尋找與LINQ的界限

Question

這段代碼工作正常，但我想知道是否沒有辦法將它寫入一個LINQ表達式中，所以數據庫服務器可以完成一次傳遞，而不是實現一個結果集，然後循環通過它，這是我的代碼將產生。

var logs = from AssetLog log in dc.AssetLogs 
      where log.AssetId == assetId && log.Recorded >= start && log.Recorded <= finish 
      select log; 
    return new GetInteractionBoundsResult() 
    { 
    N = logs.Max(log => log.Latitude), 
    S = logs.Min(log => log.Latitude), 
    W = logs.Min(log => log.Longitude), 
    E = logs.Max(log => log.Longitude) 
    }; 

所以，LINQ大師，你會怎麼寫上面，使其產生在數據庫或多或少這樣的：

SELECT MIN(Latitude) S, MAX(Latitude) N, MIN(Longitude) W, MAX(Longitude) E 
FROM ASSETLOG WHERE etc etc 

Answer 1

當然，只是欺騙你的LINQ提供程序，以爲它仍然有工作查詢到了最後：

var logs = from asset in dc.Assets 
      where asset.AssetId == assetId 
      let g = asset.AssetLogs 
       .Where(log => log.Recorded >= start && log.Recorded <= finish) 
      select new GetInteractionBoundsResult 
       { 
        N = g.Max(log => log.Latitude), 
        S = g.Min(log => log.Latitude), 
        W = g.Min(log => log.Longitude), 
        E = g.Max(log => log.Longitude) 
       }; 
    return logs.Single(); 

一個GROUP BY可能有更好的表現比加入，上面的查詢會產生：

var logs = from log in dc.AssetLogs 
      where log.AssetId == assetId && 
       log.Recorded >= start && log.Recorded <= finish 
      group log by log.AssetId into g 
      select new GetInteractionBoundsResult 
       { 
        N = g.Max(log => log.Latitude), 
        S = g.Min(log => log.Latitude), 
        W = g.Min(log => log.Longitude), 
        E = g.Max(log => log.Longitude) 
       }; 
    return logs.Single(); 

Answer 2

這將需要一個相當複雜的LINQ提供程序，但子查詢可能工作：

var res = from asset in dc.Assets 
      where log.AssetId == assetId 
      let logs = (from AssetLog log in asset.AssetLogs 
         where log.Recorded >= start && log.Recorded <= finish 
         select log) 
      select new GetInteractionBoundsResult() 
      { 
       N = logs.Max(log => log.Latitude), 
       S = logs.Min(log => log.Latitude), 
       W = logs.Min(log => log.Longitude), 
       E = logs.Max(log => log.Longitude) 
      }; 
    return res.Single();