Open File dialog based on button click

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Thanks to all who shared code. but I'm facing the foll exception when I use openfiledialog

"Current thread must be set to single thread apartment (STA) mode before OLE calls can be made.
Ensure that your Main function has STAThreadAttribute marked on it. This exception is only raised if a
debugger is attached to the process"

Thanks in advance

Posted 10-Aug-11 20:16pm

CyborgForever

Updated 10-Aug-11 20:55pm

v2

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Toniyo Jackson · Answer 1 · 2011-08-10T20:30:00

Solution 3

You have to add this.

objOpenFileDialog.Filter = "Excel files|*.xls;*.xlsx";

Posted 10-Aug-11 20:30pm

Toniyo Jackson

v2

shreya1987 · Answer 2 · 2011-08-10T20:45:00

Solution 4

Simple.

C#

OpenFileDialog fileD = new OpenFileDialog(); //create object
fileD.Filter = "Excel files|*.xls;*.xlsx"; //define filter
fileD.ShowDialog(); //show dialog
MessageBox.show(fileD.fileName); //Get uploaded file

Posted 10-Aug-11 20:45pm

shreya1987

Suresh Suthar · Answer 3 · 2011-08-10T20:20:00

Solution 1

All of your question asked on CP are of basic level.
I think nobody will provide code for this type of simple task.
And please Google your query before posting on CP and if you are unable to find any suitable solution than ask on CP.

Here is you answer[^].

Posted 10-Aug-11 20:20pm

Suresh Suthar

Comments

CyborgForever 11-Aug-11 2:26am

I tried the code on my own but it throws some exception so I thought some issues in my code

"Current thread must be set to single thread apartment (STA) mode before OLE calls can be made. Ensure that your Main function has STAThreadAttribute marked on it. This exception is only raised if a debugger is attached to the process"

Praveen Kullu · Answer 4 · 2011-08-10T20:24:00

Solution 2

This is all you need.

C#

private void button1_Click(...)
{
OpenFileDialog ofd = new OpenFileDialog();
ofd.filter="Excel files|*.xlsx";
ofd.ShowDialog();
}

Posted 10-Aug-11 20:24pm

Praveen Kullu

Updated 10-Aug-11 20:31pm

v2

Mohammad A Rahman · Answer 5 · 2011-08-10T20:48:00

Solution 5

It might be helpful,

How to: Open Files Using the OpenFileDialog

[^]

FileDialog.Filter Property

[^]

:)

Posted 10-Aug-11 20:48pm

Mohammad A Rahman