View ABAP CDS在时间戳字段上计数不同
在时间戳字段上使用带有DISTINCT的COUNT时出现问题。下面是我的CD视图,请帮助。 我想把计数放在ConfirmedDate字段上View ABAP CDS在时间戳字段上计数不同,view,abap,hana,cds,View,Abap,Hana,Cds,在时间戳字段上使用带有DISTINCT的COUNT时出现问题。下面是我的CD视图,请帮助。 我想把计数放在ConfirmedDate字段上 @AbapCatalog.sqlViewName: 'ZXEWMIWT' @AbapCatalog.compiler.compareFilter: true @AbapCatalog.preserveKey: true @AccessControl.authorizationCheck: #CHECK @EndUserText.label: 'In
@AbapCatalog.sqlViewName: 'ZXEWMIWT'
@AbapCatalog.compiler.compareFilter: true
@AbapCatalog.preserveKey: true
@AccessControl.authorizationCheck: #CHECK
@EndUserText.label: 'Interface View for Warehouse Task detail'
@OData.publish: true
// I* type:ddls
define view ZXEWMI_WT as select from /scwm/ordim_c as ORDIM_C {
key ORDIM_C.lgnum as WarehouseNo,
@UI.selectionField: [{ position: 1 }]
@UI.lineItem: [{ position : 1 }]
ORDIM_C.processor as Processor,
@UI.lineItem: [{ position : 2 }]
//count(distinct ORDIM_C.confirmed_at) as sum_wt
**tstmp_to_dats( ORDIM_C.confirmed_at,
abap_system_timezone( $session.client,'NULL' ),
$session.client,
'NULL' ) as ConfirmedDate**
}where processor <> ''
group by lgnum, processor, confirmed_at;
看起来您在字段ORDIM_C.confirm_at中有毫秒级的准确时间戳,但您希望按天对结果进行分组 不幸的是,GroupBy只允许您按输入表/视图中的列进行分组。它不能用于计算列 但您可以做的是首先创建一个单独的CDS视图,该视图提供表/scwm/ordim_c的字段,并将时间戳转换为日期,然后查询该视图 视图1:
define view Z_ORDIM_C_WITH_DAY as
select from /scwm/ordim_c as ORDIM_C {
key ORDIM_C.lgnum as WarehouseNo,
ORDIM_C.processor as Processor,
tstmp_to_dats( ORDIM_C.confirmed_at,
abap_system_timezone( $session.client,'NULL' ),
$session.client,
'NULL' ) as ConfirmedDate
}
视图2:
define view ZXEWMI_WT as
select from Z_ORDIM_C_WITH_DAY {
key WarehouseNo,
@UI.selectionField: [{ position: 1 }]
@UI.lineItem: [{ position : 1 }]
Processor,
@UI.lineItem: [{ position : 2 }]
count( * ) as sum_wt
ConfirmedDate
}
where processor <> ''
group by WarehouseNo, Processor, ConfirmedDate;
我有一个问题,你的问题是什么?在ConfirmedDate字段中,我需要将count与distinct放在一起,它不接受该表达式。。。这张CD是按用户显示每天的拾取次数…这里不需要区分,我身边没有EWM,但我假设confirmed_at是一个时间戳字段,因此在完全相同的秒/毫秒内不会有两个任务。请给出您的表行和预期输出,以及/scwm/ordim_c table definition>,这样在完全相同的秒/毫秒内就不会有两个任务了,您为什么这么认为?除非有主键或唯一约束,否则这是完全可能的。