OpenFileDialog for file selection

Question

4.00/5 (1 vote)

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Hello,

I have a field with a file name, and I want that when I click on that field, the file selection window (explorer) gets opened and thus select the desired file (must be an xls or xlsx file).

But when I click on the field, it opens for edit, nothing else.

in the design file, I have the following instruction:

C#

this.bomNameFld.Click += new System.EventHandler(this.bomNameFld_Click);

and the method, with the following code

private void bomNameFld_Click(object sender, EventArgs e)
{
    OpenFileDialog openFile = new OpenFileDialog();
    string file = openFile.FileName;
    bomNameFld.Text = file;
}

I have also tryed using Mouseclick and MouseEventhandler, but the result is the same.

Can anyone help me?

Thank you in advance.

David.

Posted 13-Mar-13 3:32am

NeueStudium2012

Updated 13-Mar-13 3:49am

Mike Meinz

v3

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Comments

[no name] 13-Mar-13 9:35am

"it opens for edit, nothing else" and what is it that you expect it to do instead of "nothing else"?

Pheonyx 13-Mar-13 9:36am

Can you explain your question please as it is unclear what you are trying to do.

Mike Meinz 13-Mar-13 9:44am

When you say "it opens for edit, nothing else", do you mean that the Textbox control gets focus to allow you to type into it and the OpenFileDialog window is not displayed?

sjelen 13-Mar-13 11:59am

Do not edit design file manually, it's auto generated whenever you change something in designer and you might loose your code.
Set your event handler from designer or from code in .cs file, never in .designer.cs

2 solutions

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Mike Meinz · Accepted Answer · 2013-03-13T03:48:00

You have to call the ShowDialog method after you instantiate your openFile object.

private void bomNameFld_Click(object sender, EventArgs e)
        {
            OpenFileDialog openFile = new OpenFileDialog();
            if( openFile.ShowDialog() == DialogResult.OK )
            {
               string file = openFile.FileName;
               bomNameFld.Text = file;
            }
        }

Also, you should check the Boolean result returned from the call to ShowDialog. If the user clicks OK, then True is returned. If the user clicks Cancel, then False is returned.

See OpenFileDialog.ShowDialog Method[^]

CHill60 · Accepted Answer · 2013-03-13T03:50:00

Solution 2

You've created a new OpenFileDialog but you're not doing anything with it!
Try adding

DialogResult ds = openFile.ShowDialog();

and using

if(ds == System.Windows.Forms.DialogResult.OK)

before you do anything with the result

Posted 13-Mar-13 3:50am

CHill60