我有一个Oracle查询
select max(m.id),
m.someId keep (DENSE_RANK FIRST ORDER BY m.UpdateDate desc)
from MyTable m
groupBy m.someId
对于这样的数据:
id UpdateDate someId
1 20-01-2012 10
2 20-01-2012 10
3 01-01-2012 10
4 10-02-2012 20
5 01-02-2012 20
6 01-04-2012 30
将完全归还给我:
2 10
4 20
6 30
因此,对于每个someId,它会搜索最新的updateDate并返回相应的id
。 (如果最新日期有几个ID,则需要最新的ID。)
但对于SQL服务器,此查询的工作方式是否相同?我的意思是这种结构keep (dense_rank first order by ..)
?
答案 0 :(得分:8)
我认为您的特定查询不会运行SQL Server。但是你可以做到这一点同样的结果:
SELECT id, SomeId
FROM ( SELECT *, ROW_NUMBER() OVER(PARTITION BY someId ORDER BY UpdateDate DESC, id DESC) Corr
FROM MyTable) A
WHERE Corr = 1
答案 1 :(得分:3)
我回来并回到这个问题和答案。不幸的是,在使用"窗口功能进行迁移时,有几种情况可以用于排名"变得非常复杂。那些情况是:
因此我会在答案中添加其他信息。 原始数据SQLFIDDLE:http://sqlfiddle.com/#!6/e5c6d/6
1。阅读oracle函数:
select max(m.id), m.someId keep (DENSE_RANK FIRST ORDER BY m.UpdateDate desc)
from MyTable m
groupBy m.someId
我们选择组中最大的m.id(someId,UpdateDate),其中UpdateDate最大的是组(someId)
2。直接导致无法正常工作因为错误: Column' MyTable.UpdateDate'在选择列表中无效,因为它不包含在聚合函数或GROUP BY子句中。
SELECT FIRST_VALUE(id) OVER(PARTITION BY someId ORDER BY UpdateDate DESC, id DESC) first_in_orderedset , someId
FROM MyTable
GROUP BY someId
3。改进'直接'是无效的
SELECT someId, MIN(first_in_orderedset)
FROM
( SELECT FIRST_VALUE(id) OVER(PARTITION BY someId ORDER BY UpdateDate DESC, id DESC) first_in_orderedset , someId
FROM MyTable ) t
GROUP BY someId;
4。交叉申请:
SELECT grouped.someId, orderedSet.FirstUpdateDate, maxInSet.first_in_orderedset FROM
(
SELECT mt.someId
FROM MyTable mt
GROUP BY mt.someId
) grouped CROSS APPLY
(
SELECT top 1 mt2.UpdateDate as FirstUpdateDate
FROM MyTable mt2
WHERE mt2.someId=grouped.someId
ORDER BY UpdateDate desc
) orderedSet CROSS APPLY
(
SELECT max(mt3.id) as first_in_orderedset
FROM MyTable mt3
WHERE mt3.someId=grouped.someId and mt3.UpdateDate=orderedSet.FirstUpdateDate
) maxInSet;
5. 现在让我们获取更复杂的表和更复杂的查询: ORACLE:http://sqlfiddle.com/#!4/c943c/23 SQL SERVER:http://sqlfiddle.com/#!6/dc7fb/1/0 (数据是预生成的,在两个沙箱中都是相同的 - 很容易比较结果) 表:
CREATE TABLE AlarmReports (
id int PRIMARY KEY,
clientId int, businessAreaId int , projectId int, taskId int,
process1Spent int, process1Lag int, process1AlarmRate varchar2(1) null,
process2Spent int, process2Lag int, process2AlarmRate varchar2(1) null,
process3Spent int, process3Lag int, process3AlarmRate varchar2(1) null
)
Oracle查询:
SELECT clientId, businessAreaId, projectId,
sum(process1Spent),
sum(process2Spent),
sum(process3Spent),
MIN(process1AlarmRate) KEEP (DENSE_RANK FIRST ORDER BY process1Lag DESC),
MIN(process2AlarmRate) KEEP (DENSE_RANK FIRST ORDER BY process2Lag DESC),
MIN(process3AlarmRate) KEEP (DENSE_RANK FIRST ORDER BY process3Lag DESC)
FROM AlarmReports
GROUP BY GROUPING SETS ((),(clientId),(clientId, projectId),(businessAreaId),(clientId,businessAreaId))
SQL查询:
(to be continued)
实际上我已计划将我的自定义聚合与c#一起使用。如果有人感兴趣,请与我联系...自定义聚合是这类问题的最佳解决方案,但就varchar长度来说并不是很好的。对于每个varchar长度,您将有义务创建"专门的"聚合功能
答案 2 :(得分:1)
SQL Server不支持“keep”构造,因此您需要使用子查询:
select m.*
from (select *, row_number() over (partition by m.someid ORDER BY m.UpdateDate desc) as seqnum
from MyTable m
) m
where seqnum = 1
这将找到每个m.id的第一行,其中包含最新的UpdateDate。然后它在外部查询中选择该行。请注意,此方法不需要使用组。
答案 3 :(得分:1)
这绝对有用。首先尝试,然后争辩。 当您有多个订单时,可以执行此操作(在Oracle上进行示例):
- 这一个与keep_rank
WITH a AS (SELECT 1 s1, 4 s2, 'a' c, 10 g FROM dual UNION all
SELECT 2 s1, 2 s2, 'b' c, 10 g FROM dual UNION ALL
SELECT 3 s1, 1 s2, 'c' c, 20 g FROM dual UNION ALL
SELECT 4 s1, 3 s2, 'd' c, 20 g FROM dual)
SELECT g,
MAX(c) KEEP (DENSE_RANK FIRST ORDER BY s1) s1,
MAX(c) KEEP (DENSE_RANK FIRST ORDER BY s2) s2
FROM a
GROUP BY g
- 这个没有保持dense_rank
WITH a AS (SELECT 1 s1, 4 s2, 'a' c, 10 g FROM dual UNION all
SELECT 2 s1, 2 s2, 'b' c, 10 g FROM dual UNION ALL
SELECT 3 s1, 1 s2, 'c' c, 20 g FROM dual UNION ALL
SELECT 4 s1, 3 s2, 'd' c, 20 g FROM dual)
SELECT g,
MAX(DECODE(s1, 1, c)) s1,
MAX(DECODE(s2, 1, c)) s2
FROM (SELECT g,c,
ROW_NUMBER() OVER (PARTITION BY g ORDER BY s1) s1,
ROW_NUMBER() OVER (PARTITION BY g ORDER BY s2) s2
FROM a) b
GROUP BY g
答案 4 :(得分:0)
如果有人在Postgres中寻找Oracle KEEP DENSE_RANK模拟:
CREATE TABLE myt (
"id" INTEGER,
"update_date" timestamp,
"some_id" INTEGER
);
INSERT INTO myt
("id", "update_date", "some_id")
VALUES
('1', '2012-01-20', '10'),
('2', '2012-01-20', '10'),
('3', '2012-01-01', '10'),
('4', '2012-10-02', '20'),
('5', '2012-01-02', '20'),
('6', '2012-01-04', '30');
select
some_id,
(array_agg(id order by update_date desc, id desc))[1]
from myt
group by some_id
order by some_id