASE WHEN weekno = 1 THEN sumcount ELSE 0 END) as lastweek FROM ( SELECT DATEPART(WEEK,CURRENT_TIMESTAMP) - DATEPART(WEEK,C.date) as weekno, name as appname, SUM(count) as sumcount FROM App A INNER JOIN CountTable C ON A.[id] = C.[app_id] WHERE DATEPART(WEEK,C.date) BETWEEN DATEPART(WEEK,CURRENT_TIMESTAMP) - 1 AND DATEPART(WEEK,CURRENT_TIMESTAMP) GROUP BY DATEPART(WEEK,C.date),name ) AS T GROUP BY appname

<强> SQL Fiddle

<强>输出

| appname | thisweek | lastweek | |-------------|----------|----------| | Office 2007 | 180 | 200 | | Office 2010 | 378 | 500 | | Office 2013 | 1000 | 750 |

Answer 2

您可以将此通用查询与当前工作日的变量一起使用：

DECLARE @week date = '2016-01-18';

WITH data AS (
    SELECT a.name, c.[count] 
        , w = CASE WHEN c.[date] = @week THEN 0 ELSE 1 END
    FROM @Counts c
    INNER JOIN @Apps a ON c.app_id = a.id
    WHERE [date] = @week OR [date] = DATEADD(day, -7, @week)
)
SELECT App = name, count_this_week = [0], count_prev_week = [1]
FROM data d
PIVOT (
    MAX([count])
    FOR w IN ([0], [1])
) p

<强>输出：

App         count_this_week count_prev_week
Office 2007 180             200
Office 2010 378             500
Office 2013 1000            750

您的数据：

DECLARE @Apps TABLE ([id] int, [name] varchar(11));
DECLARE @Counts TABLE([id] int, [app_id] int, [count] int, [date] date);

INSERT INTO @Apps([id], [name])
VALUES
    (1, 'Office 2007'),
    (2, 'Office 2010'),
    (3, 'Office 2013')
;

INSERT INTO @Counts([id], [app_id], [count], [date])
VALUES
    (1, 1, 200, '2016-01-11'),
    (2, 2, 500, '2016-01-11'),
    (3, 3, 750, '2016-01-11'),
    (4, 1, 180, '2016-01-18'),
    (5, 2, 378, '2016-01-18'),
    (6, 3, 1000, '2016-01-18')
;

Answer 3

SELECT * 
FROM count 
JOIN app ON app.id=count.app_id
WHERE date BETWEEN '2016-01-18' AND '2016-01-11'

按日期获取计数列表

3 个答案: