从SQL中的多个表中计数记录

Question

我想从数据库表中计算文档的不同状态。

假设表中的一个具有：

select docid, year, type, statusid 
from table1 
where docid in (1, 2) and year = '2016' 

这将返回：

docid year type statusid 
------------------------ 
1  2016 pdf 231 
1  2016 pdf 231 
1  2016 pdf 231 
1  2016 pdf 232 
1  2016 pdf 232 
1  2016 pdf 235 
1  2016 pdf 235 

，但我需要回到只有一个记录像

docid year type granted revoked deleted others 
---------------------------------------------------- 
1  2016 pdf 3  2  2  0 

而状态在不同的表

表2

statusid  status  masterid 
---------------------------------- 
231   granted  51 
232   revoked  51 
235   deleted  51 
236   others  51 

我想的是：

select 
    docid, year, type,statusid 
    case 
     when statusid = 231 
      then count(statusid) 
     else 0 
    as granted, 
    case 
     when statusid = 232 
      then count(statusid) 
     else 0 as revoked, 
    case 
     when statusid = 235 
      then count(statusid) 
     else 0 as deleted, 
    case 
     when statusid = 236 
      then count(statusid) 
     else 0 as others 
from 
    table1 
where 
    docid in (1, 2) and year = '2016' 
group by 
    docid, year, type, statusid 

但这回3行而输出应该是唯一一个所有状态数列。

Answer 1

你就近了。该case需要在参数的聚合函数，你不想statusid在group by：

select docid, year, type, 
     sum(case when statusid = 231 then 1 else 0 end) as granted, 
     sum(case when statusid = 232 then 1 else 0 end) as revoked, 
     sum(case when statusid = 235 then 1 else 0 end) as deleted, 
     sum(case when statusid = 236 then 1 else 0 end) as others 
from table1 
where docid in (1,2) and year='2016' 
group by docid, year, type 

Answer 2

可以使用旋转为同一如下：如下

Select * from (
    Select docid, [year], [type], [status] from table1 d 
    Join table2 dt on d.statusid = dt.statusid 
) a 
Pivot (count([status]) for [status] in ([granted],[revoked],[deleted],[others])) p 

输出：

+-------+------+------+---------+---------+---------+--------+ 
| docid | year | type | granted | revoked | deleted | others | 
+-------+------+------+---------+---------+---------+--------+ 
|  1 | 2016 | pdf |  3 |  2 |  2 |  0 | 
+-------+------+------+---------+---------+---------+--------+