Question

我必须处理来自我无法控制的第三方REST API的XML响应。问题是XML响应的格式不正确，如下所示：

<XML>
<Meta>
<Status>Success</Status>
<Debug/>
</Meta>
<Result>
<abc>
 <DivisionID>tttttttttt</DivisionID>
 <UserName><![CDATA[ xxx#xxxxx]]></UserName>
 <UserFirstName>xxxx</UserFirstName>
 <UserLastName>xxxx</UserLastName>
 <UserAccountType>xxxxxxx</UserAccountType>
 <UserEmail>xxxxx@xxxxx.xom</UserEmail>
 <UserAccountStatus>Active</UserAccountStatus>
</abc>
<def>
 <DivisionID/>
 <UserName><![CDATA[ xxx#xxxx]]></UserName>
 <UserFirstName>yyyy</UserFirstName>
 <UserLastName>vvvvvv</UserLastName>
 <UserAccountType>uuuuuuuuu</UserAccountType>
 <UserEmail>oooo@vvvvvv</UserEmail>
 <UserAccountStatus>Active</UserAccountStatus>
</def>
....contd
</Result>



       var requestUri = new Uri(uriString);
       HttpWebRequest httprequest =  (HttpWebRequest)WebRequest.Create(requestUri);
    var httpresponse = (HttpWebResponse)httprequest.GetResponse();
    Person people = new Person();
    List<Person> lstPerson = (from _person in         xmlDoc.Document.Element("Result").Elements("Result")
      select new Person
      {
      userName = Xdocument.Load(httpresponse.GetResponseStream()).Root.ToString(),
      userEmail = _person.Element("UserEmail").Value
      }).ToList();

我需要检索具有值＆＃34; abc＆＃34;的节点。和＆＃34; def＆＃34;并将它们存储在UserName中，UserName本身就是根节点，并且还要检索它们之间的值。所以如何做到这一点我尝试了各种各样的方法，但却无法这样做。

Answer 1

<强>更新

要使用Person的元素名称和userName的{{1}}子元素的值创建UserEmail类的列表，您可以执行以下操作：

userEmail

使用extension method

        try
        {
            // Load the XML from the external site
            XDocument xmlDoc;

            var requestUri = new Uri(uriString);
            HttpWebRequest httprequest = (HttpWebRequest)WebRequest.Create(requestUri);

            using (var httpresponse = (HttpWebResponse)httprequest.GetResponse())
            using (var stream = httpresponse.GetResponseStream())
            using (var reader = new StreamReader(stream))
            {
                xmlDoc = XDocument.Load(reader);
            }

            // Extract the name & email.
            var people = xmlDoc
                // Get the "Result" node
                .Root.Elements(xmlDoc.Root.Name.Namespace + "Result")
                // Loop through its elements
                .SelectMany(result => result.Elements())
                // Deserialize the element name and UserEmail sub-element value as a Person
                .Select(element => new Person { userName = element.Name.LocalName, userEmail = element.Element(xmlDoc.Root.Name.Namespace + "UserEmail").ValueSafe() })
                .ToList();

            // Process or return the list of people
        }
        catch (Exception ex)
        {
            // Handle any web exception encountered.
            Debug.WriteLine(ex);
            // Or rethrow if it can't be handled here
            throw;
        }

原始答案

你的问题不清楚。但如果你问的话

如果我从某些XML加载了public static class XObjectExtensions { public static string ValueSafe(this XElement element) { return element == null ? null : element.Value; } }，当列表的每个元素都有自定义元素名称时，有没有办法使用XDocument反序列化嵌入列表的部分？

然后你可以这样做：

加载XmlSerializer
导航到要反序列化的列表。
对于每个元素，请记住元素名称，然后使用类名覆盖。
使用XElement.CreateReader()为该元素创建XDocument，然后将其传递给XmlReader进行反序列化。

即：

XmlSerializer

给定XML字符串

    // Load the document
    var doc = XDocument.Parse(xml);

    var people = doc
        // Navigate to the list
        .Root.Elements("Result")
        .SelectMany(r => r.Elements())
        // Deserialize each element in the list as a KeyValuePair<string, Person>
        .Select(element =>
            {
                var name = element.Name;
                element.Name = typeof(Person).DefaultXmlElementName(); // Overwrite name with name used by serializer.
                using (var reader = element.CreateReader())
                {
                    var person = (Person)new XmlSerializer(typeof(Person)).Deserialize(reader);
                    return new KeyValuePair<string, Person>(name.LocalName, person);
                }
            })
        .ToList();

使用扩展方法：

    string xml = @"<XML>
        <Meta>
            <Status>Success</Status>
            <Debug/>
        </Meta>
        <Result>
            <abc>
                 <DivisionID>tttttttttt</DivisionID>
                 <UserName><![CDATA[ xxx#xxxxx]]></UserName>
                 <UserFirstName>xxxx</UserFirstName>
                 <UserLastName>xxxx</UserLastName>
                 <UserAccountType>xxxxxxx</UserAccountType>
                 <UserEmail>xxxxx@xxxxx.xom</UserEmail>
                 <UserAccountStatus>Active</UserAccountStatus>
            </abc>
            <def>
                 <DivisionID/>
                 <UserName><![CDATA[ xxx#xxxx]]></UserName>
                 <UserFirstName>yyyy</UserFirstName>
                 <UserLastName>vvvvvv</UserLastName>
                 <UserAccountType>uuuuuuuuu</UserAccountType>
                 <UserEmail>oooo@vvvvvv</UserEmail>
                 <UserAccountStatus>Active</UserAccountStatus>
            </def>
        </Result>
    </XML>
    ";

处理XML响应并转换为格式不正确的集合

1 个答案: