Question

我有几个Entity Framework 7（核心）实体：

> filestring
[1] "// Two-Dimensional Dynamic Array#include <iostream>using namespace std;typedef int* IntArrayPtr;int main(){  int d1, d2;  cout << \"Enter the row and col dims of the array:"

我有一个字符串数组如下：

public class Person {
   public virtual Address Address { get; set; }
   public virtual ICollection<Hobby> Hobbies { get; set; }
}

public class Address {
   public String Street { get; set; }
   public virtual Country Country { get; set; }
}

鉴于我得到了这个字符串数组：

String[] entities = new String[] { "Hobbies", "Address.Country" }

在EF6中，我可以做类似的事情：

context.Persons
  .Include(x => x.Hobbies)
  .Include(x => x.Address).ThenInclude(x => x.Country);

但在EF7 Include中不允许使用字符串。我创建了字典：

context.Persons.Include(entities[0]).Include(entities[1]);

会有类似的内容：

private readonly Dictionary<String, LambdaExpression> _properties = new Dictionary<String, LambdaExpression>();

我有扩展名：

x => x.Hobbies is for "Hobbies"
x => x.Address.Country is for "Address.Country"

我需要给予词典应用以下内容：

对于＆＃34; x =＆gt; x.Hobbies＆＃34;只是做：

public static IQueryable<T> Include<T>(this IQueryable<T> source, Dictionary<String, LambdaExpression> properties) {
}

如果表达式类似于＆＃34; x =＆gt; x.Address.Country＆＃34;添加：
```
 source.Include(x => x.Hobbies);
```

可以这样做吗？

Answer 1

不确定ThenInclude()和EF 7，但您可以在您的存储库中执行类似的操作（使用EF 6测试）：

public MyEntity GetMyEntity_EagerlyLoad(DbContext context, int id, params Expression<Func<MyEntity, object>>[] propertiesToLoad)
{
    var q = context.MyEntities.Where(m => m.ID == id);

    foreach (var prop in propertiesToLoad)
        q = q.Include(prop);

    return q.SingleOrDefault();
}

然后你可以这样称呼它：

repo.GetMyEntity_EagerlyLoad(context, id, m => m.Property1, m => m.Property2, m => m.Property1.NestedProperty)

编辑：还有另一种方法可以使用投影对EF进行急切加载。你可以这样做一个通用的存储库方法：

public MyEntity GetMyEntity_EagerlyLoad<T>(DbContext context, int id, Expression<Func<MyEntity, T>> loadingProjection, Func<T, MyEntity> resultProjection)
{
    var q = context.MyEntities.Where(m => m.ID == id);

    return q.Select(loadingProjection).AsEnumerable().Select(resultProjection).SingleOrDefault();
}

然后使用要加载的属性和希望方法返回的实体调用它：

repo.GetMyEntity_EagerlyLoad(context, id, m => new { myEntity = m, m.Property1, m.Property2, m.Property1.NestedProperty }, m => m.myEntity)

根据Lambda表达式字典应用Include和ThenInclude

1 个答案: