如何更改XML节点值

Question

我有一个XML(这正是它的样子):

<PolicyChangeSet schemaVersion="2.1" username="" description="">
    <Attachment name="" contentType="">
        <Description/>
        <Location></Location>
    </Attachment>
</PolicyChangeSet>

这是在用户的机器上.

我需要为每个节点添加值:用户名,描述,附件名称,内容类型和位置.

这是我到目前为止:

string newValue = string.Empty;
            XmlDocument xmlDoc = new XmlDocument();

            xmlDoc.Load(filePath);
            XmlNode node = xmlDoc.SelectSingleNode("/PolicyChangeSet");
            node.Attributes["username"].Value = AppVars.Username;
            node.Attributes["description"].Value = "Adding new .tiff image.";
            node.Attributes["name"].Value = "POLICY";
            node.Attributes["contentType"].Value = "content Typeeee";

            //node.Attributes["location"].InnerText = "zzz";

            xmlDoc.Save(filePath);

有帮助吗？

Answer 1

用XPath.XmlNode node = xmlDoc.SelectSingleNode("/PolicyChangeSet");选择您的根节点.

Answer 2

xmlDoc.SelectSingleNode("/PolicyChangeSet/Attachment/Description").InnerText = "My Description";
xmlDoc.SelectSingleNode("/PolicyChangeSet/Attachment/Location").InnerText = "My Location";

Answer 3

明白了这个——

xmlDoc.Load(filePath);
            XmlNode node = xmlDoc.SelectSingleNode("/PolicyChangeSet");
            node.Attributes["username"].Value = AppVars.Username;
            node.Attributes["description"].Value = "Adding new .tiff image.";

            node = xmlDoc.SelectSingleNode("/PolicyChangeSet/Attachment");
            node.Attributes["name"].Value = "POLICY";
            node.Attributes["contentType"].Value = "content Typeeee";
xmlDoc.Save(filePath);