A number of posts have indicated developers having problem getting an
OpenFileDialog to startup in a specific directory when the dialog is
shown. The first time you show the dialog, the InitialDirector y will
cause the dialog to show files in that directory. However, any further
times that the dialog is displayed, the directory that is shown will
always be the one initially used regardless of the RestoreDirector y
setting. This appears to be a behaviour on Windows 2000/XP.
To solve the problem, simply create a *new* OpenFileDialog object at
runtime each time you need to display the dialog. Do this rather than
referencing an existing object that is on your form. The object on the
form gets only instantiated once and preserves its values while at the
same time ignoring the RestoreDirector y property. So here is an
example. Place a OpenFileDialog control on your form as you usually
do. In this example, call it dlgFileSelect. When you go to display it,
here is an example of what code you would use to:
this.dlgSelectF ile = new OpenFileDialog( );
this.dlgSelectF ile.InitialDire ctory = @"c:\somedirect ory";
this.dlgSelectF ile.RestoreDire ctory = true;
this.dlgSelectF ile.ShowDialog( this);
Johann Blake
1  2330 
Geesh.
That particular dialog is always pokey for me when it's first loaded. I
suspect that it has to do with the fact that it checks out my CD and
network drives. Couldn't we just reset the initialDirector y property,
and show the same object?
Creating a new one means that I have to sit there waiting for the dialog
to be created every time.
Mike
Johann Blake wrote: A number of posts have indicated developers having problem getting an
OpenFileDialog to startup in a specific directory when the dialog is
shown. The first time you show the dialog, the InitialDirector y will
cause the dialog to show files in that directory. However, any further
times that the dialog is displayed, the directory that is shown will
always be the one initially used regardless of the RestoreDirector y
setting. This appears to be a behaviour on Windows 2000/XP.
To solve the problem, simply create a *new* OpenFileDialog object at
runtime each time you need to display the dialog. Do this rather than
referencing an existing object that is on your form. The object on the
form gets only instantiated once and preserves its values while at the
same time ignoring the RestoreDirector y property. So here is an
example. Place a OpenFileDialog control on your form as you usually
do. In this example, call it dlgFileSelect. When you go to display it,
here is an example of what code you would use to:
this.dlgSelectF ile = new OpenFileDialog( );
this.dlgSelectF ile.InitialDire ctory = @"c:\somedirect ory";
this.dlgSelectF ile.RestoreDire ctory = true;
this.dlgSelectF ile.ShowDialog( this);
Johann Blake This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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OpenFileDialog to startup in a specific directory when the dialog is
shown. The first time you show the dialog, the InitialDirectory will
cause the dialog to show files in that directory. However, any further
times that the dialog is displayed, the directory that is shown will
always be the one initially used regardless of the RestoreDirectory
setting. This appears to be...
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