Java WEB

发布日期：2018-12-11 作者：atguigu 9906人浏览

12.5 JQuery 异步请求返回JSON数据

protected void doGet(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {

List<Employee> emps = new EmployeeDaoJdbcImpl().getAllEmps();

Gson gson = new Gson();

String jsonStr = gson.toJson(emps);

response.setContentType("text/html;charset=utf-8");

PrintWriter out = response.getWriter();

out.println(jsonStr);

out.close();

}

function getJsonStr(){

//通过JQuery发送异步请求，将所有的员工信息通过json的格式返回

$.ajax({

url:"getEmpsJsonStr",

type:"post",

dataType:"json",

success:function(data){ // 会直接将后台返回的json字符串转换成js对象

var str = "<tr><th>Id</th><th>LastName</th><th>Email</th><th>Gender</th></tr>";

for(var i= 0 ;i <data.length;i++){

var emp = data[i];

str+="<tr align='center'><td>"

+emp.id+

"</td><td>"

+emp.lastName+

"</td><td>"

+emp.email+

"</td><td>"

+emp.gender+

"</td></tr>"

}

$("#tb").html(str);

}

});

}

<body>

</table>

</body>