Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
619 views
in Technique[技术] by (71.8m points)

c# - Xml Sequence deserialization with RestSharp

I have this xml feed from an API with a XML sequence.

<?xml version="1.0" encoding="UTF-8" ?>
<Function>
    <Cmd>2002</Cmd>
    <Status>1</Status>
    <Cmd>2003</Cmd>
    <Status>0</Status>
    <Cmd>2004</Cmd>
    <Status>0</Status>
    <Cmd>1012</Cmd>
    <Status>3</Status>
    <Cmd>2006</Cmd>
    <Status>0</Status>
    <Cmd>2007</Cmd>
    <Status>0</Status>
    ...
</Function>

I already tried a few options for deserialization with Restsharp. Ideally I would like to have something like the following, but it's obviously not working.

public class MyResponse
{
    public List<Setting> Settings { get; set;}
}

public class Setting
{
    public int Cmd { get; set; }
    public int Status { get; set; }
}

Thank you

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

You can use the DotNetXmlDeserializer of RestSharp to make Microsoft's XmlSerializer do the actual deserialization. Define your MyResponse class as follows, using XML serialization attributes to specify element names and also special handling for the Cmd/Status alternating sequence of elements:

[XmlRoot("Function")]
public class MyResponse
{
    [XmlIgnore]
    public List<Setting> Settings { get; set; }

    /// <summary>
    /// Proxy property to convert Settings to an alternating sequence of Cmd / Status elements.
    /// </summary>
    [Browsable(false), EditorBrowsable(EditorBrowsableState.Never)]
    [XmlAnyElement]
    public XElement[] Elements 
    {
        get
        {
            if (Settings == null)
                return null;
            return Settings.SelectMany(s => new[] { new XElement("Cmd", s.Cmd), new XElement("Status", s.Status) }).ToArray();
        }
        set
        {
            if (value == null)
                Settings = null;
            else
                Settings = value.Where(e => e.Name == "Cmd").Zip(value.Where(e => e.Name == "Status"), (cmd, status) => new Setting { Cmd = (int)cmd, Status = (int)status }).ToList();
        }
    }
}

Then deserialize as follows:

        var serializer = new DotNetXmlDeserializer();
        var myResponse = serializer.Deserialize<MyResponse>(response);

Prototype fiddle.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...