刷新div后保持打开多个div

Question

我有问题，页面刷新后我想保持div的样式display: block;。

$(document).on("tap", "a.mini_ledon-icon", function() {
  var cmd = ($(this).attr('id'));
  var p = cmd.split(" ");
  var off = p[0] + ' off';
  $.post("functions/sendcmd.php", {
    CMD: cmd
  })
  $(this).removeClass("mini_ledon-icon");
  $(this).addClass("mini_ledoff-icon");
  $(this).attr("id", off);
});
$(document).on("tap", "a.mini_ledoff-icon", function() {
  var cmd = ($(this).attr('id'));
  var p = cmd.split(" ");
  var on = p[0] + ' on';
  $.post("functions/sendcmd.php", {
    CMD: cmd
  })
  $(this).removeClass("mini_ledoff-icon");
  $(this).addClass("mini_ledon-icon");
  $(this).attr("id", on);
});
var isScrolling = false;
var lastScrollPos = 0;
var counter = 0;


$(function() {

  $('#menu_rep').on('scroll', function() {
    isScrolling = true;
    lastScrollPos = this.scrollTop;
  });

  $(document).on("click", ".submenu-item", function() {
    $(this).parent().find('.dropdown-item').toggleClass('active-dropdown');
    $('.dropdown-item').not($(this).parent().find('.dropdown-item')).removeClass('active-dropdown');
    $(this).parent().find('.submenu').toggle(150);
    $('.submenu').not($(this).parent().find('.submenu')).hide();
    return false;
  });

  refreshTimer = setInterval(agg_menu, 10000);

  function agg_menu() {
    if (!isScrolling) {
      var referer = 'indoor.php #menu_rep';
      $("#menu_rep").load(referer);
    }
    isScrolling = false;
  }
});

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<div class="navigation-wrapper" id="menu_rep">
  <div class="nav-item" id="zona_1">
    <a href="javascript:void(0);" class="submenu-item bedroom2-icon">MODULO A<em class="dropdown-item"></em></a>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="5 off" class="mini_ledoff-icon">POWER 1
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="2 off" class="mini_ledoff-icon">POWER 2
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="1 off" class="mini_ledoff-icon">POWER 3
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="6 off" class="mini_ledoff-icon">POWER 4
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
  </div>
  <div class="nav-item" id="zona_2">
    <a href="javascript:void(0);" class="submenu-item bedroom-icon">MODULO B<em class="dropdown-item"></em></a>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="14 off" class="mini_ledoff-icon">POWER 1
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="11 off" class="mini_ledoff-icon">POWER 2
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="10 off" class="mini_ledoff-icon">POWER 3
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
    <div class="submenu" style="display: none;">
      <a href="javascript:void(0);" id="15 off" class="mini_ledoff-icon">POWER 4
                                  <em class="unselected-item"></em>
                                  </a>
    </div>
  </div>
</div>

我尝试使用jquery.cookie，但是出了点问题，有人可以帮我忙吗？

Answer 1

删除此行，它应该对您有用。

$('.submenu').not($(this).parent().find('.submenu')).hide();