在 C# 中解析 JCR XML 导出答案

【问题标题】：Parse JCR XML export in C#在 C# 中解析 JCR XML 导出
【发布时间】：2019-10-14 09:08:10
【问题描述】：

我从一位客户那里收到了一些需要用 C# 处理的 XML 文件。查看结构并进行一些谷歌搜索，似乎该内容已从 JCR 导出（我根本没有使用过这个，所以我可能错了）。结构如下：

<?xml version="1.0" encoding="UTF-8"?>
<sv:node xmlns:sv="http://www.jcp.org/jcr/sv/1.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" sv:name="foodtruck-gruener-sepp">
  <sv:property sv:name="jcr:primaryType" sv:type="Name">
    <sv:value>mgnl:Event</sv:value>
  </sv:property>
  <sv:property sv:name="jcr:mixinTypes" sv:type="Name" sv:multiple="true">
    <sv:value>mgnl:hasVersion</sv:value>
  </sv:property>
  <sv:property sv:name="jcr:uuid" sv:type="String">
    <sv:value>6371704f-1cc4-4ab9-9298-43c9dd4f79ab</sv:value>
  </sv:property>
  <sv:property sv:name="name" sv:type="String">
    <sv:value>user name</sv:value>
  </sv:property>
  <sv:property sv:name="email" sv:type="String">
    <sv:value>info@somedomain.com</sv:value>
  </sv:property>
</sv:node>

我将如何解析这个？这可以通过使用XMLElementAttribute 和XmlSerializer 来完成吗？

【问题讨论】：

与处理任何 XML 文件的方式相同。您可以使用 XmlSerializer 或 LINQ to XML。如果要在 Visual Studio 中生成强类型类，请转到 Edit> Paste Special > Paste XML as classes。如果你有一个 XSD，你可以从中生成类

标签： c# xml xmlserializer jcr

【解决方案1】：

下面的代码使用 xml linq 并将结果放入字典中。代码没有解析xml的Type属性

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


namespace ConsoleApplication1
{
    class Program
    {
        const string FILENAME = @"c:\temp\test.xml";
        static void Main(string[] args)
        {
            XDocument doc = XDocument.Load(FILENAME);
            XElement node = doc.Root;
            XNamespace sv = node.GetNamespaceOfPrefix("sv");

            Dictionary<string, string> dict = doc.Descendants(sv + "property")
                .GroupBy(x => (string)x.Attribute(sv + "name"), y => (string)y.Element(sv + "value"))
                .ToDictionary(x => x.Key, y => y.FirstOrDefault());
        }
    }
}

【讨论】：

【解决方案2】：

必须创建一个具有相同类型的 XML 的类，可以通过 Visual Studio Edit>Paste Special 来反序列化它。

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "http://www.jcp.org/jcr/sv/1.0")]
[System.Xml.Serialization.XmlRootAttribute(Namespace = "http://www.jcp.org/jcr/sv/1.0", IsNullable = false)]
public partial class node
{

    private nodeProperty[] propertyField;

    private string nameField;

    /// <remarks/>
    [System.Xml.Serialization.XmlElementAttribute("property")]
    public nodeProperty[] property
    {
        get
        {
            return this.propertyField;
        }
        set
        {
            this.propertyField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute(Form = System.Xml.Schema.XmlSchemaForm.Qualified)]
    public string name
    {
        get
        {
            return this.nameField;
        }
        set
        {
            this.nameField = value;
        }
    }
}

/// <remarks/>
[System.SerializableAttribute()]
[System.ComponentModel.DesignerCategoryAttribute("code")]
[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true, Namespace = "http://www.jcp.org/jcr/sv/1.0")]
public partial class nodeProperty
{

    private string valueField;

    private string nameField;

    private string typeField;

    private bool multipleField;

    private bool multipleFieldSpecified;

    /// <remarks/>
    public string value
    {
        get
        {
            return this.valueField;
        }
        set
        {
            this.valueField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute(Form = System.Xml.Schema.XmlSchemaForm.Qualified)]
    public string name
    {
        get
        {
            return this.nameField;
        }
        set
        {
            this.nameField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute(Form = System.Xml.Schema.XmlSchemaForm.Qualified)]
    public string type
    {
        get
        {
            return this.typeField;
        }
        set
        {
            this.typeField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlAttributeAttribute(Form = System.Xml.Schema.XmlSchemaForm.Qualified)]
    public bool multiple
    {
        get
        {
            return this.multipleField;
        }
        set
        {
            this.multipleField = value;
        }
    }

    /// <remarks/>
    [System.Xml.Serialization.XmlIgnoreAttribute()]
    public bool multipleSpecified
    {
        get
        {
            return this.multipleFieldSpecified;
        }
        set
        {
            this.multipleFieldSpecified = value;
        }
    }
}

【讨论】：