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Is there any possible way to check the internet connection for each and every time of submission in JavaScript?

Prakash Pazhanisamy
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    Possible duplicate of [Detect that the Internet connection is offline?](https://stackoverflow.com/questions/189430/detect-that-the-internet-connection-is-offline) –  Oct 03 '17 at 05:15

4 Answers4

5

You can use navigator.onLine to check the either there is internet connection or not. You can also use 2 event listener to notify when network connection is connected or disconnected.

console.log(navigator.onLine);

// OR

window.addEventListener('online', () => {
   //console.log('Online');
});

window.addEventListener('offline', () => {
        //console.log('Offline');  
});
Prakash Pazhanisamy
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Arun Redhu
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  • `console.log(navigator.onLine)` is working. But `window.addEventListener` is not working. – Prakash Pazhanisamy Oct 03 '17 at 11:00
  • `window.addEventListener` will only work when your connection changes i.e when you goes from online to offline or vice-versa – Arun Redhu Oct 04 '17 at 04:59
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    In Chrome and Safari navigator.online will output true if you are connected to a LAN or a router. That does not mean you have internet access. https://developer.mozilla.org/en-US/docs/Web/API/NavigatorOnLine/onLine Try @Raed Salah solution below instead. – Poppe76 Oct 18 '19 at 12:40
3

Navigator.online is a method that checks whether there is a connection or no. However this does not check "Internet connection" as it will always output true if you are connected to a network with or without internet.

here is a snippet 

function findIp() {
  var findIP = new Promise(r => {
    var w = window,
      a = new (w.RTCPeerConnection ||
        w.mozRTCPeerConnection ||
        w.webkitRTCPeerConnection)({ iceServers: [] }),
      b = () => {};
    a.createDataChannel("");
    a.createOffer(c => a.setLocalDescription(c, b, b), b);
    a.onicecandidate = c => {
      try {
        c.candidate.candidate
          .match(
            /([0-9]{1,3}(\.[0-9]{1,3}){3}|[a-f0-9]{1,4}(:[a-f0-9]{1,4}){7})/g
          )
          .forEach(r);
      } catch (e) {}
    };
  });
  findIP
    .then(ip => $("#ipchk").html("your IP: " + ip))
    .catch(e => console.error(e));
}

//gets the network status from the browser navigator api once page is loaded
function chkstatus() {
  if (navigator.onLine) {
    console.log("online");
    $("#netchk").html("online");
    $(".dot").removeClass("offline");
    $(".dot").addClass("online");
    //print ip if there is connection
    findIp();
  } else {
    console.log("offline");
    $("#netchk").html("offline");
    $(".dot").removeClass("online");
    $(".dot").addClass("offline");
  }
}

//check status every 5 seconds
setInterval(chkstatus, 5000);

$(document).ready(function() {
  chkstatus();

  //event listener for changes in the netwrok
  window.addEventListener("offline", function(e) {
    $("#netchk").html("offline");
    $(".dot").removeClass("online");
    $(".dot").addClass("offline");
    $("#ipchk").html("your ip: ");
  });

  window.addEventListener("online", function(e) {
    console.log("online");
    $("#netchk").html("online");
    $(".dot").removeClass("offline");
    $(".dot").addClass("online");
    findIp();
  });
});
#main-content {
  height: 100vh;
  display: flex;
  align-items: center;
}
#content {
  text-align: center;
  background-image: linear-gradient(to left, #4776e6, #8e54e9);
  padding: 20px;
  border-radius: 30px;
  color: #fafafa;
  margin: 0 auto;
  font-size: 20px;
  font-family: "Courier New", Courier, monospace;
  text-transform: capitalize;
  box-shadow: 0 10px 20px rgba(41, 72, 255, 0.19),
    0 6px 6px rgba(41, 72, 255, 0.23);
}

.dot {
  height: 15px;
  width: 15px;
  border-radius: 50%;
  display: inline-block;
}
.online {
  background-color: greenyellow;
}
.offline {
  background-color: #f44336;
}
<!DOCTYPE html>
<html>
  <head>
    <meta charset="utf-8" />
    <meta http-equiv="X-UA-Compatible" content="IE=edge" />
    <title>Network Status</title>
    <meta name="viewport" content="width=device-width, initial-scale=1" />
    <link rel="stylesheet" type="text/css" media="screen" href="main.css" />
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
    <script src="main.js"></script>
  </head>
  <body>
    <div class="layout">
      <div id="main-content">
        <div id="content">
          <div><span id="netchk"></span> <span class="dot"></span></div>
          <hr />
          <div id="ipchk"></div>
        </div>
      </div>
    </div>
  </body>
</html>

try connecting through your phone hotspot and disconnecting the internet/4G. you will notice it still shows online.

this code check your network status and ip every 5 secs, here is a link to the github REPO.

The best way to check for internet connection for now is through an ajax request.

Raed Salah
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0

U can do this:

// When the internet is off
addEventListener('offline', function(){
  alert("No Internet");
});

 // When the internet is on
addEventListener('online', function(){
  alert("Back online");
});

Or you can do that:

  

alert(/*It would either say true or false*/ navigator.onLine? /* On true: */"Online":/*On false:*/"Offline");
Nicecat99
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  • A good answer will always include an explanation why this would solve the issue, so that the OP and any future readers can learn from it. – Tyler2P Jan 28 '22 at 22:34
-1
var x = confirm("Are you sure you want to submit?");
if(x){
    if(navigator.onLine == true) {
        return true;
    } else {
        alert('Internet connection is lost');
        return false; 
    }
}
else {
    return false;
}
Prakash Pazhanisamy
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