Question posted in
You can find the official documentation here.

Refresh div on JSP page with Ajax call without a response – Jquery ajax

I am aware of using AJAX calls to refresh content inside a div using the response received, but in my case I have no response but still want to refresh the div.

My application has a table called ‘Cart’, on which when I click the delete button, it removes an item from the cart. I remove the item by sending an AJAX call to the Java servlet that updates the cart in an ArrayList. As I’m removing only one object, I dont have a response, but still want to update the cart with the removed item.

I have tried the following ways:

  1. Redirect to the same page using servlet after updating cart array. This causes the page to reload.
    request.getRequestDispatcher(target).forward(request, response); where target is index.jspx

  2. Use $(..).load('index.jspx') to try and refresh the particular div, but this places the entire contents of index.jspx into the tiny div with the cart table.

Tags:

2

Answers


  1. 0

    You actually don’t have to redirect the whole thing again. Instead of redirecting, just use response.getWriter().write(<Your updated Array>); inside doPost() or doGet() depending on which you are using.

    This will send your updated array to your AJAX code, instead of refreshing the whole page.

    Login or Signup to reply.
  2. 0

    An AJAX request (Asynchronous Javascript And XML) is sent to your server. You can define a callback for your AJAX request, that is, a Javascript function that is going to be executed once the request ended. You can also use promises for this purpose. Anyway, once you know how you can execute a function AFTER the request has completed, you need to decide whether you need information from the server.

    If you are content with not receiving any meaningful answers from the server and are okay with assuming that the removal was successful, then you can simply remember which delete button was pressed, find the row which contains it and remove() it. In this case you do not need to send further requests to the server.

    Otherwise, if you need to get further information from the server (because new items might have been created/modified/removed), then simply change your server-side in order to send back a JSON about the items to the client-side and refresh your table there.

    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search