如何在 c# 中使用 openFileDialog 打開 file.txt？

Question

我必須打開並讀取.txt文件，這是我正在使用的代碼：

Stream myStream;
openFileDialog1.FileName = string.Empty; 
openFileDialog1.InitialDirectory = "F:\\";
if (openFileDialog1.ShowDialog() == DialogResult.OK) 
{
    var compareType = StringComparison.InvariantCultureIgnoreCase;
    var fileName = Path.GetFileNameWithoutExtension(openFileDialog1.FileName);
    var extension = Path.GetExtension(openFileDialog1.FileName);
    if (extension.Equals(".txt", compareType))
    {
        try 
        { 
            using (myStream = openFileDialog1.OpenFile()) 
            { 
                string file = Path.GetFileName(openFileDialog1.FileName);
                string path = Path.GetFullPath(file); //when i did it like this it's work fine but all the time give me same path whatever where my "*.txt" file is
                //Insert code to read the stream here. 
                //fileName = openFileDialog1.FileName; 
                StreamReader reader = new StreamReader(path);
                MessageBox.Show(file, "fileName");
                MessageBox.Show(path, "Directory");
            } 
        } 
        // Exception thrown: Empty path name is not legal
        catch (ArgumentException ex) 
        { 
            MessageBox.Show("Error: Could not read file from disk. " +
                            "Original error: " + ex.Message); 
        } 
    }
    else 
    {
        MessageBox.Show("Invaild File Type Selected");
    } 
} 

上面的代碼拋出一個異常，上面寫着"Empty path name is not legal" 。

我究竟做錯了什么？

Answer 1

正如hmemcpy所指出的，您的問題在以下幾行中

using (myStream = openFileDialog1.OpenFile())
{
   string file = Path.GetFileName(openFileDialog1.FileName);
   string path = Path.GetDirectoryName(file);
   StreamReader reader = new StreamReader(path);
   MessageBox.Show(file, "fileName");
   MessageBox.Show(path, "Directory");
} 

我為你分解：

/*
 * Opend the file selected by the user (for instance, 'C:\user\someFile.txt'), 
 * creating a FileStream
 */
using (myStream = openFileDialog1.OpenFile())
{
   /*
    * Gets the name of the the selected by the user: 'someFile.txt'
    */
   string file = Path.GetFileName(openFileDialog1.FileName);

   /*
    * Gets the path of the above file: ''
    *
    * That's because the above line gets the name of the file without any path.
    * If there is no path, there is nothing for the line below to return
    */
   string path = Path.GetDirectoryName(file);

   /*
    * Try to open a reader for the above bar: Exception!
    */
   StreamReader reader = new StreamReader(path);

   MessageBox.Show(file, "fileName");
   MessageBox.Show(path, "Directory");
} 

您應該做的是將代碼更改為類似

using (myStream = openFileDialog1.OpenFile())
{
   // ...
   var reader = new StreamReader(myStream);
   // ...
}

Answer 2

您希望用戶訪問 select only.txt 文件嗎？ 然后使用.Filter屬性，如下所示：

openFileDialog1.Filter = "txt files (*.txt)|*.txt";

Answer 3

您的錯誤在以下行中：

string file = Path.GetFileName(openFileDialog1.FileName);
string path = Path.GetDirectoryName(file);

在第一行file變量將只包含文件名，例如MyFile.txt ，使第二行返回一個空字符串給path變量。 在您的代碼下方，您將嘗試創建一個帶有空路徑的StreamReader ，這就是引發異常的原因。

順便說一句，這正是異常告訴你的。 如果您刪除 using 塊周圍的try..catch ，您會在 Visual Studio 中的調試期間看到它發生。

Answer 4

StreamReader 在您傳遞字符串時接受 Stream 類型的 object。 嘗試這個，

Stream myStream;

        using (myStream = openFileDialog1.OpenFile())
        {
            string file = Path.GetFileName(openFileDialog1.FileName);
            string file2 = Path.GetFileNameWithoutExtension(openFileDialog1.FileName);

            string path = Path.GetDirectoryName(openFileDialog1.FileName);

            StreamReader reader = new StreamReader(myStream);

            while (!reader.EndOfStream)
            {
                MessageBox.Show(reader.ReadLine());
            }

            MessageBox.Show(openFileDialog1.FileName.ToString());
            MessageBox.Show(file, "fileName");
            MessageBox.Show(path, "Directory");
        }