在。net中检查Internet连接的最快和最有效的方法是什么?
当前回答
public static bool HasConnection()
{
try
{
System.Net.IPHostEntry i = System.Net.Dns.GetHostEntry("www.google.com");
return true;
}
catch
{
return false;
}
}
工作
其他回答
private bool ping()
{
System.Net.NetworkInformation.Ping pingSender = new System.Net.NetworkInformation.Ping();
System.Net.NetworkInformation.PingReply reply = pingSender.Send(address);
if (reply.Status == System.Net.NetworkInformation.IPStatus.Success)
{
return true;
}
else
{
return false;
}
}
我已经看到了上面列出的所有选项,唯一可行的选项来检查互联网是否可用是“Ping”选项。 导入[DllImport("Wininet.dll")]和System.Net.NetworkInformation.NetworkInterface.GetAllNetworkInterfaces()或NetworkInterface类的任何其他变体都不能很好地检测网络的可用性。这些方法只检查网线是否插好。
“Ping选项”
if(Connection is available)返回true
if(连接不可用且网线已插入)返回false
if(网线未插入)抛出异常
的NetworkInterface
if(Internet可用)返回True
if(Internet不可用且网线已插入)返回True
if(Network Cable is Not plug in)返回false
[DllImport杂志》(“Wininet等”)。
if(Internet可用)返回True
if(Internet不可用且网线已插入)返回True
if(Network Cable is Not plug in)返回false
因此,在[DllImport("Wininet.dll")]和NetworkInterface的情况下,没有办法知道互联网连接是否可用。
多线程版本的ping:
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Net.NetworkInformation;
using System.Threading;
namespace OnlineCheck
{
class Program
{
static bool isOnline = false;
static void Main(string[] args)
{
List<string> ipList = new List<string> {
"1.1.1.1", // Bad ip
"2.2.2.2",
"4.2.2.2",
"8.8.8.8",
"9.9.9.9",
"208.67.222.222",
"139.130.4.5"
};
int timeOut = 1000 * 5; // Seconds
List<Thread> threadList = new List<Thread>();
foreach (string ip in ipList)
{
Thread threadTest = new Thread(() => IsOnline(ip));
threadList.Add(threadTest);
threadTest.Start();
}
Stopwatch stopwatch = Stopwatch.StartNew();
while (!isOnline && stopwatch.ElapsedMilliseconds <= timeOut)
{
Thread.Sleep(10); // Cooldown the CPU
}
foreach (Thread thread in threadList)
{
thread.Abort(); // We love threads, don't we?
}
Console.WriteLine("Am I online: " + isOnline.ToYesNo());
Console.ReadKey();
}
static bool Ping(string host, int timeout = 3000, int buffer = 32)
{
bool result = false;
try
{
Ping ping = new Ping();
byte[] byteBuffer = new byte[buffer];
PingOptions options = new PingOptions();
PingReply reply = ping.Send(host, timeout, byteBuffer, options);
result = (reply.Status == IPStatus.Success);
}
catch (Exception ex)
{
}
return result;
}
static void IsOnline(string host)
{
isOnline = Ping(host) || isOnline;
}
}
public static class BooleanExtensions
{
public static string ToYesNo(this bool value)
{
return value ? "Yes" : "No";
}
}
}
public static bool CheckForInternetConnection(int timeoutMs = 10000, string url = null)
{
try
{
url ??= CultureInfo.InstalledUICulture switch
{
{ Name: var n } when n.StartsWith("fa") => // Iran
"http://www.aparat.com",
{ Name: var n } when n.StartsWith("zh") => // China
"http://www.baidu.com",
_ =>
"http://www.gstatic.com/generate_204",
};
var request = (HttpWebRequest)WebRequest.Create(url);
request.KeepAlive = false;
request.Timeout = timeoutMs;
using (var response = (HttpWebResponse)request.GetResponse())
return true;
}
catch
{
return false;
}
}
我不同意有人说:“在执行任务之前检查连接有什么意义,因为检查之后连接可能会丢失”。 当然,我们作为开发人员承担的许多编程任务中都存在一定程度的不确定性,但将不确定性降低到可接受的水平是挑战的一部分。
我最近遇到了这个问题,使一个应用程序,其中包括一个映射功能,链接到一个在线瓷砖服务器。这一功能将在缺乏互联网连接时被禁用。
本页上的一些响应非常好,但确实导致了许多性能问题,例如挂起,主要是在没有连接的情况下。
以下是我最终使用的解决方案,在这些答案和我同事的帮助下:
// Insert this where check is required, in my case program start
ThreadPool.QueueUserWorkItem(CheckInternetConnectivity);
}
void CheckInternetConnectivity(object state)
{
if (System.Net.NetworkInformation.NetworkInterface.GetIsNetworkAvailable())
{
using (WebClient webClient = new WebClient())
{
webClient.CachePolicy = new System.Net.Cache.RequestCachePolicy(System.Net.Cache.RequestCacheLevel.BypassCache);
webClient.Proxy = null;
webClient.OpenReadCompleted += webClient_OpenReadCompleted;
webClient.OpenReadAsync(new Uri("<url of choice here>"));
}
}
}
volatile bool internetAvailable = false; // boolean used elsewhere in code
void webClient_OpenReadCompleted(object sender, OpenReadCompletedEventArgs e)
{
if (e.Error == null)
{
internetAvailable = true;
Dispatcher.Invoke(DispatcherPriority.Normal, new Action(() =>
{
// UI changes made here
}));
}
}