It's my fist time working with XML and I use a basic XML file format like this:
<root>
<staff id="1">
<name>name 1</name>
<phone>123456</phone>
</staff>
<staff id="2">
<name>name 2</name>
<phone>123789</phone>
<phone2>123789</phone2>
</staff>
</root>
some nodes have more elements (phone2 in this case). I want to add (or remove) an element on a node. I'm creating a WinForms in C# that work with this XML. I'm doing:
- I read the XML to have a XmlNodeList variable.
- From XmlNodeList I get the node that I want modify to a XmlNode variable.
- I modify name or phone on XmlNode
- I read again the XML file and I update the correct node with the XmlNode variable new info.
My problem is I don't know how add (or remove) the element "phone2" on my XmlNode variable.
program.cs:
public static XmlNode staff;
public static XmlNodeList xnList = GetList();
public static XmlNodeList GetList()
{
XmlNodeList xnList;
XmlDocument doc = new XmlDocument();
doc.Load(path);
xnList = doc.SelectNodes("/root/staff");
return xnList;
}
public static void GetID(string id)
{
foreach (XmlNode xn in xnList)
{
if(xn.Attributes["id"].Value == id)
{
staff = xn;
}
}
}
form1.cs
private void btnOK_Click(object sender, EventArgs e)
{
Program.staff["name"].InnerText = textBoxName.Text;
Program.staff["phone"].InnerText = textBoxPhone.Text;
if (Program.staff.SelectSingleNode("phone2") == null)
{
// here I want to create "phone2" in Program.staff if not exist
// to update XML file later.
Program.staff["phone2"].InnerText = textBoxPhone2.Text;
}
}
I don't find the correct method to do it and maybe it's not the best way to do it, but I accept suggestions...
CodePudding user response:
There are multiple ways to work with XML files. I'll show two options below.
Test.xml:
<root>
<staff id="1">
<name>Name 1</name>
<phone>123456</phone>
</staff>
<staff id="2">
<name>Name 2</name>
<phone>123457</phone>
<phone>123458</phone>
</staff>
</root>
Option 1 (LINQ to XML):
Add the following using directive:
using System.Xml.Linq;
CreateXmlLinq:
private void CreateXmlLinq(string filename)
{
XElement root = new XElement("root",
new XElement("staff", new XAttribute("id", "1"),
new XElement("name", "Name 1"),
new XElement("phone", "123456")),
new XElement("staff", new XAttribute("id", "2"),
new XElement("name", "Name 2"),
new XElement("phone", "123457"),
new XElement("phone", "123458"))
);
root.Save(filename);
}
Usage:
using (SaveFileDialog sfd = new SaveFileDialog())
{
sfd.Filter = "XML File (*.xml)|*.xml";
sfd.FileName = "Test.xml";
if (sfd.ShowDialog() == DialogResult.OK)
{
//save to file
CreateXmlLinq(sfd.FileName);
Debug.WriteLine($"Info: Saved to {sfd.FileName}");
}
}
To remove phone number 123458
where staff id = 2:
RemovePhone:
private void RemovePhone(string filename, string id, string phoneNumber)
{
//load from file
XElement root = XElement.Load(filename);
//remove specified phone number
root.Elements("staff").Where(s => s.Attribute("id").Value == id).Elements("phone").Where(p => p.Value == phoneNumber).Remove();
//save to file
root.Save(filename);
}
Option 2 (XML Serialization):
For this approach, we'll use nested classes.
Add the following using directives to each of the classes:
using System.Collections.Generic;
using System.Xml;
using System.Xml.Serialization;
You can name the classes whatever you like, I've chosen to prepend the word "Xml". Additionally for the nested classes, I've chosen to append the ancestors' names. In this case, there is only one ancestor (the parent) "root".
- XmlRoot
- XmlRootStaff: ("XmlRoot" "Staff)
XmlRoot.cs:
[XmlRoot(ElementName = "root", IsNullable = false)]
public class XmlRoot
{
[XmlElement(ElementName = "staff")]
public List<XmlRootStaff> Staff { get; set; } = new List<XmlRootStaff>();
}
XmlRootStaff.cs:
public class XmlRootStaff
{
[XmlAttribute(AttributeName = "id")]
public string Id { get; set; }
[XmlElement(ElementName = "name")]
public string Name { get; set; }
[XmlElement(ElementName = "phone")]
public List<string> Phone { get; set; } = new List<string>();
}
To deserialize the XML (read from file) we'll use the following method:
DeserializeXMLFileToObject:
public static T DeserializeXMLFileToObject<T>(string xmlFilename)
{
T rObject = default(T);
try
{
if (string.IsNullOrEmpty(xmlFilename))
{
return default(T);
}
using (System.IO.StreamReader xmlStream = new System.IO.StreamReader(xmlFilename))
{
System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(typeof(T));
rObject = (T)serializer.Deserialize(xmlStream);
}
}
catch (Exception ex)
{
System.Diagnostics.Debug.WriteLine(ex.Message);
throw ex;
}
return rObject;
}
Usage (deserialize):
private XmlRoot _root = null;
...
