通过JSON数组对象检索多个值

小编典典

通过JSON数组对象检索多个值

jsp

我通过JSON数组获取一个值，但是如何在JSON数组中存储多个值以及如何通过JavaScript检索呢？

auto.jsp：

 <script type="text/javascript">
 $(document).ready(function() {
    $("#combo").change(function() { // after onchange event it goes to combo1.jsp 
        $.getJSON('combo1.jsp', {
            count: this.value
        }, function(responseData) {
            var splitValues = responseData.name.split(/,/);

            $("#combo1").empty().append("<option>please select</option>");


            for (var idx in splitValues) {
                $("#combo1").append(
                $("<option></option>").html(splitValues[idx]).val(splitValues[idx]));
            }
        });

    });
});


    </script>

<body>
//first combo box
<select id="combo" name="count">

     <option value="">please select</option>

      <option value="a">A</option>

 </select>

//second combo box

<select id="combo1" name="combo1Val">

     // i am getting only "5" here, but i want to show 1,2,3,4,5 as in drop down list

 </select> 
</body>

combo1.jsp：

<%
String count=request.getParameter("count");// by onchange event of first combo, i am 
getting value "a" here
if(count.equalsIgnoreCase("a")){
// in my db there are 5 values i.e. 1,2,3,4,5 for single value "a", but while   
populating in second combo i am getting only one value "5", how? please ignore my db   
connection in jsp

  JSONObject arrayObj= new JSONObject(); 
// by db connection i am fetching 5 values but while printing in javascript i am  
getting only last one that is "5" in second combo, how can i populate all values  
1,2,3,4,5 as drop down items in second combo box?
// retrieveing 5 datas(1,2,3,4,5) from db where name =a
 while(rs.next()){
      t1=(String)(rs.getString(1));// there are 5 values in db relating to "a", but i  
am getting only last value i.e. "5" in second combo
       }
       arrayObj.put("name",t1);
          response.setContentType("application/json");
      response.getWriter().write(arrayObj.toString());
      }
%>

阅读 380

2020-06-08

共1个答案

小编典典

我认为您在第二个组合中只得到5，因为您在循环中犯了一个错误。你可以做

//in combo1.jsp

String selectedValue = request.getParameter("count");

Map<String, String> options = new Map<String, String>();
 //get your data from db
while(rs.next()){
    String t1=(String)(rs.getString(1));
    options.add(t1, t1);
}
String json = new Gson().toJson(options);
response.setContentType("application/json");
response.setCharacterEncoding("UTF-8");
response.getWriter().write(json);


    //to get data
    $.getJSON('combo1.jsp', {
        count: this.value
    }, function(options) {
        var dropdown2 = $('#combo1');
        $('>option', dropdown2).remove(); // Clean old options first.
        if (options) {
            $.each(opts, function(key, value) {
                dropdown2.append($('<option/>').val(key).text(value));
            });
        } else {
            dropdown2.append($('<option/>').text("Please select dropdown1"));
        }
    });

2020-06-08