Question

我正在使用汇总来获取一些聚合，并以表格形式将它们显示给用户。

但是，我想确保在我的汇总中，汇总的行数是相同的，即最大子集的数量。

我认为一个例子让我想要的更清楚，所以我在下面的oracle中设置了一个简单的例子：

create table test (
    co_name varchar2(100),
    rtype number,
    some_count number
)         ;
insert all
    into test (co_name, rtype, some_count) values ('A', 1, 5)
    into test (co_name, rtype, some_count) values ('A', 2, 6)
    into test (co_name, rtype, some_count) values ('A', 3, 7)
    into test (co_name, rtype, some_count) values ('B', 1, 8)
    into test (co_name, rtype, some_count) values ('B', 2, 9)
SELECT * FROM DUAL
;

select * from test;
SELECT
    co_name,
    rtype,
    count(some_count)
FROM test
GROUP BY ROLLUP(co_name, rtype)

这给了我以下结果：

CO_NAME RTYPE   SOME_COUNT
A       1       5
A       2       6
A       3       7
A               18
B       1       8
B       2       9
B               17
B               35

你会注意到B当然只有两行用于RTYPE - 1和2 那是因为有0行，其中CO_NAME = B AND RTYPE = 3

有没有办法让汇总在返回的结果数量上保持一致？我想要的是看到以下内容：

CO_NAME RTYPE   SOME_COUNT
A       1       5
A       2       6
A       3       7
A               18
B       1       8
B       2       9
B       3       0
B               17
                35

这里的任何建议都会非常有用，因为我希望我的应用程序是愚蠢的，只需将结果制成表格而不必考虑丢失的数据。我想要查询给我所需的一切。

谢谢！

编辑：我是一个涂料......在我上面的例子中，我想保持简单，但却犯了一个错误。而不是RTYPES是一组{1,2,3}可能的值，想象它是一组{'x'，'y'，'z'}可能的值......我确实将答案标记为答案，因为它回答了我问的问题，错误在我身上:(

Answer 1

rollup条款不会填补空白，您必须事先做好：

SQL> with rtypes(col) as(
  2    select level
  3      from ( select max(count(co_name)) as mx_num
  4               from test1
  5              group by co_name
  6            ) t
  7    connect by level <= t.mx_num
  8  )
  9  select t.co_name
 10       , r.col                   as rtype
 11       , sum(nvl(some_count, 0)) as some_count
 12    from test1 t
 13    partition by (t.co_name)
 14    right join rtypes r
 15       on (r.col = t.rtype)
 16  group by rollup(t.co_name, r.col)
 17  ;

结果：

Co_Name rtype   Some_Count
-------------------------------------- 
A       1       5 
A       2       6 
A       3       7 
A               18 
B       1       8 
B       2       9 
B       3       0 
B               17 
                35

WITH子句中的查询用于从1到3生成RTYPES（因为它发生了3

是此情况下rtype的最大数量）。在主查询中我们右外连接我们的

TEST1表，其中包含使用RTYPES子句的partition by() CTE的实际数据。

关于分区连接的

Find out more。

回答评论：

如果有字符，正如您所说的一个字符值（x，y，z或a，b，c它

并不重要），在RTYPE列中，我们可以重写CTE（WITH

中的查询

子句）生成如下字符集：

with rtypes(col) as(
  select chr(min_char + level - 1)
   from ( select max(ascii(rtype1)) as max_char
               , min(ascii(rtype1)) as min_char
            from test1
         ) 
 connect by min_char + level <=  max_char + 1
)

然后最终的查询将是：

with rtypes(col) as(
    select chr(min_char + level - 1)
      from ( select max(ascii(rtype1)) as max_char
                  , min(ascii(rtype1)) as min_char
              from test1
           ) 
   connect by min_char + level <=  max_char + 1
  )
select t.co_name
     , r.col                   as rtype
     , sum(nvl(some_count, 0)) as some_count
  from test1 t
  partition by (t.co_name)
  right join rtypes r
     on (r.col = t.rtype1)
  group by rollup(t.co_name, r.col)

结果：

Co_Name rtype   Some_Count
-------------------------------------- 
A       x       5 
A       y       6 
A       z       7 
A               18 
B       x       8 
B       y       9 
B       z       0 
B               17 
                35

Answer 2

好的，这是我能做的最好的事情：）

它会使用您现有的查询，然后是所有不存在的co_name和rtype组合中的联合。

相同数据：

create table test_rollup (
    co_name varchar2(100),
    rtype number,
    some_count number
)         ;

insert all
    into test_rollup (co_name, rtype, some_count) values ('A', 1, 5)
    into test_rollup (co_name, rtype, some_count) values ('A', 2, 6)
    into test_rollup (co_name, rtype, some_count) values ('A', 3, 7)
    into test_rollup (co_name, rtype, some_count) values ('B', 1, 8)
    into test_rollup (co_name, rtype, some_count) values ('B', 2, 9)
SELECT * FROM DUAL
;

select * from test_rollup;

使用WITH定义co_name和rtype的通用列表

WITH rtypes as 
(select distinct rtype from test_rollup
),
co_names as
(select distinct co_name from test_rollup
)
SELECT
    co_name,
    rtype,
    sum(some_count)
FROM test_rollup 
GROUP BY ROLLUP(co_name, rtype)    
[...]

最后在两个通用列表的笛卡尔联合中联合，其配对默认为0，减去已经占用的组合：

UNION
SELECT
    co_names.co_name,
    rtypes.rtype,
    0
FROM rtypes, co_names
where not exists 
  (select 1 
  from test_rollup 
  where test_rollup.co_name=co_names.co_name 
  and rtypes.rtype = test_rollup.rtype)

确保oracle汇总中的行数相同

2 个答案: