Include路径表达式必须引用定义的导航属性

Question

我正试图创建一个通用include，但我遇到了这个错误。怎么了？非常感谢。

Include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运算符属性。

_repository.FindWithIncludes(..., new List<...>
{
x => x.Property1,
x => x.Property2,
});   
public ICollection<TEntity> FindWithIncludes(Expression<Func<TEntity, bool>> currentExpression, List<Expression<Func<TEntity, object>>> propertiesToInclude)
{
using (var customContext = new TContext())
{
return customContext.Set<TEntity>().Include(x => propertiesToInclude.Select(currentProperty => currentProperty)).Where(currentExpression).ToList();
}
}

Answer 1

Include不能以这种方式使用：

.Include(x => propertiesToInclude.Select(currentProperty => currentProperty)

您所需要的相当于为列表的每个表达式调用Include：

.Include(x => x.Property1)
.Include(x => x.Property2)
...
.Include(x => x.PropertyN)

这可以用这样的代码来实现：

var query = customContext.Set<TEntity>().AsQueryable();
foreach (var property in propertiesToInclude)
query = query.Include(property); 
return query.Where(currentExpression).ToList();

或与使用Aggregate方法相同：

return propertiesToInclude
.Aggregate(customContext.Set<TEntity>().AsQueryable(), (q, p) => q.Include(p))
.Where(currentExpression).ToList();