如何在c#中使用openFileDialog打开文件.txt？

Question

我必须打开并从.txt文件中读取,这是我正在使用的代码:

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"空路径名不合法".

我究竟做错了什么？

Answer 1

正如hmemcpy所指出的,您的问题在于以下几行

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");
} 

我要打破你了:

/*
 * 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改为类似的东西

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,如下所示:

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

Answer 3

你的错误在于:

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

在第一行中,file变量将仅包含文件名,例如MyFile.txt,使第二行向path变量返回空字符串.在代码中,您将尝试StreamReader使用空路径创建一个,这就是抛出异常的原因.

顺便说一句,这正是异常告诉你的.如果删除try..catch使用块周围的内容,您会在Visual Studio中调试期间看到它.