CODEWITHSUNDEEP
javascriptjqueryajax
David
David

Reputation: 4508

javascript does not reconnect to server side if connection is lost

I have Jquery+Ajax page that connects to server gets some data, and displays it. This is done all the time, function calls itself after it completes all task and starts to perform again and again. But If komputer loses internet connection everything stops. And when connections comes back again, nothing happens. Here is javascript code. How can I iprove it so it will continue to work after connection is back again.

$(document).ready(function() {
var a=1;
var max='23';

function kiosk() 
{
    if(a<max) 
    {
        $.post('engine.php', { index: a },
        function(result) 
        {
            result = jQuery.parseJSON(result);

            $('#main').css({'display':'none'});


            $('#div_city').html(result.dest);
            $('#div_price').html(result.price);
            $('#div_price_pre').html('From&nbsp;');
            $('#div_price_after').html('&nbsp;Euro');

                $('#main').fadeIn(2000).delay(3000).fadeOut(2000, function() 
                    {

                            $('#main').hide;
                            a++;
                            kiosk();


                    });

        });
    }
    else
    {
        a=1;
        kiosk();
    }

}
kiosk();
});

Upvotes: 0

Views: 872

Answers (3)

Neeraj
Neeraj

Reputation: 8532

I would suggest rather than using $.post you use

$.ajax({ type: "post" , success : function1() , failure: function2() }.

In function2() you can call koisk again after a timeout. and the succces would be the function you have created right now.

This link will help you understand the ajax function jQuery Ajax error handling, show custom exception messages

Eg Code snippet:

function kiosk() 
{
if(a<max) 
{
    $.ajax({type : "POST" , url : 'engine.php',data: { index: a },
    success : function(result) 
    {
        result = jQuery.parseJSON(result);

        $('#main').css({'display':'none'});


        $('#div_city').html(result.dest);
        $('#div_price').html(result.price);
        $('#div_price_pre').html('From&nbsp;');
        $('#div_price_after').html('&nbsp;Euro');

            $('#main').fadeIn(2000).delay(3000).fadeOut(2000, function() 
                {

                        $('#main').hide;
                        a++;
                        kiosk();


                });

    },error : function (xhr, ajaxOptions, thrownError){ setTimeout(200,kiosk())  }

    });
}
else
{
    a=1;
    kiosk();
}

} kiosk(); });`

Upvotes: 3

Guillaume Poussel
Guillaume Poussel

Reputation: 9822

Using jQuery.post you are passing a callback which is executed only on success. You should use jQuery.ajax which has different callbacks. See documentation with complete, success and error callbacks.

You could even use statusCode to map a specific HTTP code to a custom function.

Upvotes: 0

arunes
arunes

Reputation: 3524

Script brokes because i think it throws exception. You can add your code try and catch block, and even if has errors you can continue to try.

try {
    result = jQuery.parseJSON(result);
    // do your work
} catch {
    // call function again
}

Or you can use, jquery ajax, it has a onerror event.

Upvotes: -1

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