C# OpenFileDialog 无响应

C# OpenFileDialog无响应

private Thread invokeThread;
private OpenFileDialog openFileDialog1;
private DialogResult result;

private void linkLabel1_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
    string patch = "C:\\";
    openFileDialog1 = new OpenFileDialog();
    openFileDialog1.InitialDirectory = patch;
    openFileDialog1.Filter = "xls files(*.xls;*.xlsx)|*.xls;*.xlsx";

    invokeThread = new Thread(new ThreadStart(InvokeMethod));
    invokeThread.SetApartmentState(ApartmentState.STA);
    invokeThread.Start();
    invokeThread.Join();

    if (result == DialogResult.OK)
    {
        if (openFileDialog1.FileName != "")
        {
            //获取到文件,进行处理
        }

    }
}
 
private void InvokeMethod()
{
    result = openFileDialog1.ShowDialog();
}

posted on 2022-08-23 17:38  糯米白白  阅读(64)  评论(0)    收藏  举报

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