选择顶部的行，直到特定列中的值出现两次

一尘不染

选择顶部的行，直到特定列中的值出现两次

sql

我有以下查询，在这里我尝试选择所有记录（按日期排序），直到EmailApproved = 1找到第二次为止。第二条记录EmailApproved = 1不应在何处选择。

declare @Test table (id int, EmailApproved bit, Created datetime)

insert into @Test (id, EmailApproved, Created)
values
  (1,0,'2011-03-07 03:58:58.423')
  , (2,0,'2011-02-21 04:55:52.103')
  , (3,0,'2011-01-29 13:24:02.103')
  , (4,1,'2010-10-12 14:41:54.217')
  , (5,0,'2010-10-12 14:34:15.903')
  , (6,0,'2010-10-12 10:10:19.123')
  , (7,1,'2010-08-27 12:07:16.073')
  , (8,1,'2010-08-25 12:15:49.413')
  , (9,0,'2010-08-25 12:14:51.970')
  , (10,1,'2010-04-12 16:43:44.777')

select *
  , case when Row1 = Row2 then 1 else 0 end Row1EqualRow2
from (
  select id, EmailApproved, Created
    , row_number() over (partition by EmailApproved order by Created desc) Row1
    , row_number() over (order by Created desc) Row2
  from @Test
) X
--where Row1 = Row2
order by Created desc

产生以下结果：

id  EmailApproved   Created                 Row1    Row2    Row1EqualsRow2
1   0               2011-03-07 03:58:58.423 1       1       1
2   0               2011-02-21 04:55:52.103 2       2       1
3   0               2011-01-29 13:24:02.103 3       3       1
4   1               2010-10-12 14:41:54.217 1       4       0
5   0               2010-10-12 14:34:15.903 4       5       0
6   0               2010-10-12 10:10:19.123 5       6       0
7   1               2010-08-27 12:07:16.073 2       7       0
8   1               2010-08-25 12:15:49.413 3       8       0
9   0               2010-08-25 12:14:51.970 6       9       0
10  1               2010-04-12 16:43:44.777 4       10      0

我真正想要的是：

id  EmailApproved   Created                 Row1    Row2    Row1EqualsRow2
1   0               2011-03-07 03:58:58.423 1       1       1
2   0               2011-02-21 04:55:52.103 2       2       1
3   0               2011-01-29 13:24:02.103 3       3       1
4   1               2010-10-12 14:41:54.217 1       4       0
5   0               2010-10-12 14:34:15.903 4       5       0
6   0               2010-10-12 10:10:19.123 5       6       0

注意：Row，Row2＆Row1EqualsRow2只是用于显示我的计算的工作列。

阅读 135

2021-05-05

共1个答案

一尘不染

脚步：

rn如果id未按顺序在所有行上创建一个行号。
创建一个行号，按进行approv_rn分区，EmailApproved以便我们EmailApproved = 1第二次知道何时
使用outer apply来查找的second实例的行号EmailApproved = 1
在where子句中，筛选出所有行，其中行号是>=在步骤3中找到的值。

如果有1或0EmailApproved条记录可用，则outer apply将会返回null，在这种情况下，将返回所有可用行。

; with test as
(
    select  *, 
            rn         = row_number() over (order by Created desc),
            approv_rn  = row_number() over (partition by EmailApproved 
                                                order by Created desc)
    from    @Test
)
select  *
from    test t
        outer apply
        (
            select  x.rn
            from    test x
            where   x.EmailApproved = 1
            and     x.approv_rn     = 2
        ) x
where   t.rn    < x.rn or x.rn is null
order by t.Created desc

2021-05-05