help

[sarojkumarjena]

I want to open a text file so I am using this code OpenFileDialog openFileDialog1 = new OpenFileDialog(); openFileDialog1.InitialDirectory = "c:\\" ; openFileDialog1.Filter = "txt files (*.txt)|*.txt|All files (*.*)|*.*" ; But my application contains a richtextbox where I want to show the contents of the file How to do it

[trannguyen86]

Your question not specify what do you want to do please add more info in to your question! add:You have bad title!!

[sarojkumarjena]

When I will click the open menu dialog box will appear where I will browse the text file.And contets of the text file will be shown in the richtextbox which the application contains

[AB7771]

After the user selects the text file from the OpenFile dialog use the richtextbox.loadFile() method and specify the filename in the method.

Thanks & Regards, Pramod "Everyone is a genius at least once a year"

[sarojkumarjena]

But how to store the selected file name at run time,so that I can pass the parameter to load File method

[Chatura Dilan]

if (openFileDialog1.ShowDialog() == System.Windows.Forms.DialogResult.OK) { richTextBox1.LoadFile(openFileDialog1.FileName); } You can store FileName in registry

Chatura Dilan SC Java Programmer

[saqib82]

u can use a global variable to store the name of selected file (at the event the file is being selected) and then use that variable to pass it to the method. sAqIb