【发布时间】:2017-09-30 16:05:17
【问题描述】:
这是我的查询:
select convert(date, create_timestamp) as date
,sum(count(*)) over (order by convert(date, create_timestamp)) as cumulative
from job_posting jp
group by convert(date, create_timestamp)
order by convert(date, create_timestamp)
它按日期分组并显示以下结果:
2015-09-02 1
2015-09-03 2
2015-09-04 5
2015-09-05 7
2015-09-07 14
我想创建一个新查询以返回按月份和年份分组的结果,如下所示:
2015-09
2015-10
2015-11
我正在努力寻找合适的 SQL 命令来帮助解决这个问题。此行导致问题:
,sum(count(*)) over (order by convert(date, create_timestamp)) as cumulative
【问题讨论】:
-
选择 CONVERT (varchar(10), create_timestamp, 103) AS [DD/MM/YYYY] 为完整格式,现在您可以根据需要使用它
标签: sql-server datetime group-by