如何在JavaScript中检测Internet连接是否离线?
当前回答
几乎所有主流浏览器现在都支持window.navigator. online属性,以及相应的在线和离线窗口事件。运行以下代码片段来测试它:
console.log('Initially ' + (window.navigator.onLine ? 'on' : 'off') + 'line'); window.addEventListener('online', () => console.log('Became online')); window.addEventListener('offline', () => console.log('Became offline')); document.getElementById('statusCheck').addEventListener('click', () => console.log('window.navigator.onLine is ' + window.navigator.onLine)); <button id="statusCheck">Click to check the <tt>window.navigator.onLine</tt> property</button><br /><br /> Check the console below for results:
尝试将您的系统或浏览器设置为离线/在线模式,并检查日志或window.navigator.onLine属性的值变化。
请注意Mozilla文档中的这段话:
In Chrome and Safari, if the browser is not able to connect to a local area network (LAN) or a router, it is offline; all other conditions return true. So while you can assume that the browser is offline when it returns a false value, you cannot assume that a true value necessarily means that the browser can access the internet. You could be getting false positives, such as in cases where the computer is running a virtualization software that has virtual ethernet adapters that are always "connected." Therefore, if you really want to determine the online status of the browser, you should develop additional means for checking. In Firefox and Internet Explorer, switching the browser to offline mode sends a false value. Until Firefox 41, all other conditions return a true value; since Firefox 41, on OS X and Windows, the value will follow the actual network connectivity.
(重点是我自己的)
这意味着如果window.navigator.onLine为false(或者您得到一个脱机事件),则保证您没有Internet连接。
然而,如果它是真的(或者你得到一个在线事件),它充其量只意味着系统连接到某个网络。这并不意味着你可以上网。要检查这一点,您仍然需要使用其他答案中描述的解决方案之一。
我最初打算把这篇文章作为格兰特·瓦格纳(Grant Wagner)的回答的更新,但这似乎太过编辑了,尤其是考虑到2014年的更新已经不是他的了。
其他回答
对于两种不同的情况,有两个答案:-
If you are using JavaScript on a website(i.e; or any front-end part) The simplest way to do it is: <h2>The Navigator Object</h2> <p>The onLine property returns true if the browser is online:</p> <p id="demo"></p> <script> document.getElementById("demo").innerHTML = "navigator.onLine is " + navigator.onLine; </script> But if you're using js on server side(i.e; node etc.), You can determine that the connection is lost by making failed XHR requests. The standard approach is to retry the request a few times. If it doesn't go through, alert the user to check the connection, and fail gracefully.
window.navigator.onLine
是你要找的,但这里要添加的东西很少,首先,如果它是你想要持续检查的应用程序上的一些东西(比如看看用户是否突然脱机,在这种情况下大多数时候是正确的,那么你也需要监听变化),为此你添加事件监听器到窗口来检测任何变化,为了检查用户是否脱机,你可以这样做:
window.addEventListener("offline",
()=> console.log("No Internet")
);
检查是否在线:
window.addEventListener("online",
()=> console.log("Connected Internet")
);
我不得不为一个与学校有很多合作的客户制作一个web应用程序(基于ajax),这些学校的互联网连接经常不好,我使用这个简单的功能来检测是否有连接,工作得非常好!
我使用CodeIgniter和Jquery:
function checkOnline() {
setTimeout("doOnlineCheck()", 20000);
}
function doOnlineCheck() {
//if the server can be reached it returns 1, other wise it times out
var submitURL = $("#base_path").val() + "index.php/menu/online";
$.ajax({
url : submitURL,
type : "post",
dataType : "msg",
timeout : 5000,
success : function(msg) {
if(msg==1) {
$("#online").addClass("online");
$("#online").removeClass("offline");
} else {
$("#online").addClass("offline");
$("#online").removeClass("online");
}
checkOnline();
},
error : function() {
$("#online").addClass("offline");
$("#online").removeClass("online");
checkOnline();
}
});
}
对域的ajax调用是检测您是否脱机的最简单方法
$.ajax({
type: "HEAD",
url: document.location.pathname + "?param=" + new Date(),
error: function() { return false; },
success: function() { return true; }
});
这只是给大家一个概念,还需要改进。
例如,error=404仍然意味着您在线
如何发送一个不透明的http请求到google.com与无cors?
fetch('https://google.com', {
method: 'GET', // *GET, POST, PUT, DELETE, etc.
mode: 'no-cors',
}).then((result) => {
console.log(result)
}).catch(e => {
console.error(e)
})
设置no-cors的原因是,即使在我的pc上取消网络连接时,我也收到了cors错误。所以不管有没有网络连接,我都被封锁了。添加no-cors使请求不透明,这显然似乎绕过cors,并允许我只是简单地检查我是否可以连接到谷歌。
供参考:我在这里使用fetch进行http请求。 https://www.npmjs.com/package/fetch