簡體   English   中英

如何使用Restsharp反序列化給定的xml?

[英]How to deserialize the given xml with Restsharp?

我想問一個問題,以了解如何反序列化xml對象來自api。

<Sistemrent>
<Sube id="1">
    <Sube_Kodu>ABC</Sube_Kodu>
    <Sube_Ismi>AAA BBB</Sube_Ismi>
    <Kayit_ID>1</Kayit_ID>
</Sube>
<Sube id="2">
    <Sube_Kodu>BCD</Sube_Kodu>
    <Sube_Ismi>BBB CCC</Sube_Ismi>
    <Kayit_ID>1</Kayit_ID>
</Sube>
</Sistemrent>

您可以找到我們為解析該xml而生成的以下類。

[XmlRoot(ElementName = "Sistemrent")]
public class Sistemrent
{
    [XmlElement(ElementName = "Sube")]
    public List<Sube> Sube { get; set; }
}

[XmlRoot(ElementName = "Sube")]
public class Sube
{
    [XmlElement(ElementName = "Sube_Kodu")]
    public string Sube_Kodu { get; set; }
    [XmlElement(ElementName = "Sube_Ismi")]
    public string Sube_Ismi { get; set; }
    [XmlElement(ElementName = "Kayit_ID")]
    public string Kayit_ID { get; set; }
    [XmlAttribute(AttributeName = "id")]
    public string Id { get; set; }
}

它不會將xml轉換為對象,並且需要您的幫助。

謝謝

有用。 參見下面的代碼:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.IO;
using System.Xml;
using System.Xml.Serialization;


namespace Oppgave3Lesson1
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            XmlReader reader = XmlReader.Create(FILENAME);

            XmlSerializer serializer = new XmlSerializer(typeof(Sistemrent));

            Sistemrent sistemrent = (Sistemrent)serializer.Deserialize(reader);

        }
    }

    [XmlRoot(ElementName = "Sistemrent")]
    public class Sistemrent
    {
        [XmlElement(ElementName = "Sube")]
        public List<Sube> Sube { get; set; }
    }

    [XmlRoot(ElementName = "Sube")]
    public class Sube
    {
        [XmlElement(ElementName = "Sube_Kodu")]
        public string Sube_Kodu { get; set; }
        [XmlElement(ElementName = "Sube_Ismi")]
        public string Sube_Ismi { get; set; }
        [XmlElement(ElementName = "Kayit_ID")]
        public string Kayit_ID { get; set; }
        [XmlAttribute(AttributeName = "id")]
        public string Id { get; set; }
    }

}

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM