Select the Start button, then type settings. Select Settings > Network & internet. The status of your network connection will appear at the top. Windows 10 lets you quickly check your network connection status.
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.
Sidenote: To put the entire application in an "offline" state may lead to a lot of error-prone work of handling state.. wireless connections may come and go, etc. So your best bet may be to just fail gracefully, preserve the data, and alert the user.. allowing them to eventually fix the connection problem if there is one, and to continue using your app with a fair amount of forgiveness.
Sidenote: You could check a reliable site like google for connectivity, but this may not be entirely useful as just trying to make your own request, because while Google may be available, your own application may not be, and you're still going to have to handle your own connection problem. Trying to send a ping to google would be a good way to confirm that the internet connection itself is down, so if that information is useful to you, then it might be worth the trouble.
Sidenote: Sending a Ping could be achieved in the same way that you would make any kind of two-way ajax request, but sending a ping to google, in this case, would pose some challenges. First, we'd have the same cross-domain issues that are typically encountered in making Ajax communications. One option is to set up a server-side proxy, wherein we actually ping
google (or whatever site), and return the results of the ping to the app. This is a catch-22 because if the internet connection is actually the problem, we won't be able to get to the server, and if the connection problem is only on our own domain, we won't be able to tell the difference. Other cross-domain techniques could be tried, for example, embedding an iframe in your page which points to google.com, and then polling the iframe for success/failure (examine the contents, etc). Embedding an image may not really tell us anything, because we need a useful response from the communication mechanism in order to draw a good conclusion about what's going on. So again, determining the state of the internet connection as a whole may be more trouble than it's worth. You'll have to weight these options out for your specific app.
Almost all major browsers now support the window.navigator.onLine
property, and the corresponding online
and offline
window events. Run the following code snippet to test it:
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:
Try setting your system or browser in offline/online mode and check the log or the window.navigator.onLine
property for the value changes.
Note however this quote from Mozilla Documentation:
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 afalse
value, you cannot assume that atrue
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 atrue
value; since Firefox 41, on OS X and Windows, the value will follow the actual network connectivity.
(emphasis is my own)
This means that if window.navigator.onLine
is false
(or you get an offline
event), you are guaranteed to have no Internet connection.
If it is true
however (or you get an online
event), it only means the system is connected to some network, at best. It does not mean that you have Internet access for example. To check that, you will still need to use one of the solutions described in the other answers.
I initially intended to post this as an update to Grant Wagner's answer, but it seemed too much of an edit, especially considering that the 2014 update was already not from him.
IE 8 will support the window.navigator.onLine property.
But of course that doesn't help with other browsers or operating systems. I predict other browser vendors will decide to provide that property as well given the importance of knowing online/offline status in Ajax applications.
Until that happens, either XHR or an Image()
or <img>
request can provide something close to the functionality you want.
Update (2014/11/16)
Major browsers now support this property, but your results will vary.
Quote from Mozilla Documentation:
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 afalse
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. All other conditions return atrue
value.
if(navigator.onLine){
alert('online');
} else {
alert('offline');
}
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