我试图获取列表1中的所有ID,并使用列表1中的ID,我试图获取列表2中的所有值以及基于列表2中的值的计数。
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DECLARE @Table1 AS TABLE (
id int,
l1 varchar(20)
);
INSERT INTO @Table1 VALUES
(1,'sun'),
(2,'shine'),
(3,'moon'),
(4,'light'),
(5,'earth'),
(6,'revolves'),
(7,'flow'),
(8,'fire'),
(9,'fighter'),
(10,'sun'),
(10,'shine'),
(11,'shine'),
(12,'moon'),
(1,'revolves'),
(10,'revolves'),
(2,'air'),
(3,'shine'),
(4,'fire'),
(5,'love'),
(6,'sun'),
(7,'rises');
/*
OPERATION 1
fetch all distinct ID's that has values from List 1
List1
sun
moon
earth
Initial OUTPUT1:
distinct_id list1_value
1 sun
3 moon
5 earth
10 sun
12 moon
6 sun
OPERATION2
fetch all the id, count_of_list2_values, list2_values
based on the id's that we recieved from OPERATION1
List2
shine
revolves
Expected Output:
id list1-value count_of_list2_values, list2_values
1 sun 1 revolves
3 moon 1 shine
5 earth 0 NULL
10 sun 2 shine,revolves
12 moon 0 NULL
6 sun 1 revolves
*/
我的查询: 这是我尝试过的
select id, count(l1),l1
from @table1
where id in ('shine','revolves') and id in ('sun','moon','earth')
如何实现这一目标。 我知道这应该是具有多个in的子查询。如何实现?
SQL小提琴链接: https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=7a85dbf51ca5b5d35e87d968c46300bb foo foo
答案 0 :(得分:1)
与此:
with
cte as(
select t1.id, t2.l1
from table1 t1 left join (
select * from table1 where l1 in ('shine','revolves')
) t2 on t2.id = t1.id
where t1.l1 in ('sun','moon','earth')
),
cte1 as(
select
c.id,
stuff(( select ',' + cte.l1 from cte where id = c.id for xml path(''), type).value('.', 'NVARCHAR(MAX)'), 1, 1, '') col
from cte c
)
select
id,
count(col) count_of_list2_values,
max(col) list2_values
from cte1
group by id
第一CTE给出以下结果:
id | l1
-: | :-------
1 | revolves
3 | shine
5 | null
10 | shine
10 | revolves
12 | null
6 | revolves
,第二个对这些结果进行操作以将l1
的公用分组值连接起来。
最后,我对第二个CTE的结果使用group by id
和夸张。
请参见demo
结果:
id | count_of_list2_values | list2_values
-: | --------------------: | :-------------
1 | 1 | revolves
3 | 1 | shine
5 | 0 | null
6 | 1 | revolves
10 | 2 | shine,revolves
12 | 0 | null
答案 1 :(得分:1)
如果您使用的是Sql Server 2017,则可以使用string_agg
函数和outer apply
运算符:
select
l1.id,
l1.l1,
l2.cnt as count_of_list2_values,
l2.l1 as list2_values
from @Table1 as l1
outer apply (
select
count(*) as cnt,
string_agg(tt.l1, ',') as l1
from @Table1 as tt
where
tt.l1 in ('shine','revolves') and
tt.id = l1.id
) as l2
where
l1.l1 in ('sun','moon','earth')
在以前的版本中,我不确定不为此创建特殊功能就可以汇总和计数一次。当然,您可以使用xquery
来做到这一点,但这可能有点过大(至少我不会在生产代码中这样做):
select
l1.id,
l1.l1,
l2.data.value('count(l1)', 'int'),
stuff(l2.data.query('for $i in l1 return concat(",",$i/text()[1])').value('.','nvarchar(max)'),1,1,'')
from @Table1 as l1
outer apply (
select
tt.l1
from @Table1 as tt
where
tt.l1 in ('shine','revolves') and
tt.id = l1.id
for xml path(''), type
) as l2(data)
where
l1.l1 in ('sun','moon','earth')
如果您不介意对表进行两次扫描/查找,则可以使用@forpas答案或执行以下操作:
with cte_list2 as (
select tt.l1, tt.id
from @Table1 as tt
where
tt.l1 in ('shine','revolves')
)
select
l1.id,
l1.l1,
l22.cnt as count_of_list2_values,
stuff(l21.data.value('.', 'nvarchar(max)'),1,1,'') as list2_values
from @Table1 as l1
outer apply (
select
',' + tt.l1
from cte_list2 as tt
where
tt.id = l1.id
for xml path(''), type
) as l21(data)
outer apply (
select count(*) as cnt
from cte_list2 as tt
where
tt.id = l1.id
) as l22(cnt)
where
l1.l1 in ('sun','moon','earth')
答案 2 :(得分:1)
有几种方法可以完成此操作。这是我的处理方式:
首先设置数据:
DECLARE @Table1 AS TABLE (
id int,
l1 varchar(20)
) ;
INSERT INTO @Table1 VALUES
(1,'sun'),
(2,'shine'),
(3,'moon'),
(4,'light'),
(5,'earth'),
(6,'revolves'),
(7,'flow'),
(8,'fire'),
(9,'fighter'),
(10,'sun'),
(10,'shine'),
(11,'shine'),
(12,'moon'),
(1,'revolves'),
(10,'revolves'),
(2,'air'),
(3,'shine'),
(4,'fire'),
(5,'love'),
(6,'sun'),
(7,'rises') ;
由于这是已知列表,因此请按自己的设置设置“目标”数据。 (在SQL中,表的使用总是比不固定的列表好得多。糟糕,错字!我的意思是定界的列表。)
DECLARE @Targets AS TABLE (
l2 varchar(20)
) ;
INSERT INTO @Targets VALUES
('sun'),
('moon'),
('earth') ;
操作1 获取具有列表1中的值的所有唯一ID (太阳,月亮,地球)
加入联接很容易:
SELECT Id
from @Table1 t1
inner join @Targets tg
on tg.l2 = t1.l1
操作2
获取所有ID,count_of_list2_values,list2_values
根据我们从OPERATION1收到的ID
如果我正确地遵循了所需的逻辑,那么(请先阅读“加入”注释):
SELECT
tt.Id
-- This next counts how many items in the Operation 1 list are not in the target list
-- (Spaced out, to make it easier to compare with the next line)
,sum( case when tg2.l2 is null then 1 else 0 end)
-- And this concatenates them together in a string (in later editions of SQL Server)
,string_agg(case when tg2.l2 is null then tt.l1 else null end, ', ')
from @Table1 tt
inner join (-- Operation 1 as a subquery, produce list of the Ids to work with
select t1.id
from @Table1 t1
inner join @Targets tg
on tg.l2 = t1.l1
) xx
on xx.id = tt.id
-- This is used to identify the target values vs. the non-target values
left outer join @Targets tg2
on tg2.l2 = tt.l1
-- Aggregate, because that's what we need to do
group by tt.Id
-- Order it, because why not?
order by tt.Id