javascript - jQuery无法正确返回由组成的字符串

我正在将“鲍勃叔叔的有机食品”存储在data-Iname属性中，同时检索与“鲍勃叔叔”相同的值

以下是我用来从data-Iname检索值的过程

var iname = $（this）.attr（“ data-Iname”）;

这是我到目前为止所做的！

$（函数（）{
            var ItemId = 0;

        $("#AddItems").change(function () {
            var SelectedUserId = $("#AddItems").val();
            var SelectedItem = $("#AddItems option:selected").text();
            //alert(SelectedItem);
            var ItemName = "Uncle Bob's Organic";
            items.push(SelectedItem + '|' + SelectedUserId);
            ItemId++;
            //$("#tblItems").append("<tr id=" + ItemId + "><td id=" + ItemId + ">" + SelectedItem + '  ' + "<span class= 'glyphicon glyphicon-trash' title='Delete'></span><br></td></tr>");

            $("#AddedItems").append("<div id=div_" + ItemId + ">" + SelectedItem + '  ' + "<span class= 'remove glyphicon glyphicon-trash' id=remove_" + ItemId + " title='Delete' data-Iname='" + ItemName.toString() + "'></span><br></div>");

        });
    });

    $('.container').on('click', '.remove', function () {
        var iname = $(this).attr("data-Iname");
        var id = this.id;
        var split_id = id.split("_");
        var deleteindex = split_id[1];
        for (var i = 0; i < items.length; i++)
        {
            if (iname == items[i])
            {
                alert(items[i])
                items.splice(i, 1);
            }
        }
        alert(items);
        $("#div_" + deleteindex).remove();

    });

最佳答案

试试这个代替

var ItemName = "Uncle Bob's Organic";
$("#AddedItems").append(
  "<div id=div_" +
    ItemId +
    ">" +
    SelectedItem +
    " " +
    "<span class= 'remove glyphicon glyphicon-trash' id=remove_" +
    ItemId +
    " title='Delete' data-Iname=\"" +
    ItemName.toString() +
    "\"></span><br></div>"
);

所以您的输出类似于data-Iname ='Uncle Bob's Organic'..我只是用双引号替换了外部引号