Question

当我尝试打开.txt文件时，它只在textbox中显示其位置。我没有想法:(希望你能帮助我......

代码：

private void OpenItem_Click(object sender, EventArgs e)
{
    openFileDialog1.ShowDialog();
    System.IO.StringReader OpenFile = new System.IO.StringReader(openFileDialog1.FileName);
    richTextBox1.Text = OpenFile.ReadToEnd();
    OpenFile.Close();
}

Answer 1

我使用File.OpenText()方法来阅读文本文件。您还应该使用using语句来正确处理对象。

if(openFileDialog1.ShowDialog() == DialogResult.OK)
{
    try
    {
        // Make sure a file was selected
        if ((myStream = openFileDialog1.OpenFile()) != null) {
            // Open stream
            using (StreamReader sr = File.OpenText(openFileDialog1.FileName)) 
            {
                // Read the text
                richTextBox1.Text = sr.ReadToEnd();
            }
        }
    }
    catch (Exception ex)
    {
        MessageBox.Show("An error occured: " + ex.Message);
    }
}

Answer 2

使用File.ReadAllText

richTextBox1.Text = File.ReadAllText(openFileDialog1.FileName);

Answer 3

StringReader从您传递给它的字符串中读取字符 - 在本例中为文件的名称。如果要阅读文件内容，请使用StreamReader：

var OpenFile = new System.IO.StreamReader(openFileDialog1.FileName);
richTextBox1.Text = OpenFile.ReadToEnd();

Answer 4

这很容易。这是你需要做的：

1）将using System.IO;放在命名空间之上。

2）创建一个新方法：

    public static void read()
    {
         StreamReader readme = null;

         try
         {
              readme = File.OpenText(@"C:\path\to\your\.txt\file.txt");
              Console.WriteLine(readme.ReadToEnd());
         }
         // will return an invalid file name error
         catch (FileNotFoundException errorMsg)
         {
              Console.WriteLine("Error, " + errorMsg.Message);
         }
         // will return an invalid path error
         catch (Exception errorMsg)
         {
             Console.WriteLine("Error, " + errorMsg.Message);
         }
         finally
         {
              if (readme != null)
              {
                  readme.Close();
              }
         }
    }

3）在您的主要方法中调用它：read();

4）你已经完成了！

在c＃中打开文本文件

4 个答案: