Sql 累计金额不起作用

我试图得到整个列的累积和。下面是我的数据样本

DT                             ETB_DT
Friday 00:00:00.0000000 Friday 04:00:00.0000000
Friday 00:00:00.0000000 Friday 08:00:00.0000000
Friday 00:00:00.0000000 Friday 12:00:00.0000000
Friday 00:00:00.0000000 Friday 16:00:00.0000000
Friday 00:00:00.0000000 Friday 20:00:00.0000000

No_of_records   
22             
41  
86  
65  
25  

我使用了下面的代码来获取累积列，该列应该添加到前一行

选择*
，累计=总和（无记录）
从没有累计4
按DT分组，ETB_DT，无记录
按DT订购

但是，这给了我以下输出

DT                             ETB_DT
Friday 00:00:00.0000000 Friday 04:00:00.0000000
Friday 00:00:00.0000000 Friday 08:00:00.0000000
Friday 00:00:00.0000000 Friday 12:00:00.0000000
Friday 00:00:00.0000000 Friday 16:00:00.0000000
Friday 00:00:00.0000000 Friday 20:00:00.0000000

No_of_records   cumulative
22                  22
41                  41
86                  86
65                  65
25                  25

下面是我想要的输出

DT                             ETB_DT
Friday 00:00:00.0000000 Friday 04:00:00.0000000
Friday 00:00:00.0000000 Friday 08:00:00.0000000
Friday 00:00:00.0000000 Friday 12:00:00.0000000
Friday 00:00:00.0000000 Friday 16:00:00.0000000
Friday 00:00:00.0000000 Friday 20:00:00.0000000

No_of_records   cumulative
22                  22
41                  63
86                  149
65                  214
25                  239

---

你们有什么想法吗？非常感谢您的帮助。

这是累积总和的语法：

Select DT, ETB_DT, No_of_records
       sum(No_of_records) over (order by DT, ETB_DT) as cumulative
From withoutcumulative4
Order by DT;

带有order by的

over

子句是使总和累积的原因。没有给它命名

---

据我所知，您不需要

分组方式

这是累积和的语法：

Select DT, ETB_DT, No_of_records
       sum(No_of_records) over (order by DT, ETB_DT) as cumulative
From withoutcumulative4
Order by DT;

带有order by的

over

子句是使总和累积的原因。没有给它命名

据我所知，您不需要

分组方式

您可以使用：

Select *
       ,cumulative = sum(No_of_records) over (order by ETB_DT) 
 From  withoutcumulative4
  Order by DT

您可以使用：

Select *
       ,cumulative = sum(No_of_records) over (order by ETB_DT) 
 From  withoutcumulative4
  Order by DT