骨料日期時間的日期和時間列拆分

Question

我有以下查詢

from booking in query 
join ba in Context.BookingAddresses on booking.Id equals ba.BookingId into collections 
let firstCollection = (from d in collections where d.AddressType == BookingAddressType.Collection select d.RequestedDate).Min()        
where 
EntityFunctions.TruncateTime(queryArgs.DateFrom.Value) <= EntityFunctions.TruncateTime(firstCollection) && 
EntityFunctions.TruncateTime(queryArgs.DateTo.Value) >= EntityFunctions.TruncateTime(firstCollection) 
select booking; 

在let子句我真的需要一個合併DateOnly和時間跨度值的最小值 d.RequestedDate [DateOnly] + d .RequestedDateTimeFrom [時間跨度]看起來像這樣的DB：

 
RequestedDate : 2013-06-01 
RequestedDateTimeFrom : 13:50 

這並不編譯：

let firstCollection = (from d in collections where d.AddressType == BookingAddressType.Collection select d.RequestedDate + d.RequestedDateTimeFrom) 

編輯：在此期間，我想到了一種不同的方法，它實際上解決了我的主要問題，即如果有更多的日期時間同時具有日期和時間值，我將它們與第三個序列列進行比較。所以，把它歸結爲簡單的排序是這樣的：

from booking in query 
join ba in Context.BookingAddresses on booking.Id equals ba.BookingId into collections 
let firstCollection = collections.OrderBy(c => c.RequestedDate).ThenBy(c => c.RequestedFromTime).ThenBy(c => c.Sequence).FirstOrDefault() 
//(from d in collections where d.AddressType == BookingAddressType.Collection select d.RequestedDate).Min() 
where 
EntityFunctions.TruncateTime(queryArgs.DateFrom.Value) <= EntityFunctions.TruncateTime(firstCollection.RequestedDate) && 
EntityFunctions.TruncateTime(queryArgs.DateTo.Value) >= EntityFunctions.TruncateTime(firstCollection.RequestedDate) 
select booking; 

Answer 1

這也許不是最優雅的方式（和未經考驗的，是誠實的），但我會嘗試的時間跨度的鱗片SqlFunctions.DateAdd與SqlFunctions.DatePart

SqlFunctions.DateAdd("hh", 
         SqlFunctions.DatePart("hh", d.RequestedDateTimeFrom), 
         SqlFunctions.DateAdd("mi", 
              SqlFunctions.DatePart("mi", d.RequestedDateTimeFrom), 
              d.RequestedDate); 

（你也可以使用EntityFunctions.AddHours和EntityFunctions.AddMinutes代替SqlFunctions.DateAdd）