EF 4.1 代码优先：EF 生成的 sql 和自定义 sql 的区别答案

【问题标题】：EF 4.1 code-first: difference between EF generated sql and custom sqlEF 4.1 代码优先：EF 生成的 sql 和自定义 sql 的区别
【发布时间】：2011-10-13 21:50:44
【问题描述】：

我有一个关于实体框架生成的sql和手写的问题。在我的项目中，我有一些实体（它们对于这个 Q 并不重要），举个简单的例子，当我使用这段代码时：

var query = context.Employees.Select(e => new {
                PersonalCode = e.PersonelCode,
                Fname = e.Person.Fname,
                Family = e.Person.Family,
                Email = e.Person.Emails
            });

生成的sql是这样的：

SELECT 
[Project1].[EmployeeID] AS [EmployeeID], 
[Project1].[EmployeeID1] AS [EmployeeID1], 
[Project1].[PersonID] AS [PersonID], 
[Project1].[EmployeeID2] AS [EmployeeID2], 
[Project1].[PersonID1] AS [PersonID1], 
[Project1].[PersonelCode] AS [PersonelCode], 
[Project1].[Fname] AS [Fname], 
[Project1].[Family] AS [Family], 
[Project1].[C1] AS [C1], 
[Project1].[EmailID] AS [EmailID], 
[Project1].[Mail] AS [Mail]
FROM ( SELECT 
    [Extent1].[EmployeeID] AS [EmployeeID], 
    [Extent1].[PersonelCode] AS [PersonelCode], 
    [Join1].[PersonID] AS [PersonID], 
    [Join1].[Fname] AS [Fname], 
    [Join1].[EmployeeID] AS [EmployeeID1], 
    [Join3].[PersonID] AS [PersonID1], 
    [Join3].[Family] AS [Family], 
    [Join3].[EmployeeID] AS [EmployeeID2], 
    [Join5].[EmailID1] AS [EmailID], 
    [Join5].[Mail] AS [Mail], 
    CASE WHEN ([Join5].[EmailID2] IS NULL) THEN CAST(NULL AS int) ELSE 1 END AS [C1]
    FROM    [dbo].[Employees] AS [Extent1]
    LEFT OUTER JOIN  (SELECT [Extent2].[PersonID] AS [PersonID], [Extent2].[Fname] AS [Fname], [Extent3].[EmployeeID] AS [EmployeeID]
        FROM  [dbo].[Persons] AS [Extent2]
        LEFT OUTER JOIN [dbo].[Employees] AS [Extent3] ON [Extent2].[PersonID] = [Extent3].[EmployeeID] ) AS [Join1] ON [Extent1].[EmployeeID] = [Join1].[PersonID]
    LEFT OUTER JOIN  (SELECT [Extent4].[PersonID] AS [PersonID], [Extent4].[Family] AS [Family], [Extent5].[EmployeeID] AS [EmployeeID]
        FROM  [dbo].[Persons] AS [Extent4]
        LEFT OUTER JOIN [dbo].[Employees] AS [Extent5] ON [Extent4].[PersonID] = [Extent5].[EmployeeID] ) AS [Join3] ON [Extent1].[EmployeeID] = [Join3].[PersonID]
    LEFT OUTER JOIN  (SELECT [Extent6].[EmailID] AS [EmailID2], [Extent6].[PersonID] AS [PersonID], [Extent7].[EmailID] AS [EmailID1], [Extent7].[Mail] AS [Mail]
        FROM  [dbo].[EmailsForPersons] AS [Extent6]
        INNER JOIN [dbo].[Emails] AS [Extent7] ON [Extent6].[EmailID] = [Extent7].[EmailID] ) AS [Join5] ON [Join5].[PersonID] = [Extent1].[EmployeeID]
)  AS [Project1]
ORDER BY [Project1].[EmployeeID] ASC, [Project1].[EmployeeID1] ASC, [Project1].[PersonID] ASC, [Project1].[EmployeeID2] ASC, [Project1].[PersonID1] ASC, [Project1].[C1] ASC

但是通过这段代码：

SELECT     Employees.PersonelCode, Persons.Fname, Persons.Family, Emails.Mail
FROM         Employees 
                        LEFT OUTER JOIN     -- or: INNER JOIN
                        Persons ON Employees.EmployeeID = Persons.PersonID 
                                LEFT OUTER JOIN
                                EmailsForPersons ON Persons.PersonID = EmailsForPersons.PersonID 
                                            LEFT OUTER JOIN
                                            Emails ON EmailsForPersons.EmailID = Emails.EmailID

我会给出同样的结果！这些代码有什么区别？哪一个有更高的性能和更高的速度？

【问题讨论】：

标签： sql sql-server performance tsql entity-framework

【解决方案1】：

您可以对这两个查询进行分析和采样，看看哪个表现更好？

另见How to clean & optimise code generated by WCF OData service?

EF 生成的 SQL 非常通用，需要在各种情况下工作。无论出于何种原因，它都非常冗长。它通常有一个SELECT [Col1] FROM (SELECT [Col1] ...) 嵌套结构，以及许多用于比较的CAST 语句。

这样做是为了确保最大的兼容性和最小化某人的棘手查询无法被翻译的可能性，还是因为生成 SQL 的代码更加清晰和简单，我们只能猜测。这是实体框架团队内部做出的设计决策。

坦率地说，我根本不会担心这一点除非您使用查询分析器并排测试这两个查询的性能。我希望两者之间的差异非常小。

如果生成的查询的性能较差，那么最简单的模式是在存储过程中编写逻辑并让 EF 调用存储过程。这会将所有控制权从 EF 手中拿走。

【讨论】：