Question

i am trying search data in xml file. if found, it will popup MessageBox and display all the data found.

this is my code.

DataView dv;
        DataSet ds = new DataSet();
        ds.ReadXml("C:\\Users\\HDAdmin\\Documents\\SliceEngine\\SliceEngine\\bin\\Debug\\saya.xml");
        dv = new DataView(ds.Tables[0]);
        dv.Sort = "Name";
        int index = dv.Find("Name");
        if (index == -1)
        {
            MessageBox.Show("Item Not Found");
        }
        else
        {
            MessageBox.Show(dv[index]["Name"].ToString()); 
        }

but it always said the item not found.

then i tried to do this.

    XmlDocument xml = new XmlDocument();            
            xml.Load("C:\\Users\\HDAdmin\\Documents\\SliceEngine\\SliceEngine\\bin\\Debug\\saya.xml");
            XmlNodeList xnList = xml.SelectNodes("/Patient/Patient/Name");
            foreach (XmlNode xn in xnList)
            {
                string name = xn["Name"].InnerText;
                listBox21.Items.Add(name);
}

for this code, i tried to put it into the listbox. by doing this, it said that it is a null object.

below is my xml file.

    <Patient>
       <Patient>
         <Level>0</Level>
         <Name>w</Name>
         <Gender>0</Gender>
      </Patient>
   </Patient>

can anybody help me with this.

Answer 1

You have to consider your code is ok! But here’s the problem:

xn["Name"].InnerText

Becase xn represents /Patient/Patient/Name and you just need to do:

xn.InnerText

to get its value.