Can you explain why you don’t want to touch your Program.cs file? This is exactly where you change the start-up form.
Change the:
Application.Run(new Form1());
to:
Application.Run(new Form4());
Secondly, you can set the filters on Open- and SaveFileDialog using the Filter property. Set it to a value like this:
XML Files|*.xml
Or for text:
Text Files|*.txt
Edited to add:
To do this from another form:
class Form1 {
Form2 form2;
void Form1_Load(object sender, EventArgs e) {
form2 = new Form2();
}
void rdb1_CheckedChanged(object sender, EventArgs e) {
if (rdb1.Checked)
form2.openFileDialog1.Filter = "XML Files|*.xml";
else
form2.openFileDialog1.Filter = "Text Files|*.txt";
}
}
Make sure you have set the Modifiers property of the openFileDialog1 on the Form2 designer to “Public” or “Internal” to allow access to it from outside the class itself.
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solved How to call a form when there are 3 forms in a project and data Transfer between the forms in C#?