Javascript – How to call ajax again when user click back button to go back last webpage

ajaxhtmljavascriptjquery

Below is my code..

HTML Code

<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<div class="body">
    <div class="dropdown_div">
        <select id="q_type" class="dropdown" onchange="getSubject(this.value)">
            <option>Question1</option>
            <option>Question2</option>  
        </select>
    </div>
    <div class="dropdown_div">
        <select id="q_subject" class="dropdown">
            <option>Subject1</option>
        </select>
    </div>
</div>

JS Code

function getSubject(val){

  $("option", $("#q_subject")).remove();
  var option = "<option>Subject</option>";
  $("#q_subject").append(option);

    $.ajax({
      url: "api.path",
      type: 'POST',
      dataType: 'json',
      data: {id: id},
      async: true,
      cache: false,
      success: function(response) {
                    alert("Hi");
          $("option", $("#q_subject")).remove();
          var option = "<option>Subject1</option>"; 
          option += "<option value=1234>Subject2</option>"; 
          $("#q_subject").append(option); 
      }
    });
}

How do I use pushState into my code and let user can click back button to return last page and then still see the ajax data?

Best Answer

First of all, you should save data received from ajax request to browser local storage. Afterwards, in order to show ajax result when browser "back" button was fired, you should bind statements that you are calling in ajax.success() method to window onpopstate event. To omit code duplication, it`s better to use a declared function instead of anonymous one. 

function success(response) {
    alert("Hi");
    $("option", $("#q_subject")).remove();
    var option = "<option>Subject1</option>"; 
    option += "<option value=1234>Subject2</option>"; 
    $("#q_subject").append(option); 
}

Save data to localstorage and call success function:

 $.ajax({
      url: "api.path",
      type: 'POST',
      dataType: 'json',
      data: {id: id},
      async: true,
      cache: false,
      success: function(response) {
          localStorage.setItem("response", response);
          success(response);
      }
    });

Call success() when "back" button was fired:

window.onpopstate = function (e) {
    var res = localStorage.getItem('response');         
    success(res);
}
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