使用LINQ从XML读取对象

Question

我正在尝试从XML读取并输入对象列表。 我的XML是

<User>
    <Name>John</Name>
    <Role>Admin</Role>
    <QAList>
        <QA>
            <Question> Question 1 </Question>
            <Answers>
                <Answer> Answer 1 </Answer>
                <Answer> Answer 2 </Answer>
                <Answer> Answer 3 </Answer>
            </Answers>
        </QA>
    </QAList>
</User>

这是我的对象类：

public class ListQuestionAnswers: INotifyPropertyChanged
{
    private List<QuestionAnswer> _questionAnswer;
    public List<QuestionAnswer> QuestionAnswerList
    {
        get { return _questionAnswer; }
        set { _questionAnswer = value; }
    }

    private string _name;
    public string Name
    {
        get { return _name; }
        set { _name = value; }
    }

    private string _role;
    public string Role
    {
        get { return _role; }
        set { _role = value; }
    }
}

public class QuestionAnswer
{
    public string _question;
    public List<AnswerList> _answers;

    public string Question
    {
        get { return _question; }
        set { _question = value; }
    }

    public List<AnswerList> Answers
    {
        get { return _answers; }
        set { _answers = value; }
    }
}

public class AnswerList
{
    private string _answer;

    public string Answer
    {
        get { return _answer; }
        set {_answer = value; }
    }
}

我可以获得名称和角色，但是如何获得质量检查清单？ 我尝试在查询中编写一个选择，但似乎没有用。

XDocument xmlDoc = XDocument.Load(@"C:\Test\Sample.xml");
var files = (from x in xmlDoc.Elements("User")
             select new TestCasesViewModel
             {
                 Name = (string)x.Element("Name").Value ?? string.Empty,
                 Role = (string)x.Element("Role").Value ?? string.Empty
             }).ToList();

Answer 1

您可以执行以下操作：

var files = (from x in xmlDoc.Elements("User")
             select
             new TestCasesViewModel
                  {
                      Name = (string)x.Element("Name") ?? string.Empty,
                      Role = (string)x.Element("Role") ?? string.Empty,
                      QuestionAnswerList = 
                          x.Descendants("QA")
                          .Select(q => new QuestionAnswer
                          {
                              Question = (string)q.Element("Question"),
                              Answers = q.Descendants("Answer").Select(a => new AnswerList { Answer = (string)a}).ToList()

                          }).ToList()
                   }).ToList();

请注意，如果您正在使用显式强制转换，则不需要访问Value属性，使用显式强制转换的整个目的是避免可能的异常，如果您使用Value属性，则显式强制转换将变得多余，因为如果找不到元素无论如何都会抛出异常...

Answer 2

在这种情况下，我发现将复杂的LINQ表达式拆分为多个子表达式非常有帮助，即使只是为了帮助开发和调试也是如此。

在不进一步了解真实xml数据的结构的情况下，这里有一些想法：

XDocument xmlDoc = XDocument.Load(@"C:\Test\Sample.xml");

var users = xmlDoc.Elements("User");

foreach(var user in users)
{
    var qaList = user.Element("QAList");

    foreach(var qa in qaList.Elements("QA"))
    {
        //at this point, you should have access the <Question> node e.g.
        string question = qa.Element("Question").Value;

        //and the list of answers would be similar to before
        var answers = qa.Element("Answers");

        foreach(var answer in answers.Elements("Answer"))
        {
            //now answer.Value should be the text of the answer
        } 
    }        
}

同样，这不是“最干净”的方法，当然还有更简洁的方法可以在LINQ中表达它。 关键是，您有更多的断点/调试选项，可以浏览正在使用的对象。