javascript 在 JS 中检查互联网连接的最快方法
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Quickest way to check internet connectivity in JS
提问by gaganHR
I have this code in a html file that checks for connectivity, but the problem here is that it takes about 10 seconds to throw the alert message box to indicate the connection is lost. I want to know if there's a much faster way to inform the user that the connection is lost without having to wait. Strictly JS thanks...
我在检查连接的 html 文件中有此代码，但这里的问题是抛出警报消息框以指示连接丢失需要大约 10 秒。我想知道是否有一种更快的方法来通知用户连接丢失而无需等待。严格的JS谢谢...
JS code:
JS代码：
<script language="JavaScript">
function SubmitButton()
{
if(navigator.onLine)
{
document.getElementById('SubmitBtn').style.visibility='visible';
}
else
{
document.getElementById('SubmitBtn').style.visibility='hidden';
}
}
function Online()
{
var status=false;
status= navigator.onLine;
if(status!= true)
{
alert('Network connectivity is lost, please try again later');
}
}
</script>
Calling it in html file here:
在 html 文件中调用它：
<INPUT name="ccMCA1input" type="checkbox" onclick="ccMCA1.ccEvaluate(),Online()" value=False>
采纳答案by Rob W
navigator.onLineis the only built-in property which can be checked (and not reliable btw).
You could create a XHR request to a reliable server, and check whether any response is being received.
navigator.onLine是唯一可以检查的内置属性（顺便说一句，不可靠）。
您可以向可靠的服务器创建 XHR 请求，并检查是否收到任何响应。
回答by Joseph
You could periodically request a 1x1 gif image from a(ny) server, making sure you use the cache buster methodto avoid caching. You could use the onloadand onerrorevents.
您可以定期从（纽约）服务器请求 1x1 gif 图像，确保使用缓存破坏器方法来避免缓存。您可以使用onload和onerror事件。
var isOnline = (function isOnline(){
// Create a closure to enclose some repeating data
var state = false;
var src = 'gif_url?t=' + Date.now();
var function onLoad = function(){state = true;}
var function onError = function(){state = false;}
// Inside the closure, we create our monitor
var timer = setInterval(function(){
var img = new Image();
img.onload = onLoad;
img.onerror = onError;
img.src = src;
}, 10000);
// Return a function that when called, returns the state in the closure
return function(){return state;};
}());
//Use as
var amIOnline = isOnline();
回答by Dhwanil Shah
Consider checking out the following URLs:
考虑查看以下 URL：
