Question

我正在尝试使用openFileDialog，它一直工作到今天早上，当我做出我认为是一个简单的改变....

我将过滤器和初始目录属性从硬编码字符串更改为应用程序设置，这就是错误发生的地方。从我能说的一切都应该没问题......我会发布新旧代码.... / p>

新代码

    private void btnOpenFile_Click(object sender, RoutedEventArgs e)
    {
        OpenFileDialog ofDialog = new OpenFileDialog();
        ofDialog.Filter = Properties.Settings.Default.openFileFilter;
        ofDialog.FilterIndex = 3;
        ofDialog.Multiselect = false;
        ofDialog.InitialDirectory = Properties.Settings.Default.openFileInitialDirectory;

        bool? fileSelected = ofDialog.ShowDialog();

        if(fileSelected == true)
        {
            selectedFileTxtBx.Text = ofDialog.FileName;
        }

应用程序设置

 Properties.Settings.Default.openFileFilter; = Exe (.exe)|*.exe|MSI (.msi)|*.msi| All (*.*)|*.*
 Properties.Settings.Default.openFileInitialDirectory; = \\\\UNC\\PATH

旧代码

    private void btnOpenFile_Click(object sender, RoutedEventArgs e)
    {
        OpenFileDialog ofDialog = new OpenFileDialog();
        ofDialog.Filter = "Exe (.exe)|*.exe|MSI (.msi)|*.msi| All (*.*)|*.*";
        ofDialog.FilterIndex = 3;
        ofDialog.Multiselect = false;
        ofDialog.InitialDirectory = "\\\\UNC\\PATH";

        bool? fileSelected = ofDialog.ShowDialog();

        if(fileSelected == true)
        {
            selectedFileTxtBx.Text = ofDialog.FileName;
        }
    }

Answer 1

如果内存正常，Properties.Settings.Default.openFileInitialDirectory实际上应该设置为\\UNC\PATH，因为字符串已经被转义。

打开文件对话框...值不在预期范围内

1 个答案: