How import from xml file to DataGridView

Question

Hello How import from xml file to DataGridView? I study 4 days about xml writing and reading and about DataGridView but cant undestand how? Please help me. Maybe show example.

Crish · Answer

create a Xml File and call the Aspx page This Code Help full to use XmlDataDocument xmlDatadoc = new XmlDataDocument (); xmlDatadoc.DataSet.ReadXml( C:\books.xml ); DataSet ds = new DataSet ( Books DataSet ); ds = xmlDatadoc.DataSet; dataGridView1.DataSource = ds.DefaultViewManager; dataGridView1.DataMember = Book ;

Sam Hobbs · Answer

And if you want more suggestions just look at the articles in this web site.

Mahesh Chand · Answer

First you load your XML File into a DataSet object and then bind DataSet to a DataGridView control. Something like this: string yourXmlFileWithFullPath = @C:\PutFullPath\XmlFileName.xml; DataSet newDataSet = new DataSet( New DataSet ); ds.ReadXml(xmlFilename); DataGridView1.DataSource = ds;

Mike Jonson · Answer

Yes work. Now undestand more. Thanks)) You are the best

Sam Hobbs · Answer

Is it working now?

Mike Jonson · Answer

Sure! I will now try.

Sam Hobbs · Answer

So what about the other changes that were suggested? Please don't ignore those.

Mike Jonson · Answer

fDebit db=new fDebit()its another form where is the my DataGridView. But method call in Form1 ( private void Form1_Load(object sender EventArgs e). To access the DataGridView I created form class object fDebit

Sam Hobbs · Answer

I assume that using ds.Tables[0] for the data source would work. I know that using the DataMember property works for this because I tested it. I know that the program can work with the changes I described since I tested it. For me the DataGridView got the data when I made the changes. What I did is to use a console program to do a quick test and the following is most of it. Form f = new Form(); DataGridView dgv = new DataGridView(); dgv.Dock = DockStyle.Fill; f.Controls.Add(dgv); DataSet ds = new DataSet(Amounts); ds.ReadXml(filePath); dgv.DataSource = ds; dgv.DataMember = amount; Application.Run(f);

Mahesh Chand · Answer

Also check if your ds.Tables[0] has any data in it. That will ensure if the XML is loaded or not.

Mahesh Chand · Answer

Is fDebit db=new fDebit(); is a new Form? Are you trying to open a new form there? If it is a new form (window) then on first Form you need to call fDebit.Show() and then write your XML load code on the fDebit's Form Load. Also change this.dataGridView_Debit.DataSource = ds.Tables[0];

Sam Hobbs · Answer

First note that the XML you posted has some minor errors such as extra dashes (-). I assume that the actual data is not like that. Your code has  fDebit db=new fDebit();  in it. I don't know what that is supposed to do but I doubt that it does what you intend it to. So the DataGridView that you are using ( db.dataGridView_Debit ) is not the one you see in the form. Finally you are not setting the DataMember property of the DataGridView. Set it to the table name (amount). If you look at the articles and samples again I am sure you will see that you need to do that also.

Mike Jonson · Answer

I have next xml file: - - xsx saxsx 23432 System.Windows.Forms.DateTimePicker Value: 11.05.2011 13:38:45 - 324324 ewdwex 323 System.Windows.Forms.DateTimePicker Value: 11.05.2011 0:00:00 And i use this code: private void Form1_Load(object sender EventArgs e) { //if file not found if (!File.Exists(filePath)) { //create new file createXMLDocument(filePath); } else { //import from xml file to DataGridView DataSet ds = new DataSet(Amounts); ds.ReadXml(filePath); fDebit db=new fDebit(); db.dataGridView_Debit.DataSource = ds; But .dataGridView is clear. Whe is problem please help me.

Mike Jonson · Answer

thanks

Mahesh Chand · Answer

If you look at DataSet class in MSDN you will learn about DataSource and DataMember properties. http://msdn.microsoft.com/en-us/library/system.data.dataset_properties.aspx

Mike Jonson · Answer

Thanks guys but what is this: dataGridView1.DataSource = ds.DefaultViewManager; dataGridView1.DataMember = Book;