Hello,
I have 3 forms in my project form1 , form2 , form3 and it was running smoothly now i have added one more in my project form4 . The first three forms are already linked up through ShowDialog().
I don't want to touch Program.cs file.
How can i call form 4 first as start up form ? Earlier form 1 was the first form to appear in my project.
Also i have 2 radio buttons in my form1 rdb1 and rdb2. In my form2 i have openFileDialog attached to a button Select . Now i want when the user selects rdb1 in form1 then the filter of openFileDialog in form2 should open files with only ".XML" as extension and when rdb2 is selected in Form1 then in Form2 only ".TXT" files can be opened.
I am unable to find the syntax for this in intellisense can you please help out?
Thanks in advance..