using (OpenFileDialog ofd = new OpenFileDialog())
{
ofd.Filter = "XML File (*.xml)|*.xml";
ofd.FileName = "Test.xml";
if (ofd.ShowDialog() == DialogResult.OK)
{
//deserialize
_root = HelperXml.DeserializeXMLFileToObject<XmlRoot>(ofd.FileName);
}
}
To serialize the XML (write to file) we'll use the following method:
SerializeObjectToXMLFile:
public static void SerializeObjectToXMLFile(object obj, string xmlFilename)
{
try
{
if (string.IsNullOrEmpty(xmlFilename))
{
return;
}//if
System.Xml.XmlWriterSettings settings = new System.Xml.XmlWriterSettings();
settings.OmitXmlDeclaration = false;
settings.Indent = true;
settings.NewLineHandling = System.Xml.NewLineHandling.Entitize;
using (System.Xml.XmlWriter xmlWriter = System.Xml.XmlWriter.Create(xmlFilename, settings))
{
//specify namespaces
System.Xml.Serialization.XmlSerializerNamespaces ns = new System.Xml.Serialization.XmlSerializerNamespaces();
ns.Add(string.Empty, "urn:none");
//create new instance
System.Xml.Serialization.XmlSerializer serializer = new System.Xml.Serialization.XmlSerializer(obj.GetType());
//write XML to file
serializer.Serialize(xmlWriter, obj, ns);
}
}
catch (Exception ex)
{
System.Diagnostics.Debug.WriteLine(ex.Message);
throw ex;
}
}
Usage (serialize):
private XmlRoot _root = null;
...
using (SaveFileDialog sfd = new SaveFileDialog())
{
sfd.Filter = "XML File (*.xml)|*.xml";
sfd.FileName = "Test.xml";
if (sfd.ShowDialog() == DialogResult.OK)
{
//create new instance
_root = new XmlRoot();
//add data
_root.Staff.Add(new XmlRootStaff() { Id = "1", Name = "Name 1", Phone = new List<string>() { "123456" } });
_root.Staff.Add(new XmlRootStaff() { Id = "2", Name = "Name 2", Phone = new List<string>() { "123457", "123458" } });
//serialize - save to file
SerializeObjectToXMLFile(_root, sfd.FileName);
Debug.WriteLine($"Info: Saved to {sfd.FileName}");
}
}
To remove phone number 123458
where staff id = 2:
private void RemovePhone(string id, string phoneNumber)
{
if (_root != null)
{
for (int i = 0; i < _root.Staff.Count; i )
{
if (_root.Staff[i].Id == id)
{
//remove
_root.Staff[i].Phone.Remove(phoneNumber);
break;
}
}
}
}
Resources:
- LINQ to XML overview
- XElement.Save Method
- How to delete specific nodes from an XElement?
- XML serialization
- Examples of XML Serialization
CodePudding user response:
Finally I solved changing:
program.cs
public static XmlDocument doc = new XmlDocument(); // ADDED THIS HERE
public static XmlNode staff;
public static XmlNodeList xnList = GetList();
public static XmlNodeList GetList()
{
XmlNodeList xnList;
// REMOVED XmlDocument from HERE
doc.Load(path);
xnList = doc.SelectNodes("/root/staff");
return xnList;
}
public static void GetID(string id)
{
foreach (XmlNode xn in xnList)
{
if(xn.Attributes["id"].Value == id)
{
staff = xn;
}
}
}
form1.cs
private void btnOK_Click(object sender, EventArgs e)
{
Program.staff["name"].InnerText = textBoxName.Text;
Program.staff["phone"].InnerText = textBoxPhone.Text;
if (Program.staff.SelectSingleNode("phone2") == null)
{
XmlElement elem = Program.doc.CreateElement("phone2");
elem.InnerText = textBoxPhone2.Text;
Program.staff.AppendChild(elem);
}
}