[Tony Green]

PHP код:

static void Job1(Args _args)
{
    Dialog              dialog;
    DialogField         df_filename;
    ;

    dialog      = new Dialog("File");

    df_filename = dialog.AddField(typeid(FilenameOpen),"File: ");

    dialog.Run();
    if(dialog.ClosedOk())
    {
        info(df_filename.value());
    }
}