SQL group by后根据多个条件获取单行

Question

需要帮助创建以下案例的查询：

假设我有一个 Table 包含以下记录

Name      Date        Time   Category CategoryKey
John      10/20/2012  10:00  Low       2
Sam       10/20/2012  10:00  High      4
Harry     10/20/2012  10:00  Medium    1
Michael   10/20/2012  10:00  Grey      3
Rob       10/22/2013  11:00  Low       2
Marry     10/23/2014  12:00  Low       2
Richard   10/23/2014  12:00  Grey      3
Jack      10/24/2015  1:30   High      4

然后，如果同一日期和时间有多个名称，则根据以下逻辑强制select只记录一个，并在满足以下任一条件时停止。

If Category is Medium then take name 
Else If Category is High then take name
Else If Category is Low then take name
Else If Category is Grey then take name

这样最终的结果就是

Name      Date        Time   Category CategoryKey

Harry     10/20/2012  10:00  Medium    1
Rob       10/22/2013  11:00  Low       2
Marry     10/23/2014  12:00  Low       2
Jack      10/24/2015  1:30   High      4

Answer 1

最简单的方法是row_number():

select t.*
from (select t.*,
             row_number() over (partition by date, time
                                order by (case category when 'Medium' then 1 when 'High' then 2 when 'Low' then 3 when 'Grey' then 4 else 5 end)
                               ) as seqnum
      from t
     ) t
where seqnum = 1;

这里可以方便的使用字符串函数：

             row_number() over (partition by date, time
                                order by charindex(category, 'Medium,High,Low,Grey')
                               ) as seqnum

这适用于您的情况，但您需要注意所有值都包含在内，并且 none "contain" 另一个值。

SQL group by后根据多个条件获取单行

SQL get single row based on multiple condition after group by

sql

tsql

sql-server

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