简体   繁体   中英

C# Get all text from xml node including xml markup using Visual studio generated class

Using xml to c# feature in visual studio converted the below xml markup into C# class.

<books>
<book name="Book-1">
<author>
<name>Author-1<ref>1</ref></name>
</author>
</book>
<book name="Book-2">
<author>
<name>Author-1<ref>1</ref></name>
</author>
</book>
</books>

The converted class contains

[System.Xml.Serialization.XmlTypeAttribute(AnonymousType = true)]
    public partial class booksBookAuthorName
    {
        private byte refField;     // 1

        private string[] textField; // 2

What we need is Author-1<ref>1</ref> as the output when reading the author name, by keeping tags.

We have tried https://msdn.microsoft.com/en-us/library/system.xml.serialization.xmltextattribute(v=vs.110).aspx attribute. But no hope.

Any clues ??

You can use [XmlAnyElement("name")] to capture the XML of the <name> node of each author into an XmlElement (or XElement ). The precise text you want is the InnerXml of that element:

[XmlRoot(ElementName = "author")]
public class Author
{
    [XmlAnyElement("name")]
    public XmlElement NameElement { get; set; }

    [XmlIgnore]
    public string Name
    {
        get
        {
            return NameElement == null ? null : NameElement.InnerXml;
        }
        set
        {
            if (value == null)
                NameElement = null;
            else
            {
                var element = new XmlDocument().CreateElement("name");
                element.InnerXml = value;
                NameElement = element;
            }
        }
    }
}

This eliminates the need for the booksBookAuthorName class.

Sample fiddle showing deserialization of a complete set of classes corresponding to your XML, generated initially from http://xmltocsharp.azurewebsites.net/ after which Author was modified as necessary.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM