采用BackgroundWorker,在后台执行程序,往往比开新线程更简单和更加安全。
简单的使用方法
System.ComponentModel.BackgroundWorker bw = new System.ComponentModel.BackgroundWorker();
//定义需要在子线程中干的事情
bw.DoWork += new System.ComponentModel.DoWorkEventHandler(bw_DoWork);
//定义执行完毕后需要做的事情
bw.RunWorkerCompleted += new System.ComponentModel.RunWorkerCompletedEventHandler(bw_RunWorkerCompleted);
//开始执行 bw.RunWorkerAsync();
static void bw_RunWorkerCompleted(object sender, System.ComponentModel.RunWorkerCompletedEventArgs e)
{
MessageBox.Show("Complete"+Thread.CurrentThread.ManagedThreadId.ToString());
}
static void bw_DoWork(object sender, System.ComponentModel.DoWorkEventArgs e)
{
MessageBox.Show(Thread.CurrentThread.ManagedThreadId);
}
我改写了一个通知页面处理的方法,不需要结果,通知了主程序就继续前进,保证效率
/// <summary>
/// 访问网页(异步方式),只需激活网页,不要结果
/// </summary>
/// <param name="url">网页地址</param>
/// <returns></returns>
public static void Notify(string url)
{
BackgroundWorker bw = new BackgroundWorker(); //定义一个背景对象
//定义需要在子线程中干的事情
bw.DoWork += new System.ComponentModel.DoWorkEventHandler(bw_DoWork);
bw.RunWorkerAsync(url);
}
/// <summary>
/// 后台执行:打开网页
/// </summary>
/// <param name="sender"></param>
/// <param name="e"></param>
static void bw_DoWork(object sender, System.ComponentModel.DoWorkEventArgs e)
{
string url = (string)e.Argument;
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(url);
string content = "";
try
{
HttpWebResponse response = (HttpWebResponse)request.GetResponse();
if (response.StatusCode == HttpStatusCode.OK) //防止无响应
{
StreamReader reader = new StreamReader(response.GetResponseStream(), System.Text.Encoding.GetEncoding("GB2312"));
content = reader.ReadToEnd();
reader.Close();
}
log4net.LogManager.GetLogger(System.Reflection.MethodBase.GetCurrentMethod().DeclaringType).Debug("异步方式打开网页:" + url+",返回长度:"+content.Length.ToString());
}
catch (Exception ex)
{
log4net.LogManager.GetLogger(System.Reflection.MethodBase.GetCurrentMethod().DeclaringType).Error("[Notify]打开网页" + url + "出错:" + ex.Message);
}
}
执行效果不错,这是跟踪的情况:
可以看到,假如同时发生多个访问的话,系统会分配不同的线程执行(前面的是线程号)。