如何在C＃中使用openFileDialog打开文件.txt？

如何解决如何在C＃中使用openFileDialog打开文件.txt？

| 我必须打开并从

.txt

文件中读取文件，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\");
    } 
} 

上面的代码引发异常，指出“空路径名不合法”。 我究竟做错了什么？     

解决方法

        正如hmemcpy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\");
} 

我将为您分解：

/*
 * 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,\"Directory\");
} 

您应该做的是将代码更改为类似

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

    ,        您要用户仅选择.txt文件吗？ 然后使用.Filter属性，如下所示：

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

    ,        您的错误所在：

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

在第一行中，“ 7”变量将仅包含文件名，例如

MyFile.txt

，使第二行将空字符串返回到

path

变量。再往下看，您将尝试创建带有空路径的

StreamReader

，这将引发异常。 顺便说一句，这正是异常告诉您的内容。如果删除using块周围的ѭ11，您会发现它在Visual Studio调试期间发生。     ,        当您向对象传递字符串时，StreamReader接受对象的Stream类型。 尝试这个，

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\");
        }