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1 /* 2 数据如下: 3 name val memo 4 a 2 a2(a的第二个值) 5 a 1 a1--a的第一个值 6 a 3 a3:a的第三个值 7 b 1 b1--b的第一个值 8 b 3 b3:b的第三个值 9 b 2 b2b2b2b2 10 b 4 b4b4 11 b 5 b5b5b5b5b5 12 */ 13 --创建表并插入数据: 14 create table tb(name varchar(10),val int,memo varchar(20)) 15 insert into tb values(‘a‘, 2, ‘a2(a的第二个值)‘) 16 insert into tb values(‘a‘, 1, ‘a1--a的第一个值‘) 17 insert into tb values(‘a‘, 3, ‘a3:a的第三个值‘) 18 insert into tb values(‘b‘, 1, ‘b1--b的第一个值‘) 19 insert into tb values(‘b‘, 3, ‘b3:b的第三个值‘) 20 insert into tb values(‘b‘, 2, ‘b2b2b2b2‘) 21 insert into tb values(‘b‘, 4, ‘b4b4‘) 22 insert into tb values(‘b‘, 5, ‘b5b5b5b5b5‘) 23 go 24 25 --一、按name分组取val最大的值所在行的数据。 26 --方法1: 27 select a.* from tb a where val = (select max(val) from tb where name = a.name) order by a.name 28 --方法2: 29 select a.* from tb a where not exists(select 1 from tb where name = a.name and val > a.val) 30 --方法3: 31 select a.* from tb a,(select name,max(val) val from tb group by name) b where a.name = b.name and a.val = b.val order by a.name 32 --方法4: 33 select a.* from tb a inner join (select name , max(val) val from tb group by name) b on a.name = b.name and a.val = b.val order by a.name 34 --方法5 35 select a.* from tb a where 1 > (select count(*) from tb where name = a.name and val > a.val ) order by a.name 36 /* 37 name val memo 38 ---------- ----------- -------------------- 39 a 3 a3:a的第三个值 40 b 5 b5b5b5b5b5 41 */ 42 43
写法6
SELECT *
FROM (
SELECT * ,
ROW_NUMBER() OVER ( PARTITION BY name ORDER BY val DESC ) rid
FROM tb
) AS t
WHERE rid = 1
如果上述存在一个name组中有两个以上相同的val,则查询方法1-5不正确。但是可以对方法1-5进行去重处理
例如:
insert into tb values(‘a‘, 2, ‘a2(a的第二个值)‘)
insert into tb values(‘a‘, 3, ‘a1--a的第一个值‘)
insert into tb values(‘a‘, 3, ‘a3:a的第三个值‘)
转自:http://www.cnblogs.com/zfanlong1314/p/3393946.html
【转】sql server 获取每一个类别中值最大的一条数据
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原文地址:http://www.cnblogs.com/crazybottle/p/4553309.html