这节课讲怎么处理json数据以及怎么给前台传数据
要处理json数据，需要一个第三方json处理的jar包，目前主流的有gson,fastjson,jackson，我们用fastjson，jar包已经准备好了。
然后我的idea好像出了bug，需要在ProjectStructure里也引一下才能用...不知道你们有没有这个bug
JsonTest.java
package com.test;
import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.JSONObject;
import com.bean.User;
import org.junit.Test;
public class JsonTest {
@Test
public void testJson()
{
//可以看到fastjson是阿里出品的
//处理json无非两种，字符串转对象，对象转字符串
//用单元测试测吧
User user = new User("xiaoye","123456");
//对象转字符串
String jsonStr = JSON.toJSONString(user);
System.out.println(jsonStr);
//字符串转对象
//JSON.parse()得到的对象的类型是JSONObject
Object jsonObj = JSON.parse(jsonStr);
System.out.println("jsonObj: ");
System.out.println(jsonObj.toString());
//这就是json对象和字符串之间的互转
}
}AjaxServlet.java:
package com.servlet;
import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.JSONObject;
import com.bean.User;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import java.io.IOException;
import java.io.PrintWriter;
public class AjaxServlet extends HttpServlet {
@Override
protected void service(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException {
//这里写怎么给前台返回json数据
User user = new User("xiaoye","123456");
String jsonString = JSON.toJSONString(user);
PrintWriter writer = resp.getWriter();
writer.write(jsonString);
//通过response.getWriter()后writer.write()即可给前台返回数据
}
}web.xml:
<servlet>
<servlet-name>AjaxServlet</servlet-name>
<servlet-class>com.servlet.AjaxServlet</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>AjaxServlet</servlet-name>
<url-pattern>/ajax</url-pattern>
</servlet-mapping>ajax.jsp:
<%@ page contentType="text/html;charset=UTF-8" language="java" %>
<html>
<head>
<title>Title</title>
</head>
<body>
<script src="/jquery.js"></script>
<script>
$.ajax({
url:"/ajax",
type:"get",
success:function (data) {
console.log(data);
}
});
</script>
</body>
</html>
