Question

鉴于以下内容：

public class Person
{
  public int ID { get; set;}
  public string Name { get; set;}
  public IQueryable<Pet> Pets { get;set; }
}

public class Pet
{
  public int id { get; set; }
  public int OwnerId { get; set; }
  public string Name { get; set; }
}

public class SearchCriteria
{
  public string PersonName {get; set; }
  public List<string> PetNames {get; set;}
}

使用IQueryable进行搜索时，实施所有人的选择

public List<Person> GetWithPets(SearchCriteria search)
{
     var people = (from p in context.People
                   where p.Name == search.PersonName
                   select new Person{
                          ID = p.ID,
                        Name = p.Name,
                        Pets = (from pt in context.Pets
                                where pt.OwnerId == p.ID
                                select new Pet {
                                   id = pt.ID,
                                 OwnerId = pt.OwnerId,
                                 Name = pt.Name
                                }).AsQueryable
                   }).AsQueryable();

       foreach(var str in search.PetNames)
       {
            people = people.Where(o=>o.Pets.Any(p=>p.Name == str));
       }
  return people.ToList();
}

我的问题是，无论搜索名称的foreach，在返回的人员列表中，即使有，宠物也是null 与那个人有关的宠物，我哪里出错？

修改

public class Person
{
    public int ID { get; set; }
    public string Name { get; set; }
    public IQueryable<Animal> Pets { get; set; }
}

public class Animal
{
    public int id { get; set; }
    public int? OwnerId { get; set; }
    public string Name { get; set; }
}

public class SearchCriteria
{
    public string PersonName { get; set; }
    public List<string> PetNames { get; set; }
}



class Program
{
    public static List<Person> GetWithPets(SearchCriteria search)
    {
        using (DatabaseEntities context = new DatabaseEntities())
        {
            var people = (from p in context.Peoples
                          where p.Name == search.PersonName
                          select new Person
                          {
                              ID = p.ID,
                              Name = p.Name,
                              Pets = (from pt in context.Pets
                                      where pt.OwnerID == p.ID
                                      select new Animal
                                      {
                                          id = pt.ID,
                                          OwnerId = pt.OwnerID,
                                          Name = pt.Name
                                      }).AsQueryable()
                          }).AsQueryable();

            foreach (var str in search.PetNames)
            {
                people = people.Where(o => o.Pets.Any(p => p.Name == str));
            }
            return people.ToList();
        }
    }

Answer 1

如果 Person和Pet是模型的实体，如果 Person.Pets是{{1}的导航属性实体和如果您想拥有包含所有完整Pet实体的完整Person实体，并引用您的评论......

我的方法应该返回名称相等的人员列表 search.PersonName＆amp;他们所有的宠物，但只有那些拥有它的人在search.PetNames
中有这些名字的宠物

......你可以用这个：

Pet

Answer 2

你可以试试这样的事吗

var people = (from p in context.People
                   where p.Name == search.PersonName
                   select new Person{
                          ID = p.ID,
                        Name = p.Name,
                        Pets = (from pt in context.Pets
                                where pt.OwnerId == p.ID
                                select new Pet {
                                   id = pt.ID,
                                 OwnerId = pt.OwnerId,
                                 Name = pt.Name
                                })
                   }).Include("Pets").AsQueryable();

Answer 3

或者你可以试试这样的东西

foreach(var str in search.PetNames)
       {
            people.Concat(people.Where(o=>o.Pets.Any(p=>p == str)).Include('Pet'));
       }

Answer 4

将AsQueryable()添加到Pets集合：

Pets = (from pt in context.Pets  
        where pt.OwnerId == p.ID  
        select new Pet {  
            id = pt.ID,  
            OwnerId = pt.OwnerId,  
            Name = pt.Name  
        }).AsQuerable()

列出IQueryable对象内部

4 个答案: