Sql 如何使用分析获取唯一ID的汇总汇总？

我的重复表有重复的ID，但我需要唯一ID的摘要统计信息

 Detail_id   code   book   tree
----------- ------ ------ ------
  1          BR54   COOK   PINE
  1          BR55   COOK   PINE
  1          BR51   COOK   PINE
  2          BR55   COOK   MAPL
  2          BR60   COOK   MAPL
  3          BR61   FORD   PINE
  3          BR54   FORD   PINE
  3          BR55   FORD   PINE

这是我的问题，也是关于

预期结果：

COOK_TOTAL FORD_TOTAL PINE_TOTAL MAPL_TOTL
---------- ---------- ---------- ----------
  2          1          2         1

---

只需删除

else

子句，即可修改查询以获得所需内容：

select count(case detail_book when 'COOK' THEN 1 end) as cook_total,
       count(case detail_book when 'FORD' THEN 1 end) as ford_total,
       count(case detail_tree when 'PINE' THEN 1 end) as pine_total,
       count(case detail_book when 'MAPL' THEN 1 end) as mapl_total
from detail_records;

没有

else

的

case

的默认值为

NULL

，因此

count（）

有效。就个人而言，对于这种类型的聚合，我更喜欢

sum（）

：

select sum(case when detail_book = 'COOK' THEN 1 else 0 end) as cook_total,
       sum(case when detail_book = 'FORD' THEN 1 else 0 end) as ford_total,
       sum(case when detail_tree = 'PINE' THEN 1 else 0 end) as pine_total,
       sum(case when detail_book = 'MAPL' THEN 1 else 0 end) as mapl_total
from detail_records;

---

您可以使用分析函数和内联视图来避免重复计数问题：

select sum(case when detail_book = 'COOK' and book_cntr = 1 then 1 else 0 end) as cook_total,
       sum(case when detail_book = 'FORD' and book_cntr = 1 then 1 else 0 end) as ford_total,
       sum(case when detail_tree = 'PINE' and tree_cntr = 1 then 1 else 0 end) as pine_total,
       sum(case when detail_tree = 'MAPL' and tree_cntr = 1 then 1 else 0 end) as mapl_total
  from (select d.*,
               row_number() over(partition by detail_book, detail_id order by detail_book, detail_id) as book_cntr,
               row_number() over(partition by detail_tree, detail_id order by detail_tree, detail_id) as tree_cntr
          from detail_records d) v

---

Fiddle:

此答案基于您的示例，您可以使用该示例通过获取详细信息簿和详细信息树的不同ID来避免重复ID

请在这里检查结果

---

除了分析函数，我可能会使用一种方法，首先“展平表”（

union all

），然后旋转结果：

select *
  from (
    select   detail_book i
    from     detail_records
    group by detail_id, detail_book
  union all
    select   detail_tree
    from     detail_records
    group by detail_id, detail_tree
)
  pivot(count(i) for i in ('COOK', 'FORD', 'PINE', 'MAPL'))
;

---

我认为这里不需要分析函数：

SELECT COUNT(DISTINCT CASE WHEN detail_book = 'COOK' THEN detail_id END) AS cook_total
     , COUNT(DISTINCT CASE WHEN detail_book = 'FORD' THEN detail_id END) AS ford_total
     , COUNT(DISTINCT CASE WHEN detail_tree = 'PINE' THEN detail_id END) AS pine_total
     , COUNT(DISTINCT CASE WHEN detail_tree = 'MAPL' THEN detail_id END) AS mapl_total
  FROM detail_records;

当值不匹配时，

CASE

语句返回NULL；这些不算在内

---

顺便说一句，在您的查询中，您试图将

detail\u book

匹配到

MAPL

，而我认为您想要匹配

detail\u tree

，我的查询反映了这一点。

似乎这不是用户想要得到的（尽管这是我最初的解释）。我相信他是在为每个条件计算不同的

detail\u id

值，而不是计算每个条件发生的行。是的，贾斯汀，这是正确的。此线程（）中的Fiddle#29生成正确的COOK#u总数。我想将类似的查询捕获到一个查询中。在我的答案中使用了detail\u id。。您可以检查它是否适用于you@zundarz . . . 然后将

detail\u id

添加到

select

和

groupbydetail\u id

。

select *
  from (
    select   detail_book i
    from     detail_records
    group by detail_id, detail_book
  union all
    select   detail_tree
    from     detail_records
    group by detail_id, detail_tree
)
  pivot(count(i) for i in ('COOK', 'FORD', 'PINE', 'MAPL'))
;

select *
from (
select decode(detail_book,'FORD','FORD_TOTAL','COOK','COOK_TOTAL','MAPL','MAPL_TOTAL','PINE','PINE_TOTAL','OTHER') i, 
    count(distinct detail_id) j
from detail_records
group by detail_book
union all
select DECODE(detail_tree,'FORD','FORD_TOTAL','COOK','COOK_TOTAL','MAPL','MAPL_TOTAL','PINE','PINE_TOTAL','OTHER') i, 
    count(distinct detail_id) j
from detail_records
group by detail_tree
)
  pivot(sum(j) for i in ('COOK_TOTAL', 'FORD_TOTAL', 'PINE_TOTAL', 'MAPL_TOTAL','OTHER'))
;

SELECT COUNT(DISTINCT CASE WHEN detail_book = 'COOK' THEN detail_id END) AS cook_total
     , COUNT(DISTINCT CASE WHEN detail_book = 'FORD' THEN detail_id END) AS ford_total
     , COUNT(DISTINCT CASE WHEN detail_tree = 'PINE' THEN detail_id END) AS pine_total
     , COUNT(DISTINCT CASE WHEN detail_tree = 'MAPL' THEN detail_id END) AS mapl_total
  FROM detail_records;