[أسامة]

اولآ تم تجربة الكود بهذا الشكل

كود :

       Dim xmlInitial As String = "<Test>" & vbCr & vbLf & "<test name='ahmed' age='20'>" & vbCr & vbLf & "  </test>" & vbCr & vbLf & " </Test>"
       MsgBox(xmlInitial)
       Dim xmlUser As String = "<test name='mohamed' age='20'>" & vbCr & vbLf & " </test>"
       Dim originalXml As New XmlDocument()
       Dim targetNode As String = "descendant::*[name(.) ='test'] [@age='20']"
       'originalXml.LoadXml(Path)
       originalXml.LoadXml(xmlUser)
       Dim editNode As XmlNode = originalXml.SelectSingleNode(targetNode)
       Dim fragment As XmlDocumentFragment = originalXml.CreateDocumentFragment()
       fragment.InnerXml = xmlUser
       editNode.ParentNode.ReplaceChild(fragment, editNode)
       'originalXml.Save(Path)
       MsgBox(originalXml.OuterXml)

 لاكن عند التعديل ليعمل بهذا الشكل

كود :

       Dim xmlUser As String = "<test name='mohamed' age='20'>" & vbCr & vbLf & " </test>"
       Dim originalXml As New XmlDocument()
       originalXml.LoadXml(Path)
       Dim editNode As XmlNode = originalXml.SelectSingleNode("//Test//test[@name='Ahmed'] [@age='20']")
       Dim fragment As XmlDocumentFragment = originalXml.CreateDocumentFragment()
       fragment.InnerXml = xmlUser
       editNode.ParentNode.ReplaceChild(fragment, editNode)
       originalXml.Save(Path)

يعطي خطأ فى originalXml.LoadXml(Path)

كود :

Dim Path As String = Application.StartupPath + "\test.xml"

غالبآ يمكن الخطأ من ملف xml , فى المرفق