Question

我尝试获取SQL查询以返回一组基于int值设置为1的值

 Date           Name                Split         ID 
2014-09-02    Harry Potter         1            23
2014-09-02    Harry Potter         1            434
2014-09-02    Michael Jinks        0            24
2014-09-02    Sam Smith            1            12
2014-09-02    Sam Smith            1            244
2014-09-02    Kelly Jane           0            124
2014-09-03    Harry Potter         1            23
2014-09-03    Harry Potter         1            434

我希望它只返回用户使用Harry Potter的每条记录中的一个值，如果split设置为“1”，则忽略第二个记录ID

如果Split =“1”，它必须是if语句，然后查找最高记录并返回值，但我找不到任何东西可以做到这一点。

我试过了

select  distinct * from LOG where split = 1

应该返回类似这样的内容

Date           Name                Split         ID 
2014-09-02    Harry Potter         1            23
2014-09-02    Michael Jinks        0            24
2014-09-02    Sam Smith            1            12
2014-09-02    Kelly Jane           0            124
2014-09-03    Harry Potter         1            23

Answer 1

这个怎么样？

create table #temp(
    [date] smalldatetime,
    name varchar(100),
    split int,
    id int
)
insert into #temp
select '2014-09-02', 'Harry Potter', 1, 23 union all
select '2014-09-02', 'Harry Potter', 1, 434 union all
select '2014-09-02', 'Michael Jinks', 0, 24 union all
select '2014-09-02', 'Sam Smith', 1, 12 union all
select '2014-09-02', 'Sam Smith', 1, 244 union all
select '2014-09-02', 'Kelly Jane', 0, 124 union all
select '2014-09-03', 'Harry Potter', 1, 23 union all
select '2014-09-03', 'Harry Potter', 1, 434

-- Start
;with cte as(
    select
        *,
        row_number() over(partition by name order by [date], id) as rn -- For each name display first record with earliest date and lowest id
        --row_number() over(partition by name, [date] order by id) as rn -- For each name/date combination display first record with lowest id
    from #temp -- replace with your table name
)
select
    [date],
    name,
    split,
    id
from cte
where
    split = 1
    and rn = 1
-- End
drop table #temp

Answer 2

如果您感兴趣的ID始终是最低ID，那么

SELECT Date, Name, Split, MIN(ID) 
FROM log
GROUP BY Date, Name, Split

更新新表格问题的答案。

SELECT Date, Name, Split, MIN(ID) as ID
INTO tablename
FROM log
GROUP BY Date, Name, Split

SQL Query仅根据值返回记录集

2 个答案: