首先，一個類里，有個linkLabel1

private OpenFileDialog openFileDialog1;
private DialogResult result;

private void linkLabel1_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
??????? {

??????????? openFileDialog1 = new OpenFileDialog();
??????????? string patch = Application.StartupPath + "\\LOG\\";
??????????? openFileDialog1.InitialDirectory = patch;
??????????? openFileDialog1.Filter = "xls files (*.xls)|*.xls";

????????????result = openFileDialog1.ShowDialog();

??????????? if (result == DialogResult.OK)
??????????? {
??????????????? if (openFileDialog1.FileName != "")
??????????????? {
??????????????????? Process.Start(openFileDialog1.FileName);
??????????????? }
???????????????
??????????? }

???????????
??????? }

就會報?
在可以調用 OLE 之前，必須將當前線程設置為單線程單元(STA)模式。請確保您的 Main 函數帶有 STAThreadAttribute 標記。 只有將調試器附加到該進程才會引發此異常。

在測試小程序里沒有問題，當移到大程序里就這樣的問題了?？赡苁蔷€程多的原因。解決辦法就是添加線程，代碼如下

private Thread invokeThread;

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

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

??????????? if (result == DialogResult.OK)
??????????? {
??????????????? if (openFileDialog1.FileName != "")
??????????????? {
??????????????????? Process.Start(openFileDialog1.FileName);
??????????????? }

??????????? }
??????? }

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

問題得到解決

轉載于:https://www.cnblogs.com/verna/archive/2011/02/15/1955276.html

總結

以上是生活随笔為你收集整理的OpenFileDialog 类的ShowDialog() 错误的解决的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。