如何使用分析获取唯一 ID 的汇总总数？答案

【问题标题】：How to get summary totals on unique IDs using analytics?如何使用分析获取唯一 ID 的汇总总数？
【发布时间】：2015-03-29 12:10:58
【问题描述】：

我的重复表有重复的 ID，但我想要关于唯一 ID 的汇总统计信息。

 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

这是我在SQLFiddle上的查询

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

想要的结果：

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

【问题讨论】：

标签： sql oracle distinct analytics

【解决方案1】：

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

小提琴： http://sqlfiddle.com/#!4/889a8/31/0

【讨论】：

【解决方案2】：

我认为您在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;

CASE 语句在值不匹配时返回 NULL；那些不算在内。

Updated SQL Fiddle here. 顺便说一句，在您的查询中，当我认为您想匹配 detail_tree 时，您尝试将 detail_book 匹配到 MAPL，而我的查询反映了这一点。

【讨论】：

【解决方案3】：

此答案基于您的示例 http://sqlfiddle.com/#!4/889a8/29，您可以使用该示例通过获取 detail_book 和 detail_tree 的不同 ID 来避免重复 ID

请在此处查看结果http://sqlfiddle.com/#!4/889a8/44

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

【讨论】：

【解决方案4】：

您可以通过删除else 子句来修改您的查询以获得您想要的：

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，我更喜欢 sum() 进行这种类型的聚合：

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;

【讨论】：

看来这不是用户想要达到的目的（尽管这是我最初的解释）。我相信他正在尝试计算每个条件的不同 detail_id 值，而不是尝试计算每个条件发生的行。
是的，贾斯汀说的没错。此线程中的小提琴 #29 (sqlfiddle.com/#!4/889a8/29) 为 COOK_TOTAL 生成正确的数字。我想将类似的查询捕获到一个查询中。
在我的回答中使用了 detail_id .. 你可以检查它是否适合你
@zundarz 。 . .然后你可以将detail_id 添加到select 和group by detail_id。

【解决方案5】：

除了分析函数之外，我可能会使用一种方法，首先“展平表格”(union all)，然后透视结果：

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'))
;

sql fiddle

【讨论】：

【解决方案6】：

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'))
;

【讨论】：