包含路径表达式必须引用 type.in 预加载中定义的导航属性

The Include path expression must refer to a navigation property defined on the type.in eager loading(包含路径表达式必须引用 type.in 预加载中定义的导航属性)

本文介绍了包含路径表达式必须引用 type.in 预加载中定义的导航属性的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我尝试包含这样的匿名类型:除了 CompanyTitle ,PeriodTypeName )

I try to include anonymous type like this : I want all incomelist attributes in addition to CompanyTitle ,PeriodTypeName )

 var incomeList = ctx.IncomeLists.Include(i => new
                {
                    CompanyTitle = i.CompanyId.ToString() + "/" + i.Company.CompanyName,
                    PeriodTypeName = i.ListPeriods.Select(lp => lp.PeriodType.PeriodTypeName)
                }).ToList()

<小时>

我的模型部分是这样的:


My model section like this :

但我得到以下异常:

包含路径表达式必须引用导航属性在类型上定义.使用虚线路径进行参考导航属性和用于集合导航的 Select 运算符特性.参数名称:路径

The Include path expression must refer to a navigation property defined on the type. Use dotted paths for reference navigation properties and the Select operator for collection navigation properties. Parameter name: path

结果应该是 Gridview 的数据源.

推荐答案

您不能使用 Include 来选择这样的数据.Include 用于加载相关数据.您应该使用 Include 加载您的实体,然后选择您想要的.请记住从 CompanyId 中删除 .ToString().EF 会为你做这件事.您的查询应如下所示:

You cannot use Include to select data like this. Include is used to load related data. You should load your entities using Include then select what you want. Remember to remove .ToString() from CompanyId. EF will do it for you. Your query should look like this:

var incomeList = ctx.IncomeLists
    .Include(i => i.Company)
    .Include(i => i.ListPeriods.Select(lp => lp.PeriodType))
    .Select(i => new 
    {
        CompanyTitle =  i.CompanyId + "/" + i.Company.CompanyName,
        PeriodTypeNames = i.ListPeriods.Select(lp => lp.PeriodType.PeriodTypeName)
    })
    .ToList();

这篇关于包含路径表达式必须引用 type.in 预加载中定义的导航属性的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持编程学习网!

本文标题为:包含路径表达式必须引用 type.in 预加载中定义的导航属性

基础教程推荐