从文本文件导入C# ListBox崩溃

Question

我正在使用此代码将文本文件导入到我的ListBox

        OpenFileDialog openFileDialog1 = new OpenFileDialog();
        openFileDialog1.Filter = "Text Files|*.txt";
        openFileDialog1.Title = "Select a Text file";
        openFileDialog1.FileName = "";
        DialogResult result = openFileDialog1.ShowDialog();
        if (result == DialogResult.OK)
        {
            string file = openFileDialog1.FileName;

            string[] text = System.IO.File.ReadAllLines(file);
            foreach (string line in text)
            {
                listBox2.Items.Add(line);

            }
            listBox2.Items.Add("");
        }

它适用于10行左右的小文本文件，但当我尝试导入更大的列表(4-5兆字节)时，程序没有响应，它崩溃了。

有什么帮助吗？

Answer 1

使用C#中的BufferedStream类可以提高性能。

http://msdn.microsoft.com/en-us/library/system.io.bufferedstream.aspx

Answer 2

通过使用此命令：

string[] text = System.IO.File.ReadAllLines(file);
listBox1.Items.AddRange(text);

而不是这样：

string[] text = System.IO.File.ReadAllLines(file);
foreach (string line in text)
{
       listBox2.Items.Add(line);
}

您将使执行速度提高至少10-15倍，因为您不会在每次插入项时都使listBox无效。我已经用几千条线进行了测量。

如果您的文本行数太多，瓶颈也可能是ReadAllLines。即使我不明白你为什么要插入这么多行，用户能找到他/她需要的行吗？

编辑 OK然后我建议你使用BackgroundWorker，下面是代码：

首先初始化BackGroundWorker：

 BackgroundWorker bgw;
        public Form1()
        {
            InitializeComponent();
            bgw = new BackgroundWorker();
            bgw.DoWork += new DoWorkEventHandler(bgw_DoWork);
            bgw.RunWorkerCompleted += new RunWorkerCompletedEventHandler(bgw_RunWorkerCompleted);
        }

然后在你的方法中调用它：

private void button1_Click(object sender, EventArgs e)
        {
            if (!bgw.IsBusy)
            {
                OpenFileDialog openFileDialog1 = new OpenFileDialog();
                openFileDialog1.Filter = "Text Files|*.txt";
                openFileDialog1.Title = "Select a Text file";
                openFileDialog1.FileName = "";
                DialogResult result = openFileDialog1.ShowDialog();
                if (result == DialogResult.OK)
                {
                    string file = openFileDialog1.FileName;
                    listView1.BeginUpdate();
                    bgw.RunWorkerAsync(file);
                }
            }
            else
                MessageBox.Show("File reading at the moment, try later!");
        }


        void bgw_RunWorkerCompleted(object sender, RunWorkerCompletedEventArgs e)
        {
            listView1.EndUpdate();
        }
        void bgw_DoWork(object sender, DoWorkEventArgs e)
        {
            string fileName = (string)e.Argument;
            TextReader t = new StreamReader(fileName);
            string line = string.Empty;
            while ((line = t.ReadLine()) != null)
            {
                string nLine = line;
                this.Invoke((MethodInvoker)delegate { listBox1.Items.Add(nLine); });
            }
        }

它会在读取时添加每一行，您将拥有响应式UI，并且行在完成加载之前不会影响listBox。

Answer 3

可以使用流来存储数据：

class Test
{

public static void Main()
{
    string path = @"c:\temp\MyTest.txt";

    //Create the file. 
    using (FileStream fs = File.Create(path))
    {
        AddText(fs, "This is some text");
        AddText(fs, "This is some more text,");
        AddText(fs, "\r\nand this is on a new line");
        AddText(fs, "\r\n\r\nThe following is a subset of characters:\r\n");

        for (int i=1;i < 120;i++)
        {
            AddText(fs, Convert.ToChar(i).ToString());

        }
    }

    //Open the stream and read it back. 
    using (FileStream fs = File.OpenRead(path))
    {
        byte[] b = new byte[1024];
        UTF8Encoding temp = new UTF8Encoding(true);
        while (fs.Read(b,0,b.Length) > 0)
        {
            Console.WriteLine(temp.GetString(b));
        }
    }
}

private static void AddText(FileStream fs, string value)
{
    byte[] info = new UTF8Encoding(true).GetBytes(value);
    fs.Write(info, 0, info.Length);
}

}

然后您的事件处理程序

 privateasyncvoid Button_Click(object sender, RoutedEventArgs e)
    {
        UnicodeEncoding uniencoding = new UnicodeEncoding();
        string filename = @"c:\Users\exampleuser\Documents\userinputlog.txt";

        byte[] result = uniencoding.GetBytes(UserInput.Text);

        using (FileStream SourceStream = File.Open(filename, FileMode.OpenOrCreate))
        {
            SourceStream.Seek(0, SeekOrigin.End);
            await SourceStream.WriteAsync(result, 0, result.Length);
        }
    }