我是这种语言的完全初学者,目前我正在尝试创建一个简单的文本编辑应用程序,但在尝试打开文件时遇到上述错误。 下面的代码,任何帮助将不胜感激。
private void mnuOpen_Click(object sender, EventArgs e)
{
string chosenFile = "";
openFD.Title = "Select a text file";
openFD.InitialDirectory = System.Environment.GetFolderPath(Environment.SpecialFolder.Personal);
openFD.FileName = "Choose a file..";
openFD.Filter = "Text Files|*.txt|All Files|*.*";
if (openFD.ShowDialog() != DialogResult.Cancel)
{
contentBox.LoadFile(chosenFile, RichTextBoxStreamType.PlainText);
}
}
异常本身发生在这里:
contentBox.LoadFile(chosenFile, RichTextBoxStreamType.PlainText);
答案 0 :(得分:4)
您需要将chosenFile
设置为OpenFileDialog控件中所选文件的名称:
private void mnuOpen_Click(object sender, EventArgs e)
{
string chosenFile = "";
openFD.Title = "Select a text file";
openFD.InitialDirectory = System.Environment.GetFolderPath(Environment.SpecialFolder.Personal);
openFD.FileName = "Choose a file..";
openFD.Filter = "Text Files|*.txt|All Files|*.*";
if (openFD.ShowDialog() != DialogResult.Cancel)
{
// Assign the selected filename to chosenFile
chosenFile = openFD.FileName;
contentBox.LoadFile(chosenFile, RichTextBoxStreamType.PlainText);
}
}