Question

我有一个示例数据如下：

field1  onlineoffline   trxndate
--------------------------------
F1       Offline        2018-04-03
F1       Online         2018-04-04
F1       Online         2018-04-05
F1       Offline        2018-04-06
F1       Offline        2018-04-07
F1       Offline        2018-04-08
F1       Offline        2018-04-09
F1       Online         2018-04-10
F1       Offline        2018-04-11

我需要根据trxndate（在本例中为2）获得在线第一次在线后的在线记录的数量，我还要根据trxndate计算第一次在线记录后的所有记录（7 in这个例子）在一个查询中。

我尝试使用row_number（）但是没有达到目的。有没有办法得到这个。

Answer 1

这是你想要的吗？

select count(*) as num_records,
       sum(case when onlineoffline = 'online' then 1 else 0 end) as num_online_record
from (select t.*,
             min(case when onlineoffline = 'online' then trxndate end) over () as min_online_td
      from t
     ) t
where trxndate > min_online_td;

子查询计算第一个在线记录的日期。查询的其余部分只会执行您想要的计数。

Answer 2

Relative Self

Answer 3

确定第一个在线日期，然后进行2次计数。

;WITH MinimumOnlineDate AS
(
    SELECT
        MinDate = MIN(T.trxndate)
    FROM
        YourTable AS T
    WHERE
        T.onlineoffline = 'Online'
)
SELECT
    TotalAfterFirstOnline = COUNT(1),
    TotalOnlineAfterFirstOnline = COUNT(CASE WHEN T.onlineoffline = 'Online' THEN 1 END)
FROM
    YourTable AS T
    INNER JOIN MinimumOnlineDate AS M ON T.trxndate > M.MinDate

Answer 4

其他选项与 correlation subquery

有关

select count(*) as online_counts, max(t1.counts) as no_of_onlines
from table t
join (
    select distinct min(trxndate) over() trxndate, count(*) over (order by field1) Counts 
    from table t
    where onlineoffline = 'Online' and trxndate <> (
          select min(trxndate) from table where onlineoffline = 'Online' and field1 = t.field1) 
)t1 on t.trxndate >= t1.trxndate

Answer 5

select
    sum(case when onlineoffline = 'Online' then 1 else 0 end) online,
    count(*) all 
from    table
where   trxdate > (select top 1 trxdate from table where onlineoffline = 'Online' order by trxdate)

根据sql server中的条件计算行数

5 个答案: