我有以下格式的XML:
(XmlDocument)JsonConvert.DeserializeXmlNode(requestBody, "root");
<root>
<NumberActa>20659</NumberActa>
<DegreeDate>09/10/2018</DegreeDate>
<StudentList>
<CostsCenter>ABK015q</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>LISSET MARCELA</Names>
</StudentList>
<StudentList>
<CostsCenter>ABCDE</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>MARCELA</Names>
</StudentList>
</root>
我有一个room元素,我需要添加<DA><DE></DE></DA>
并以以下格式添加它,而不是该元素
<DA>
<DE>
<NumberActa>20659</NumberActa>
<DegreeDate>09/10/2018</DegreeDate>
<StudentList>
<CostsCenter>ABK015q</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>LISSET MARCELA</Names>
</StudentList>
<StudentList>
<CostsCenter>ABCDE</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>MARCELA</Names>
</StudentList>
</DE>
</DA>
如何添加元素?
答案 0 :(得分:2)
您可以创建一个新节点并向其附加新节点,然后为每个子节点将该子节点附加到 节点:
string source= @" <root>
<NumberActa>20659</NumberActa>
<DegreeDate>09/10/2018</DegreeDate>
<StudentList>
<CostsCenter>ABK015q</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>LISSET MARCELA</Names>
</StudentList>
<StudentList>
<CostsCenter>ABCDE</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>MARCELA</Names>
</StudentList>
</root>";
var document = new XmlDocument();
document.LoadXml(source);
XmlNode oldRoot = document.SelectSingleNode("root");
XmlNode newRoot = document.CreateElement("DA");
XmlNode second = document.CreateElement("DE");
document.ReplaceChild(newRoot, oldRoot);
newRoot.AppendChild(second);
foreach (XmlNode childNode in oldRoot.ChildNodes)
{
second.AppendChild(childNode.CloneNode(true));
}
var result = document.InnerXml;
结果:
<DA>
<DE>
<NumberActa>20659</NumberActa>
<DegreeDate>09/10/2018</DegreeDate>
<StudentList>
<CostsCenter>ABK015q</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>LISSET MARCELA</Names>
</StudentList>
<StudentList>
<CostsCenter>ABCDE</CostsCenter>
<DocumentType>C.C h.</DocumentType>
<Names>MARCELA</Names>
</StudentList>
</DE>
</DA>
答案 1 :(得分:2)
谢谢:),我在下面的答案中找到了它,现在工作正常。
XmlDocument XmldocNew = new XmlDocument();
XmlElement newRoot = XmldocNew.CreateElement("DE");
XmldocNew.AppendChild(newRoot);
//newRoot.InnerXml = doc.DocumentElement.InnerXml;
newRoot.InnerXml = doc.DocumentElement.OuterXml;