使用Include

Question

我有以下代码

public class Question
{
    [Key]
    [DatabaseGenerated(DatabaseGeneratedOption.Identity)]
    public long Id { get; set; }

    [Required]
    public string QuestionText { get; set; }

    public virtual ICollection<QuestionOption> QuestionOptions { get; set; }
}

public class QuestionOption
{
    [Key]
    [DatabaseGenerated(DatabaseGeneratedOption.Identity)]
    public long Id { get; set; }

    [Required]
    public string OptionText { get; set; }

    [ForeignKey("QuestionId")]
    public Question Question { get; set; }
    public long QuestionId { get; set; }

    public virtual ICollection<QuestionOptionAnswer> QuestionOptionAnswers { get; set; }
}

public class QuestionOptionAnswer
{
    [Key]
    [DatabaseGenerated(DatabaseGeneratedOption.Identity)]
    public long Id { get; set; }

    [ForeignKey("QuestionOptionId")]
    public QuestionOption QuestionOption { get; set; }
    public long QuestionOptionId { get; set; }
}

和以下查询来检索问题：

var dbResult = await (from question in context.Question.Include(x => x.QuestionOptions).ThenInclude(x => x.QuestionOptionAnswers)
                                  where question.Id == id
                                  select question).FirstOrDefaultAsync();

一切正常，查询返回我真正需要的东西，除了我可以在QuestionOption中看到已经拥有的父Entity（Question）的事实。我遇到的情况是，一个问题有10个问题选项，每个选项都有问题父级实体返回。这是不希望的，因为我得到的东西已经多次出现。如何防止这种情况发生？

Answer 1

您不会多次获取父对象，每个父对象从数据库中作为一行返回，并且为每个QuestionOption创建一个对象。

尽管它在多个地方都被引用，包括您的 def binary_search(seq,item): """It uses non recursive method to search the item in the given seq. It returns the position of item if found, None otherwise""" left_index=0 right_index=len(seq)-1 while left_index <= right_index: #stop searching when left_index > right_indext mid_index=(right_index + left_index)//2 #find the mid point if seq[mid_index]==item: return mid_index elif seq[mid_index]>item: right_index = mid_index -1 #if mid point ele > search ele, move right pointer else: left_index = mid_index + 1 #if mid point ele < search ele, move left pointer return None a=[1,2,3,4,5,6,7,8,9] print(binary_search(a,6)) ，但对性能的影响很小。