Question

我需要从两个表中获取信息，如下所示，但是通过shift，目前它们基本上运行相同的查询4次，每次轮班运行2次。对于第一班，他们抓住所有不等于DELC的纸箱编号，并且第二次查询所有纸箱EQUAL到DELC。问题是我们想要不同的纸箱编号，如果纸箱在第一班上处理了一半并在第二班完成，即使我们做的不同，纸箱也会出现两次，因为它对每个查询都是唯一的。

有没有办法一起运行所有4个查询并对整个数据做出明显的分析？

1 Day old 1st shift = DELC
select count(distinct a.Barcode)
from [RL_Ship].[dbo].[mSCAN] as a inner join  
[RL_Ship].dbo].wmsInboundQueue] on
a.barcode = substring(message,26,20)
Where BagToteFlag = 'Y' and direction = 'Send'
and timeStamp >= '2016-06-14 03:00:00' 
and timeStamp < '2016-06-14    15:00:00' 
and substring(message,64,4) = 'DELC'
and SUBSTRING(rawdata,21,20) > '0'

1 Day old 1st shift <> DELC
select count(distinct a.Barcode)
from [RL_Ship].[dbo].[mSCAN] as a inner join    
[RL_Ship].dbo].wmsInboundQueue] on
a.barcode = substring(message,26,20) 
Where a.BagToteFlag = 'Y' and a.direction = 'Send' 
and a.timeStamp >= '2016-06-14 03:00:00'
and a.timeStamp < '2016-06-14 15:00:00'
and substring(message,64,4) <> 'DELC'
and SUBSTRING(rawdata,21,20) > '0'

1 Day old 2nd shift = DELC
select count(distinct a.Barcode)
from [RL_Ship].[dbo].[mSCAN] as a inner join [RL_Ship].[dbo].
[wmsInboundQueue] on
a.barcode = substring(message,26,20) 
Where a.BagToteFlag = 'Y' and a.direction = 'Send' 
and a.timeStamp >= '2016-06-14 15:00:00' 
and a.timeStamp < '2016-06-15 03:00:00'
and substring(message,64,4) = 'DELC'
and SUBSTRING(rawdata,21,20) > '0'

1 Day old 2nd shift <> DELC
select count(distinct a.Barcode)
from [RL_Ship].[dbo].[mSCAN] as a inner join [RL_Ship].[dbo].
[wmsInboundQueue] on
a.barcode = substring(message,26,20)
Where a.BagToteFlag = 'Y' and a.direction = 'Send'
and a.timeStamp >= '2016-06-14 15:00:00' 
and a.timeStamp < '2016-06-15 03:00:00' 
and substring(message,64,4) <> 'DELC'
and SUBSTRING(rawdata,21,20) > '0'

Answer 1

如果我理解了你的问题，那么第一个查询我会尝试得到你想要的结果

from django.contrib.auth.hashers import make_password

User.objects.bulk_create([
    User(
        username=name,
        email='sample@email.co.uk',
        password=make_password('Sample&Password!'),
        is_active=True,
    ) for name in f.read().splitlines()
])

Answer 2

如果我执行联合并取消计数部分，我会得到正确的数字，因为所有查询都会删除重复项，但如果我执行联合并且计数已关闭，则不会删除重复项。 ......

我实际上需要每个查询的总计数，因为它们是不同的班次/时间范围。

做一个计数然后来自的答案是好的，但我无法让它工作，2我认为这只会给我一个倒计数好吗？

我尝试了以下操作，但它给了我以下错误：

第36行，第15行，第1行，第36行关键字“select”附近的语法不正确。

编辑：好的，我通过在结尾添加'as table1'来实现这一点，但是我怀疑它只给了我一个总数，但没有重复，这很好，但我需要计算4个查询中的每一个。有什么想法？

select count(distinct Barcode) from
(select distinct a.barcode
from [RL_Ship].[dbo].[mSCAN] as a inner join [RL_Ship].[dbo].
[wmsInboundQueue] on
a.barcode = substring(message,26,20)
Where BagToteFlag = 'Y' and direction = 'Send'
and timeStamp >= '2016-06-14 03:00:00' and timeStamp < '2016-06-14
15:00:00' 
and substring(message,64,4) = 'DELC'
and SUBSTRING(rawdata,21,20) > '0'
union
select distinct a.Barcode
from [RL_Ship].[dbo].[mSCAN] as a inner join [RL_Ship].[dbo].
[wmsInboundQueue] on
a.barcode = substring(message,26,20) 
Where a.BagToteFlag = 'Y' and a.direction = 'Send' 
and a.timeStamp >= '2016-06-14 03:00:00'
and a.timeStamp < '2016-06-14 15:00:00'
and substring(message,64,4) <> 'DELC'
and SUBSTRING(rawdata,21,20) > '0'
union
select distinct a.Barcode
from [RL_Ship].[dbo].[mSCAN] as a inner join [RL_Ship].[dbo].
[wmsInboundQueue] on
a.barcode = substring(message,26,20) 
Where a.BagToteFlag = 'Y' and a.direction = 'Send' 
and a.timeStamp >= '2016-06-14 15:00:00' 
and a.timeStamp < '2016-06-15 03:00:00'
and substring(message,64,4) = 'DELC'
and SUBSTRING(rawdata,21,20) > '0'
union
select distinct a.Barcode
from [RL_Ship].[dbo].[mSCAN] as a inner join [RL_Ship].[dbo].
[wmsInboundQueue] on
a.barcode = substring(message,26,20)
Where a.BagToteFlag = 'Y' and a.direction = 'Send'
and a.timeStamp >= '2016-06-14 15:00:00' 
and a.timeStamp < '2016-06-15 03:00:00' 
and substring(message,64,4) <> 'DELC'
and SUBSTRING(rawdata,21,20) > '0')

SQL不同的多个查询组合

2 个答案: