SelectMany应用于3个列表

时间:2017-03-21 14:50:09

标签: c# entity-framework linq linq-to-entities entity-framework-core

我需要创建一个对象列表,OrganizationEnrolment,使用3个Entity Framework实体(User,EnrolmentType和OrganizationEnrolment)创建:

List<OrganizationEnrolment> organizationEnrolments = 
  context.Organizations
  .SelectMany(x => context.Users)
  .SelectMany(x => context.EnrolmentTypes)
  .Select(y => new OrganizationEnrolment {
     EnrolmentType = enrolmentType
     Organization = organization,
     User = user
   })

我的问题是在让SelectMany如何从3个连接表中获取enrolmentType,组织和用户之后?请注意以下代码:

EnrolmentType = enrolmentType,
Organization = organization,
User = user

无效,因为我没有变量enrolmentType,organization和user。

2 个答案:

答案 0 :(得分:4)

这是你在找什么?

var query= from o in context.Organizations
           from u in context.Users
           from t incontext.EnrolmentTypes
           select new OrganizationEnrolment {
                      EnrolmentType = t,
                      Organization = o,
                      User = u
                   };

使用方法语法将是:

var query= context.Organizations.SelectMany(o=>context.EnrolmentTypes
                                .SelectMany(t=>context.Users
                                .Select(u=>new OrganizationEnrolment {
                                                   EnrolmentType = t,
                                                   Organization = o,
                                                    User = u
                                                  })));

答案 1 :(得分:1)

您可以使用查询语法来获取所有这些变量:

from o in context.Organizations
from u in context.Users
from et in context.EnrolmentTypes
select new OrganizationEnrolment {
     EnrolmentType = et
     Organization = o,
     User = u
   }

每个本地范围变量将在select语句中可见。

Lambda语法(使用EF6检查):

context.Organizations.SelectMany(
          o => context.Users.SelectMany(
              u => context.EnrolmentTypes.Select(
                  et => new OrganizationEnrolment {
                       EnrolmentType = et
                       Organization = o,
                       User = u
                  })
              )
         )