如何在c＃中使用openFileDialog打开文件.txt？

Question

我都开并从.txt文件读取，这里是我使用的代码：

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"); 
    } 
} 

上面的代码抛出一个异常，说：“空路径名称是不合法的”。

我在做什么错？

Answer 1

正如指出的hmemcpy，你的问题是下面几行

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"); 
} 

我要打破你：

/* 
* 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"); 
} 

你应该做的是将代码cahnge到像

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

Answer 2

您希望用户只选择.txt文件吗？ 然后使用.Filter属性，像：

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

Answer 3

你的错误是在线路：

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

在第一行的file变量将只包含文件名，例如MyFile.txt，使第二行返回一个空字符串到path变量。再往下看你的代码，你将尝试创建一个空路径的StreamReader，这引发了异常。

顺便说一下，这正是异常告诉你的。如果您在使用块周围删除try..catch，则在Visual Studio的调试过程中您会看到它发生。

Answer 4

StreamReader接受Stream类型的对象whi你传递给它一个字符串。 试试这个，

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"); 
     }