How to access fields and values from an XML Database

Question

I have an XML Database that I am loading into a Dataset object in a C# 2008 Windows Forms project. I need to know how to access the individual fields and values. The actual database is not in my control so I have to use it as given. And I need to be able to offer portions of the values for editing while hiding others. So I need to how to access each individual field and value. Here is a very tiny mocked up version of the XML code: If there is an easier way than the Dataset object to process the database please let me know. Any help would be most appreciated. Thanks Bruce

Sam Hobbs · Answer

I admit that the XmlSerializer class and related classes are very frustrating to understand. I really think the documentation is very inadequate compared to what it could be. Once we understand a few basics it can be very productive. The following code works with the sample data you provided. First I created a separate file for the data definition. using System; using System.Xml; using System.Xml.Serialization; using System.Collections.Generic; namespace _72570 { // The following XmlType tells the system to not look for // ArrayOfDatabaseItem as the tag [XmlType(database)] public class Database : List { } public class DatabaseItem { [XmlAttribute] public String name; [XmlElement(DatabaseSubitem)] public List Subitems; } public class DatabaseSubitem { [XmlAttribute] public String id; [XmlElement(DatabaseField)] public List Fields; } public class DatabaseField { [XmlAttribute] public String name; [XmlAttribute] public String value; } } Then the code is: using System; using System.IO; using System.Xml; using System.Xml.Serialization; using System.Collections.Generic; using System.Diagnostics; namespace _72570 { class Program { const String filename = Database.xml; static void Main(string[] args) { Database db; XmlSerializer serializer; serializer = new XmlSerializer(typeof(Database)); FileInfo fi = new FileInfo(filename); if (!fi.Exists) { Console.WriteLine(File not found); return; } FileStream fs = new FileStream(filename FileMode.Open); try { db = (Database)serializer.Deserialize(fs); } catch (Exception ex) { Console.WriteLine(Deserialize exception: {0}
Inner exception: {1} ex.Message ex.InnerException.Message); return; } fs.Dispose(); // To prevent a sharing exception serializer = new XmlSerializer(typeof(Database)); foreach (DatabaseItem di in db) { Console.WriteLine(di.name); foreach (DatabaseSubitem dsi in di.Subitems) { Console.WriteLine(	{0} dsi.id); foreach (DatabaseField df in dsi.Fields) Console.WriteLine(		{0}={1} df.name df.value); } } } } }

Jayhawks Rock · Answer

Thanks for the response and this does give me access to the DatabaseItem name field but I can't see how to access any of the DatabaseSubitem id fields or the DatabaseField name and value fields. Can you point me to that next level? Thanks again Bruce

Sam Hobbs · Answer

Since I don't know for sure what your data looks like I can't say for sure that you can use the XmlSerializer class but there is a good chance you can. Note however that it will take time to learn about use of the XmlSerializer class but it is useful to learn it. The following console program is a skeleton of what you would need. This sample will only read in the names of each DatabaseItem. You probably can use the XmlSerializer class to read in all of the data but as I said I don't know the format of all your data. Note that the InnerException member of an exception can be null and you need to check for null there but I forgot to modify the sample code for that. Otherwise you can use this code as-is to try out use of the XmlSerializer class. It is one way that can make things easy. using System; using System.IO; using System.Xml; using System.Xml.Serialization; using System.Collections.Generic; using System.Diagnostics; namespace SampleProject { public class DatabaseItem { [XmlAttribute] public String name; } class Program { const String filename = Database.xml; static void Main(string[] args) { List LinksList; XmlSerializer serializer; XmlAttributeOverrides xOver = new XmlAttributeOverrides(); XmlAttributes xAttrs = new XmlAttributes(); xAttrs.XmlType = new XmlTypeAttribute(database); xOver.Add(typeof(List ) xAttrs); serializer = new XmlSerializer(typeof(List ) xOver); FileInfo fi = new FileInfo(filename); if (!fi.Exists) { Console.WriteLine(File not found); return; } FileStream fs = new FileStream(filename FileMode.Open); try { LinksList = (List )serializer.Deserialize(fs); } catch (Exception ex) { Console.WriteLine(Deserialize exception: {0}
Inner exception: {1} ex.Message ex.InnerException.Message); return; } fs.Dispose(); // To prevent a sharing exception foreach (DatabaseItem di in LinksList) { Console.WriteLine(di.name); } } } }

Amit Choudhary · Answer

hi friend Using dataset to manipulate the data from an xml file is itself and easiest way other wise you have to manipulate that xml tags using Your programming logic that itself have its complexity .... so just try Dataset it has lots of functions and properties you can use to manipulate its Data....