jquery - 选择正确的itemID jquery

Question

我附加了这个字符串，列出了 mycart.html 中的项目

 t = "<div class='items'><img src='"+ serverURL() + "/images/" + 
arr[i].itemImage + "' class='image'><div class='deletebtn'><input 
type='image' id='deletebtn' src='images/deletebtn.jpg' class='deletebtn'> 
</div><div class='itemName'><b> "+arr[i].itemName+"</b></div> <div 
class='itemID'>Item ID : " +arr[i].itemID+" </div><div class='itemSize'> 
<b>"+ arr[i].itemSize+"</b></div><div class='itemPrice'>$"+ 
arr[i].itemPrice +"</div></div>"

如果单击“deletebtn”，我的目标是删除该项目。如何获取正确的 itemID，以便能够将值传输到我的 json 对象？

 function deleteitem(){

    var url = serverURL() + "/deletemycartitems.php";
    var  email = localStorage.getItem("email");
   //var itemID = ;
  var JSONObject = {
    "email": email,
    "itemID": itemID,

  };

  $.ajax({
   url: url,
   type: 'GET',
   data: JSONObject,
   dataType:'json',
   contentType: "application/json; charset=utf-8",
   success: function (arr) {
  _getmycart(arr);
  },
  error: function () {
   alert("Failed");
  }
});


}

标签： jqueryhtml

Answer 1

您需要将您的 itemid 作为 data-itemid 存储在按钮中

<input 
type='image' id='deletebtn' src='images/deletebtn.jpg' class='deletebtn' data-itemid='"+arr[i].itemID+"'> 

然后单击删除按钮，尝试检索此 ID 为

let itemId = $(this).data('itemid');