单击按钮时遵循 select 选项 url

follow select option url on click on button

我有一个 select 输入类型,其中我有许多具有 URL 值的选项。我还有一个按钮,显示是否有任何选项被 selected。但是如果我选择并按下按钮,我想要一种方法,它应该会带我到 URL。我不擅长JS。这是我所做的。谢谢

<form class="center" name="jump">
  <select name="menu">
    <option value="">States</option>
    <option value="https://siteurl.com/product/alabama/">Alabama</option>
    <option value="https://siteurl.com/product/alaska-advance-directive-for-health-care/">Alaska</option>
    <option value="https://siteurl.com/product/arizona-advance-directive-for-health-care/">Arizona</option>
    <option value="https://siteurl.com/product/arkansas-advance-directive-for-health-care/">Arkansas</option>
    <option value="https://siteurl.com/product/california-advance-directive-for-health-care/">California</option>
    <option value="https://siteurl.com/product/colorado-advance-directive-for-health-care/">Colorado</option>
  </select>
  <br>

  <input class="box" type="button" style="display: inline-block;">

</form>

<script type="text/javascript">
$(document).ready(function(){

   $("select").change(function(){
        $(this).find("option:selected").each(function(){
        if($(this).attr("value")==""){

                $(".box").hide();
            }
            else{
                $(".box").show();
      $('.box').on('click', function () {
          var url = $(this).attr("value"); 
          if (url) { 
              window.location = url; 
          }
          return false;
      });

            }
        });
    }).change();
});



</script>

尝试

<script type="text/javascript">
$(document).ready(function(){

 $("select").change(function(){
    $(this).find("option:selected").each(function(){
    if($(this).attr("value")==""){

            $(".box").hide();
        }
        else{
            $(".box").show();
            var self=this;
  $('.box').on('click', function () {
      var url = $(self).attr("value"); 
      if (url) { 
          window.location = url; 
      }
      return false;
  });

        }
    });
}).change();
});
</script>

我认为它会工作检查这个 link here