为什么在使用openFileDialog时我没有看到我改变的所有设置？

Question

在form1的顶部我做了

OpenFiledialog openFileDialog1 = new OpenFiledialog();

然后:

private void changeWorkingDirectoryToolStripMenuItem_Click(object sender, EventArgs e)
{
    DialogResult result = openFileDialog1.ShowDialog();
    openFileDialog1.Filter =  
"BMP|*.bmp|GIF|*.gif|JPG|*.jpg;*.jpeg|PNG|*.png|TIFF|*.tif;*.tiff|"
   + "All Graphics Types|*.bmp;*.jpg;*.jpeg;*.png;*.tif;*.tiff";
    openFileDialog1.InitialDirectory = @"c:\";
    openFileDialog1.Multiselect = true;
    if (result == DialogResult.OK)
    {
        string[] files = openFileDialog1.FileNames;

        try
        {
            if (files.Length > 0)
            {
                label6.Text = files.Length.ToString();
                label6.Visible = true;
                string directoryPath = Path.GetDirectoryName(files[0]);
                label12.Text = directoryPath;
                label12.Visible = true;
            }
        }
        catch (IOException)
        {
        }
    }
}

但是当我点击按钮时,init目录是文件,而不是c:我根本看不到过滤器,但我看到了所有文件.这就像我没有影响的设置.

Answer 1

ShowDialog 是模态的,所以它等待用户单击确定/取消.

只有这样,其余的代码才会运行.您应该只ShowDialog在完成属性设置后调用:

openFileDialog1.Filter = "BMP|*.bmp|GIF|*.gif|JPG|*.jpg...";
openFileDialog1.InitialDirectory = @"c:\";
openFileDialog1.Multiselect = true;

DialogResult result = openFileDialog1.ShowDialog();
if (result == DialogResult.OK)