Question

这是我的ajax请求

var id = $(this).attr('data-id');

$.ajax({
    type: 'POST',
    url: '../include/residents.jsp',
    dataType: "json",
    data: {
        id: id,
    },
    success: function(data) {},
    error: function(data) {}
}).done(function() {
});

这就是resident.jsp

<%@ page import="java.io.*,java.util.*, javax.servlet.*,java.text.*" %>
<%@ page import="javax.swing.*" %>
<%@page import="java.text.DateFormat"%>
<%@page import="java.text.SimpleDateFormat"%>
<%@page import="java.sql.PreparedStatement"%>
<%@ page language="java"%>
<%@page import="java.sql.ResultSet"%>
<%@page import="java.sql.Statement"%>
<%@ include file="dbconfig.jsp" %>

<% 
    String id = request.getParameter("id");


    try {

        conn.setAutoCommit(false);
        PreparedStatement preparedStatement = null;
        Map map = new HashMap();
        String add_supplemental = "SELECT * FROM residence_tb WHERE ri_id = ?"; 

            preparedStatement = conn.prepareStatement(add_supplemental);
            preparedStatement.setString(1,id);
            out.println("before");
            ResultSet rs = preparedStatement.executeQuery();
            while (rs.next()) {
                    String lname = rs.getString("ri_lname");
                    String fname = rs.getString("ri_fname");
                    String mname = rs.getString("ri_mname");
                    String suffix = rs.getString("ri_suffix");  
                    String bdate = rs.getString("ri_bdate");
                    String age = rs.getString("ri_age");    
                    String gender = rs.getString("ri_gender");
                    String status = rs.getString("ri_status");  
                    String mobileno = rs.getString("ri_mobileno");
                    String telephoneno = rs.getString("ri_telephoneno");    
                    String province = rs.getString("ri_province");  
                    String municipality = rs.getString("ri_municipality");  
                    String brgy = rs.getString("ri_brgy");  
                    String street = rs.getString("ri_street");  


                    map.put("lname", lname);
                    map.put("fname", fname);


            }

          out.println(map); 
//        rst.close();
//        stmt.close();

        conn.commit();
        conn.close();

    } catch(Exception e) {
        e.getMessage();
        e.printStackTrace();
    }
%>

当我out.println(map);得到

时，它已经很好了

{fname=Juan, lname=Dela Cruz}

在回复中。当我尝试return map时，我得到了

不兼容类型：意外返回值

如何返回map以便我可以成功使用它或在ajax中完成功能

Answer 1

请从响应中返回一个JSON字符串。 JSON是浏览器中的地图。

Map<String, Object> jsonMap = new HashMap<String, Object>();
jsonMap.put( "Status", "Success" );
jsonMap.put( "Rows", 100 );
ObjectMapper mapper = new ObjectMapper();
mapper.getFactory().configure(JsonGenerator.Feature.ESCAPE_NON_ASCII, true);        
String jsonStr = mapper.writeValueAsString( jsonMap );
response.getOutputStream().print( jsonStr );

ObjectMapper是Jackson / fastxml的一个类。此代码将JSON返回给浏览器。

response.getOutputStream().print("{\"result\":[{ \"lname\": \"Smith\", \"fname\": \"John\"}, { \"lname\": \"Smith1\", \"fname\": \"John1\" }, { \"lname\": \"Smith2\", \"fname\": \"John2\" }]}");

在浏览器中：

var res = JSON.parse(request.responseText)
console.log(res.result);

如何从ajax请求返回地图数据

1 个答案: