how to use open file dialog box in c#

Question

2.20/5 (3 votes)

See more:

Hello ,
Please demonstrate to me how i can use the open file dialog box.
I need to as the user to select a file , one the user selects the file i needs to get its file path. how can i do this.

Posted 30-Jun-10 18:33pm

asjadazeez

Add a Solution

5 solutions

Solution 1

Hi,

here is some code

C#

if (openFileDialog1.ShowDialog() == DialogResult.OK)
{
    MessageBox.Show(openFileDialog1.FileName);
}

If you need to get the path only, just search for the last "\" in the FileName.

Hope This helps

Posted 30-Jun-10 18:38pm

JF2015

Comments

William Winner 1-Jul-10 0:50am

or you could use the Path methods in System.IO.

Solution 2

Create object of OpenFileDialog

OpenFileDialog opDialog = new OpenFileDialog();

Handle the FilePk event

opDialog.FileOk += new CancelEventHandler(opDialog_FileOk);

show the dialog

opDialog.ShowDialog();

handler function

C#

void opDialog_FileOk(object sender, CancelEventArgs e)
    {
      string file_Name = opDialog.FileName;
    }

Posted 30-Jun-10 18:42pm

Vineeth P Joseph

Updated 30-Jun-10 18:43pm

v2

Solution 3

Well, doing this [^]gave me these:
How to: Open Files Using the OpenFileDialog Component[^]
File Open Dialog box in C#[^]
There are many more if needed.

Posted 30-Jun-10 18:46pm

Sandeep Mewara

Solution 5

i hope this will work .
if you want only the selected file name then use the following code

 if OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK  then 
 messagebox.show(openfiledialog1.filename)
end if

if you want a selected file path then use following code

dim folder as string
dim files as array
if OpenFileDialog1.ShowDialog = Windows.Forms.DialogResult.OK  then
  folder = OpenFileDialog1.FileName
  files = Directory.GetFiles(folder)
end if

Posted 30-Jun-10 21:46pm

curious_mind

Comments

ridoy 22-Dec-14 4:08am

is it for c#?

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

sainath437 · Accepted Answer · 2010-06-30T18:47:00

Hi..
Before asking this question you must have googled it ....

OpenFile DialogControl

<pre lang="C++">
OpenFileDialog ofd = new OpenFileDialog();
 string filename = "";
 string path = "";        
  if (ofd.ShowDialog() == DialogResult.OK)      
    {                  
                filename = System.IO.Path.GetFileName(ofd.FileName);   
                  path = System.IO.Path.GetDirectoryName(ofd.FileName);     
     }     
     MessageBox.Show(filename, "Filename");      
    MessageBox.Show(path, "Directory");