选择顶行，直到特定列中的值出现两次

Posted 2023-03-31

【中文标题】选择顶行，直到特定列中的值出现两次

【英文标题】：Select top rows until value in specific column has appeared twice

【发布时间】：2020-05-04 03:31:14

【问题描述】：

我有以下查询，我试图选择所有记录，按日期排序，直到第二次找到EmailApproved = 1。不应选择EmailApproved = 1 的第二条记录。

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;

这会产生以下结果：

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 和 Row1EqualsRow2 只是显示我的计算的工作列。

【参考方案1】：

步骤：

在所有行上创建一个行号rn，以防id 不按顺序排列。 创建一个行号approv_rn，由EmailApproved 分区，这样我们就知道第二次EmailApproved = 1 的时间了 使用outer apply 查找EmailApproved = 1 的second 实例的行号 在where 子句中，过滤掉行号为>= 的所有行，在步骤3 中找到值。 如果有 1 或 0 个 EmailApproved 记录可用，则 outer apply 将返回 null，在这种情况下返回所有可用行。

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;