rank() 一组项目按计数(*)
rank() a group of items by count(*)
我在使用 Oracle 分析函数时遇到了一些问题,需要帮助。
这是一个通用示例:
create table test (item varchar2(10), value varchar2(10));
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value2');
insert into test values ('item1','value2');
insert into test values ('item3','value2');
insert into test values ('item3','value2');
insert into test values ('item3','value2');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value2');
insert into test values ('item5','value2');
insert into test values ('item5','value2');
select item, value, count(*) c,
sum(count(*)) over () total,
sum(count(*)) over (partition by item) total_by_item,
dense_rank() over (order by count(*) desc) dense_rank
from test
group by item, value
order by 5 desc;
查询结果为:
ITEM VALUE C TOTAL TOTAL_BY_ITEM DENSE_RANK
---------- ---------- -- ---------- -------------- ----------
item5 value1 7 20 10 1
item5 value2 3 20 10 3
item1 value2 2 20 7 4
item1 value1 5 20 7 2
item3 value2 3 20 3 3
如何获得 TOTAL_BY_ITEM 排名的项目?所以它看起来像这样:
ITEM VALUE C TOTAL TOTAL_BY_ITEM WHAT_I_NEED
---------- ---------- -- ---------- -------------- -----------
item5 value1 7 20 10 1
item5 value2 3 20 10 1
item1 value2 2 20 7 2
item1 value1 5 20 7 2
item3 value2 3 20 3 3
是否可以在没有另一个连接或子查询的情况下实现这一点?我觉得这是可能的。我自然认为它必须是这样的:dense_rank(count(*)) over(按项目划分),就像我用来获取第 5 列的分析 SUM 一样,但它不起作用。
我认为这不是您要搜索的内容,但仅供参考,无需子查询,您可以使用 MODEL 子句获得相同的结果:
select item, value, c, total, total_by_item, what_i_need
from test
group by item, value
model
dimension by (row_number() over (order by null) d)
measures (
item, value,
count(*) c,
sum(count(*)) over () total,
sum(count(*)) over (partition by item) total_by_item,
1 what_i_need
)
rules (
what_i_need[any] = dense_rank() over (order by total_by_item[cv()] desc)
)
order by 5 desc;
我认为如果没有子查询,您将无法实现它。
我在使用 Oracle 分析函数时遇到了一些问题,需要帮助。 这是一个通用示例:
create table test (item varchar2(10), value varchar2(10));
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value1');
insert into test values ('item1','value2');
insert into test values ('item1','value2');
insert into test values ('item3','value2');
insert into test values ('item3','value2');
insert into test values ('item3','value2');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value1');
insert into test values ('item5','value2');
insert into test values ('item5','value2');
insert into test values ('item5','value2');
select item, value, count(*) c,
sum(count(*)) over () total,
sum(count(*)) over (partition by item) total_by_item,
dense_rank() over (order by count(*) desc) dense_rank
from test
group by item, value
order by 5 desc;
查询结果为:
ITEM VALUE C TOTAL TOTAL_BY_ITEM DENSE_RANK
---------- ---------- -- ---------- -------------- ----------
item5 value1 7 20 10 1
item5 value2 3 20 10 3
item1 value2 2 20 7 4
item1 value1 5 20 7 2
item3 value2 3 20 3 3
如何获得 TOTAL_BY_ITEM 排名的项目?所以它看起来像这样:
ITEM VALUE C TOTAL TOTAL_BY_ITEM WHAT_I_NEED
---------- ---------- -- ---------- -------------- -----------
item5 value1 7 20 10 1
item5 value2 3 20 10 1
item1 value2 2 20 7 2
item1 value1 5 20 7 2
item3 value2 3 20 3 3
是否可以在没有另一个连接或子查询的情况下实现这一点?我觉得这是可能的。我自然认为它必须是这样的:dense_rank(count(*)) over(按项目划分),就像我用来获取第 5 列的分析 SUM 一样,但它不起作用。
我认为这不是您要搜索的内容,但仅供参考,无需子查询,您可以使用 MODEL 子句获得相同的结果:
select item, value, c, total, total_by_item, what_i_need
from test
group by item, value
model
dimension by (row_number() over (order by null) d)
measures (
item, value,
count(*) c,
sum(count(*)) over () total,
sum(count(*)) over (partition by item) total_by_item,
1 what_i_need
)
rules (
what_i_need[any] = dense_rank() over (order by total_by_item[cv()] desc)
)
order by 5 desc;
我认为如果没有子查询,您将无法实现它。