这里的第一篇文章非常简单。
我一直在研究在我正在开发的应用程序中简化一些复杂的查询,并且我在下面稍微摸了一下。
所以说我有这两个类:
域实体" EmailRecipient" (与EF代码优先使用,因此期望使用相同的列名生成SQL表。)
public class EmailRecipient
{
public Guid Id { get; set; }
public string FriendlyName { get; set; }
public string ExchangeName { get; set; }
public string Surname { get; set; }
public string Forename { get; set; }
public string EmailAddress { get; set; }
public string JobTitle { get; set; }
public virtual List<SentEmail> SentEmails { get; set; }
}
和一个简单的JSON序列化类叫做#34; EmailLite&#34;定义为
public class EmailLite
{
public string EmailAddress { get; set; }
public Guid Id { get; set; }
public string FriendlyName { get; set; }
}
在我的专业EF6(.1.3)DbContext中,我有一个名为EmailRecipients的DbSet。
所以自然地对EmailRecipients执行这个linq表达式
EmailRecipients.Select(x => new EmailLite
{
Id = x.Id,
EmailAddress = x.EmailAddress,
FriendlyName = x.FriendlyName
});
生成的SQL是
SELECT
1 AS [C1],
[Extent1].[Id] AS [Id],
[Extent1].[EmailAddress] AS [EmailAddress],
[Extent1].[FriendlyName] AS [FriendlyName]
FROM [dbo].[EmailRecipients] AS [Extent1]
为什么我这样做:
Func<EmailRecipient, EmailLite> projectionFunction = x => new EmailLite
{
Id = x.Id,
EmailAddress = x.EmailAddress,
FriendlyName = x.FriendlyName
};
EmailRecipients.Select(projectionFunction);
我是否获得了以下(完整)SQL生成的内容:
SELECT
[Extent1].[Id] AS [Id],
[Extent1].[FriendlyName] AS [FriendlyName],
[Extent1].[ExchangeName] AS [ExchangeName],
[Extent1].[Surname] AS [Surname],
[Extent1].[Forename] AS [Forename],
[Extent1].[EmailAddress] AS [EmailAddress],
[Extent1].[JobTitle] AS [JobTitle],
[Extent1].[SubscribedOn] AS [SubscribedOn]
FROM [dbo].[EmailRecipients] AS [Extent1]
非常感谢任何帮助!
干杯, 饱和
答案 0 :(得分:3)
IQueryable<T>.Select()
以Expression<Func<T,TOut>>
作为参数,您实际使用的函数是IEnumerable<T>.Select()
,它接受一个委托。因此,您告诉EF,从那一刻开始,您使用IEnumerable
而不是IQueryable
,其余查询将在内存中执行=&gt;你正在获取所有列。
EmailRecipients <-- in memory from here on --> .Select(projectionFunction);
您需要做的就是将projectionFunction
更改为表达式,它将起作用:
Expression<Func<EmailRecipient, EmailLite>> projectionFunction = x => new EmailLite
{
Id = x.Id,
EmailAddress = x.EmailAddress,
FriendlyName = x.FriendlyName
};