Question

我有一个这样的表：

date       | page
-----      |-----
2018-01-01 | good
2018-01-01 | good
2018-01-01 | good
2018-01-01 | bad
2018-01-02 | good

如何按日期按页面列中的值进行组织：

date       | good | bad
-----      |------|----
2018-01-01 | 3    | 1
2018-01-02 | 1    | 0

Answer 1

您需要条件聚合：

select date, 
       sum(case when page = 'good' then 1 else 0 end) as good, 
       sum(case when page = 'bad' then 1 else 0 end) as bad
from table t
group by date;

但是，MySQL具有以下简称：

select date, 
       sum( page = 'good') as good, 
       sum( page = 'bad' ) as bad
from table t
group by date;

Answer 2

这可以工作，只需在sum函数中添加一个case：

select date, 
       sum( case when page = 'good' then 1 else 0 end) as good, 
       sum( case when page = 'bad' then 1 else 0 end ) as bad 
from table t 
group by date

Answer 3

将pivot表用作

 create table #pivot(date date,page varchar(20))
 insert into #pivot values
 ('2018-01-01','good')
 ,('2018-01-01','good')
 ,('2018-01-01','good')
 ,('2018-01-01','bad')
 ,('2018-01-02','good')

 select date,good,bad from(  
 select date, page from #pivot 
 ) f
 pivot
( count(page) for page in (good,bad)
) p

Answer 4

我不确定您使用的平台是什么，但是如果是在Oracle中，它将执行相同的操作。

select date, sum(decode(page,'good',1,0)) as good, sum(decode(page,'bad',1,0)) as bad              
from table t                                                                       
group by date;

SQL枢轴列名称和按日期组织

4 个答案: