4

我们已经安装了 .NET 4.5，所以我们不应该受到像以前版本的 EF 中这样的已知错误的影响。实体框架正在生成错误的 SQL。我有以下查询

from a in _practiceRepository.Adjustments       
                                  select new AdjustmentAndFinancialInformation
                                             {
                                                 IsReport = true,
                                                 BillingProvider = a.InvoiceReceivable.Invoice.BillingProvider.User.UserName,
                                                 ServiceLocation = a.InvoiceReceivable.Invoice.Encounter.ServiceLocation.Name,
                                                 BillingOrganization = a.InvoiceReceivable.Invoice.BillingProvider.BillingOrganization.Name,
                                                 FinancialTypeGroup = "",
                                                 AdjustmentTypeName = a.AdjustmentType.Name,
                                                 AdjustmentTypeShortName = a.AdjustmentType.ShortName,
                                                 IsDebit = a.AdjustmentType.IsDebit,
                                                 IsCash = a.AdjustmentType.IsCash,
                                                 FinancialInformationTypeName = "",
                                                 FinancialInformationTypeShortName = "",
                                                 DateTimePosted = a.PostedDateTime,
                                                 PatientLastName = a.InvoiceReceivable.Invoice.Encounter.Patient.LastName,
                                                 PatientFirstName = a.InvoiceReceivable.Invoice.Encounter.Patient.FirstName,
                                                 PatientId = a.InvoiceReceivable.Invoice.Encounter.Patient.Id,
                                                 Amount = a.Amount,
                                                 InvoiceDateTime = a.InvoiceReceivable.Invoice.DateTime
                                             };

它会生成以下带有大量 LEFT JOINS 的 SQL（这会使性能降低大约 10 倍）。

我已经多次验证所有关系都是不可为空的。

    SELECT TOP (10) 
[Project1].[AdjustmentTypeId] AS [AdjustmentTypeId], 
[Project1].[C1] AS [C1], 
[Project1].[UserName] AS [UserName], 
[Project1].[Name] AS [Name], 
[Project1].[Name1] AS [Name1], 
[Project1].[Name3] AS [Name2], 
[Project1].[Name2] AS [Name3], 
[Project1].[ShortName] AS [ShortName], 
[Project1].[IsDebit] AS [IsDebit], 
[Project1].[IsCash] AS [IsCash], 
[Project1].[C2] AS [C2], 
[Project1].[C3] AS [C3], 
[Project1].[PostedDateTime] AS [PostedDateTime], 
[Project1].[LastName] AS [LastName], 
[Project1].[FirstName] AS [FirstName], 
[Project1].[PatientId] AS [PatientId], 
[Project1].[Amount] AS [Amount], 
[Project1].[DateTime] AS [DateTime]
FROM (SELECT
    [Extent1].[Amount] AS [Amount], 
    [Extent1].[PostedDateTime] AS [PostedDateTime], 
    [Extent1].[AdjustmentTypeId] AS [AdjustmentTypeId], 
    [Extent5].[UserName] AS [UserName], 
    [Extent8].[Name] AS [Name], 
    [Extent11].[Name] AS [Name1], 
    [Extent12].[Name] AS [Name2], 
    [Extent12].[ShortName] AS [ShortName], 
    [Extent12].[IsCash] AS [IsCash], 
    [Extent12].[IsDebit] AS [IsDebit], 
    [Extent13].[Name] AS [Name3], 
    [Extent16].[LastName] AS [LastName], 
    [Extent19].[FirstName] AS [FirstName], 
    [Extent21].[PatientId] AS [PatientId], 
    [Extent22].[DateTime] AS [DateTime], 
    cast(1 as bit) AS [C1], 
    N'' AS [C2], 
    N'' AS [C3]
    FROM                      [model].[Adjustments] AS [Extent1]
    LEFT OUTER JOIN [model].[InvoiceReceivables] AS [Extent2] ON [Extent1].[InvoiceReceivableId] = [Extent2].[Id]
    LEFT OUTER JOIN [model].[Invoices] AS [Extent3] ON [Extent2].[InvoiceId] = [Extent3].[Id]
    LEFT OUTER JOIN [model].[Providers] AS [Extent4] ON [Extent3].[BillingProviderId] = [Extent4].[Id]
    LEFT OUTER JOIN [model].[Users] AS [Extent5] ON ([Extent4].[UserId] = [Extent5].[Id]) AND ([Extent5].[__EntityType__] IN (N'User',N'Doctor'))
    LEFT OUTER JOIN [model].[Invoices] AS [Extent6] ON [Extent2].[InvoiceId] = [Extent6].[Id]
    LEFT OUTER JOIN [model].[Encounters] AS [Extent7] ON [Extent6].[EncounterId] = [Extent7].[Id]
    LEFT OUTER JOIN [model].[ServiceLocations] AS [Extent8] ON [Extent7].[ServiceLocationId] = [Extent8].[Id]
    LEFT OUTER JOIN [model].[Invoices] AS [Extent9] ON [Extent2].[InvoiceId] = [Extent9].[Id]
    LEFT OUTER JOIN [model].[Providers] AS [Extent10] ON [Extent9].[BillingProviderId] = [Extent10].[Id]
    LEFT OUTER JOIN [model].[BillingOrganizations] AS [Extent11] ON ([Extent10].[BillingOrganizationId] = [Extent11].[Id]) AND ([Extent11].[__EntityType__] IN (N'BillingOrganization',N'PersonBillingOrganization'))
    LEFT OUTER JOIN [model].[AdjustmentTypes] AS [Extent12] ON [Extent1].[AdjustmentTypeId] = [Extent12].[Id]
    LEFT OUTER JOIN [model].[FinancialTypeGroups] AS [Extent13] ON [Extent12].[FinancialTypeGroupId] = [Extent13].[Id]
    LEFT OUTER JOIN [model].[Invoices] AS [Extent14] ON [Extent2].[InvoiceId] = [Extent14].[Id]
    LEFT OUTER JOIN [model].[Encounters] AS [Extent15] ON [Extent14].[EncounterId] = [Extent15].[Id]
    LEFT OUTER JOIN [model].[Patients] AS [Extent16] ON [Extent15].[PatientId] = [Extent16].[Id]
    LEFT OUTER JOIN [model].[Invoices] AS [Extent17] ON [Extent2].[InvoiceId] = [Extent17].[Id]
    LEFT OUTER JOIN [model].[Encounters] AS [Extent18] ON [Extent17].[EncounterId] = [Extent18].[Id]
    LEFT OUTER JOIN [model].[Patients] AS [Extent19] ON [Extent18].[PatientId] = [Extent19].[Id]
    LEFT OUTER JOIN [model].[Invoices] AS [Extent20] ON [Extent2].[InvoiceId] = [Extent20].[Id]
    LEFT OUTER JOIN [model].[Encounters] AS [Extent21] ON [Extent20].[EncounterId] = [Extent21].[Id]
    LEFT OUTER JOIN [model].[Invoices] AS [Extent22] ON [Extent2].[InvoiceId] = [Extent22].[Id]
)  AS [Project1]
ORDER BY [Project1].[DateTime] ASC

如何让实体框架正常工作并生成正确的 SQL？

4

1 回答 1

0

我用下面的代码做了一个测试：

 using (TestEntities context = new TestEntities())
        {
            var qry = from a in context.Adjustments
                      join b in context.InvoiceReceivables on a.ID equals b.AdjustmentID
                      join c in context.Invoices on b.ID equals c.InvoiceRecievableID
                      select new { a.Name, IRName = b.Name, IName = c.Name};


            var list = qry.ToList();
        }

它生成了以下 SQL：

SELECT 
[Extent1].[ID] AS [ID], 
[Extent1].[Name] AS [Name], 
[Extent2].[Name] AS [Name1], 
[Extent3].[Name] AS [Name2]
FROM   [dbo].[Adjustments] AS [Extent1]
INNER JOIN [dbo].[InvoiceReceivable] AS [Extent2] ON [Extent1].[ID] = [Extent2].[AdjustmentID]
INNER JOIN [dbo].[Invoice] AS [Extent3] ON [Extent2].[ID] = [Extent3].[InvoiceRecievableID]

看起来您的模型有 5 个具有 1:1 关系的表，并且您使用了每个类一个表的方法。您可以考虑每个层次结构的表。

于 2013-11-20T18:06:22.810 回答