【问题标题】:Pass json to dropdown将 json 传递给下拉列表
【发布时间】:2014-03-23 18:13:55
【问题描述】:

如何在下拉选项中将列表存储在此 servlet 中?如果 Age 的值将存储为下拉选项的值并且名称将显示在下拉列表中,那么等效的 jquery 是什么

public class testServlet extends HttpServlet {
protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
        response.setContentType("application/json");
        PrintWriter out = response.getWriter();
                List<Student> studlist = new ArrayList<Student>();
            Student s1 = new Student();
            s1.setName("Student 1");
            s1.setAge(10);
            Student s2 = new Student();
            s2.setName("Student 2");
            s2.setAge(14);
            Gson gson = new GsonBuilder().setPrettyPrinting().create();
        String json = gson.toJson(studlist);
        out.print(json);
}
}

试一试:

$(document).ready(function() {
    $('#student').ready(function(event) {  
       $.get('studentServlet',function(responseJson) {   
        var $select = $('#student');                                                 
           $.each(responseJson, function(studlist) {               
               $('<option>').val(this.['age']).text(this['name']).appendTo($select);      
                });
        });
    });
}); 

【问题讨论】:

    标签: java jquery json jsp servlets


    【解决方案1】:
    $(document).ready(function() {
        $('#student').ready(function(event) {  
           $.get('studentServlet',function(responseJson) {   
            var $select = $('#student');                                                 
               $.each(responseJson, function(studlist) {               
                   $('<option>').val(this.['age']).text(this['name']).appendTo($select);      
                    });
            });
        });
    }); 
    

    【讨论】:

      猜你喜欢
      • 1970-01-01
      • 1970-01-01
      • 2011-08-13
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      • 1970-01-01
      相关资源
      最近更新 更多