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Is there any way to check reachability test of server in jQuery

Is there any way to check reachability test of server .If application not able to connect server then it show a alert? is there any thing method to check.. I check the internet connection .If there is no connection then i am showing a alert .But if there is an connection but there is no reachability of server than how can i handle this.? I am checking like this connection status..!!

setInterval(function () {
    connectionStatus = navigator.onLine ? 'online' : 'offline';
    if(connectionStatus=="offline"){
        // alert("There is no connection");
    }
}, 100);



$.ajax({url: "192.168.12.171",
        dataType: "jsonp",
        statusCode: {
            200: function (response) {
                alert('status 200');
            },
            404: function (response) {
                alert('status  404 ');
            }
        }                        
 });
like image 854
user2563256 Avatar asked Jul 26 '13 09:07

user2563256


1 Answers

To test if your server is up you will need to connect to it and check status massage:

$.ajax('LINK GOES HERE', {
  statusCode: {
    404: function() {
      alert('Not working');
    },
    200: function() {
      alert('Working');
    }
  }
});

Working jsFiddle example: http://jsfiddle.net/Gajotres/PMrDn/47/

$.ajax({url: "http://api.themoviedb.org/2.1/Movie.search/en/json/23afca60ebf72f8d88cdcae2c4f31866/The Goonies",
        dataType: "jsonp",
        statusCode: {
            200: function (response) {
                alert('status 200');
            },
            404: function (response) {
                alert('status  404 ');
            }
        }                        
 });

EDIT :

Use this:

$.ajax({url: "http://192.168.12.171",
        type: "HEAD",
        timeout:1000,
        statusCode: {
            200: function (response) {
                alert('Working!');
            },
            400: function (response) {
                alert('Not working!');
            },
            0: function (response) {
                alert('Not working!');
            }              
        }
 });

Working example: http://jsfiddle.net/Gajotres/PMrDn/48/

like image 117
Gajotres Avatar answered Nov 12 '22 13:11

Gajotres