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
请求头错误
$.ajax({
url: /your_url,
type: "POST or GET",
data: your_data,
success: function(result){
//do stuff
},
error: function(xhr, status, error) {
//detect if user is online and avoid the use of async
$.ajax({
type: "HEAD",
url: document.location.pathname,
error: function() {
//user is offline, do stuff
console.log("you are offline");
}
});
}
});
几乎所有主流浏览器现在都支持window.navigator. online属性,以及相应的在线和离线窗口事件。运行以下代码片段来测试它:
console.log('Initially ' + (window.navigator.onLine ? 'on' : 'off') + 'line'); window.addEventListener('online', () => console.log('Became online')); window.addEventListener('offline', () => console.log('Became offline')); document.getElementById('statusCheck').addEventListener('click', () => console.log('window.navigator.onLine is ' + window.navigator.onLine)); <button id="statusCheck">Click to check the <tt>window.navigator.onLine</tt> property</button><br /><br /> Check the console below for results:
尝试将您的系统或浏览器设置为离线/在线模式,并检查日志或window.navigator.onLine属性的值变化。
请注意Mozilla文档中的这段话:
In Chrome and Safari, if the browser is not able to connect to a local area network (LAN) or a router, it is offline; all other conditions return true. So while you can assume that the browser is offline when it returns a false value, you cannot assume that a true value necessarily means that the browser can access the internet. You could be getting false positives, such as in cases where the computer is running a virtualization software that has virtual ethernet adapters that are always "connected." Therefore, if you really want to determine the online status of the browser, you should develop additional means for checking. In Firefox and Internet Explorer, switching the browser to offline mode sends a false value. Until Firefox 41, all other conditions return a true value; since Firefox 41, on OS X and Windows, the value will follow the actual network connectivity.
(重点是我自己的)
这意味着如果window.navigator.onLine为false(或者您得到一个脱机事件),则保证您没有Internet连接。
然而,如果它是真的(或者你得到一个在线事件),它充其量只意味着系统连接到某个网络。这并不意味着你可以上网。要检查这一点,您仍然需要使用其他答案中描述的解决方案之一。
我最初打算把这篇文章作为格兰特·瓦格纳(Grant Wagner)的回答的更新,但这似乎太过编辑了,尤其是考虑到2014年的更新已经不是他的了。
您可以使用$.ajax()的错误回调,它在请求失败时触发。如果textStatus等于字符串"timeout",这可能意味着连接中断:
function (XMLHttpRequest, textStatus, errorThrown) {
// typically only one of textStatus or errorThrown
// will have info
this; // the options for this ajax request
}
医生说:
错误:一个函数被调用,如果请求 失败。函数被传递了3个 参数:XMLHttpRequest对象 描述错误类型的字符串 这是可选的 异常对象,如果发生。 第二个可能的值 参数(除null外)为"timeout", "error", "notmodified"和 “parsererror”。这是一个Ajax事件
例如:
$.ajax({
type: "GET",
url: "keepalive.php",
success: function(msg){
alert("Connection active!")
},
error: function(XMLHttpRequest, textStatus, errorThrown) {
if(textStatus == 'timeout') {
alert('Connection seems dead!');
}
}
});
HTML5应用程序缓存API指定导航器。onLine,目前在IE8测试版中可用,WebKit(例如。Safari)的夜间运行,并且已经在Firefox 3中得到支持
导航器等方法的问题。onLine是他们不兼容的一些浏览器和移动版本,一个选项,帮助我很多是使用经典的XMLHttpRequest方法,也预见到可能的情况下,文件存储在缓存响应XMLHttpRequest。Status大于200小于304。
这是我的代码:
var xhr = new XMLHttpRequest();
//index.php is in my web
xhr.open('HEAD', 'index.php', true);
xhr.send();
xhr.addEventListener("readystatechange", processRequest, false);
function processRequest(e) {
if (xhr.readyState == 4) {
//If you use a cache storage manager (service worker), it is likely that the
//index.php file will be available even without internet, so do the following validation
if (xhr.status >= 200 && xhr.status < 304) {
console.log('On line!');
} else {
console.log('Offline :(');
}
}
}