name adx tran_id cost ts
ck 5 125.168.10.0 33.00 1407234660
ck 5 187.18.99.00 33.32 1407234661
ck 5 125.168.10.0 33.24 1407234661
我只要这两行,第三行的tran_id和第一行的重复了,所以我 要不最后面一行去重去掉
答案1:
select
t1.tran_id
,t2.
,t2.
from
(select distinct tran_id from table) t1
join
table t2
on t1.tran_id=t2.tran_id
复制代码
评论:
如果使用distinct的话,我要把tran_id放在第一列,很丑查出来的数据
答案2:
select * from (select *,row_number() over (partition by tran_idorder by timestamp asc) num from table) t where t.num=1;
复制代码
分析:
row_number() over (partition by tran_idorder by timestamp desc) num 取num=1 的
这个是取 group by 按timestamp 排序的第一条数据
按每个 guid group 然后 按timestamp 排序 然后 加行标
然后去 行标为 1 的
附上: ROW_NUMBER() OVER函数的基本用法
语法:ROW_NUMBER() OVER(PARTITION BY COLUMN ORDER BY COLUMN)
简单的说row_number()从1开始,为每一条分组记录返回一个数字,这里的ROW_NUMBER() OVER (ORDER BY xlh DESC) 是先把xlh列降序,再为降序以后的没条xlh记录返回一个序号。
示例:
xlh row_num
1700 1
1500 2
1085 3
710 4
row_number() OVER (PARTITION BY COL1 ORDER BY COL2) 表示根据COL1分组,在分组内部根据 COL2排序,而此函数计算的值就表示每组内部排序后的顺序编号(组内连续的唯一的)
实例:
初始化数据
create table employee (empid int ,deptid int ,salary decimal(10,2))
insert into employee values(1,10,5500.00)
insert into employee values(2,10,4500.00)
insert into employee values(3,20,1900.00)
insert into employee values(4,20,4800.00)
insert into employee values(5,40,6500.00)
insert into employee values(6,40,14500.00)
insert into employee values(7,40,44500.00)
insert into employee values(8,50,6500.00)
insert into employee values(9,50,7500.00)