如何上传json格式的数据到服务器?

发布网友发布时间：2022-04-23 07:13

共5个回答

懂视网时间：2022-05-15 01:34

这次给大家带来使用JSON传输数据到服务端，的注意事项有哪些，下面就是实战案例，一起来看一下。

准备Hero.java

public class Hero { 
 private String name; 
 private int hp; 
 public String getName() { 
 return name; 
 } public void setName(String name) { 
 this.name = name; 
 } 
 public int getHp() { 
 return hp; 
 } 
 public void setHp(int hp) { 
 this.hp = hp; 
 } 
 @Override 
 public String toString() { 
 return "Hero [name=" + name + ", hp=" + hp + "]"; 
 } 
} 
public class Hero {
 private String name;
 private int hp;
 public String getName() {
 return name;
 }
 public void setName(String name) {
 this.name = name;
 }
 public int getHp() {
 return hp;
 }
 public void setHp(int hp) {
 this.hp = hp;
 }
 @Override
 public String toString() {
 return "Hero [name=" + name + ", hp=" + hp + "]";
 }
}submit.html文件
[html] view plain copy print?<!DOCTYPE html> 
<html> 
<head> 
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"> 
<title>用AJAX以JSON方式提交数据</title> 
<script type="text/javascript" src="jquery.min.js"></script> 
</head> 
<body> 
 <form > 
 名称：<input type="text" id="name"/><br/> 
 血量：<input type="text" id="hp"/><br/> 
 <input type="button" value="提交" id="sender"> 
 </form> 
 <p id="messagep"></p> 
 <script> 
 $('#sender').click(function(){ 
 var name=document.getElementById('name').value; 
 var hp=document.getElementById('hp').value; 
 var hero={"name":name,"hp":hp}; 
 var url="submitServlet"; 
 $.post( 
 url, 
 {"data":JSON.stringify(hero)}, 
 function(data) { 
 alert("提交成功，请在Tomcat控制台查看服务端接收到的数据"); 
 }); 
 }); 
 </script> 
</body> 
</body> 
</html> 
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"> 
<title>用AJAX以JSON方式提交数据</title> 
<script type="text/javascript" src="jquery.min.js"></script> 
</head> 
<body> 
 <form > 
 名称：<input type="text" id="name"/><br/> 
 血量：<input type="text" id="hp"/><br/> 
 <input type="button" value="提交" id="sender"> 
 </form> 
 <p id="messagep"></p> 
 <script> 
 $('#sender').click(function(){ 
 var name=document.getElementById('name').value; 
 var hp=document.getElementById('hp').value; 
 var hero={"name":name,"hp":hp}; 
 var url="submitServlet"; 
 $.post(
 url, 
 {"data":JSON.stringify(hero)},
 function(data) { 
 alert("提交成功，请在Tomcat控制台查看服务端接收到的数据");
 }); 
 }); 
 </script> 
</body> 
</body>
</html>

JSON.stringify函数的作用是将一个javascript对象，转换为JSON格式的字符串。

准备SubmitServlet用来接收数据

import java.io.IOException; 
import javax.servlet.ServletException; 
import javax.servlet.http.HttpServlet; 
import javax.servlet.http.HttpServletRequest; 
import javax.servlet.http.HttpServletResponse; 
import net.sf.json.JSONObject; 
public class SubmitServlet extends HttpServlet { 
 protected void service(HttpServletRequest request, HttpServletResponse response) 
 throws ServletException, IOException { 
 String data =request.getParameter("data"); 
 System.out.println("服务端接收到的数据是：" +data); 
 JSONObject json=JSONObject.fromObject(data); 
 System.out.println("转换为JSON对象之后是："+ json); 
 Hero hero = (Hero)JSONObject.toBean(json,Hero.class); 
 System.out.println("转换为Hero对象之后是："+hero); 
 } 
} 
import java.io.IOException;
import javax.servlet.ServletException;
import javax.servlet.http.HttpServlet;
import javax.servlet.http.HttpServletRequest;
import javax.servlet.http.HttpServletResponse;
import net.sf.json.JSONObject; 
public class SubmitServlet extends HttpServlet { 
 protected void service(HttpServletRequest request, HttpServletResponse response) 
 throws ServletException, IOException {
 String data =request.getParameter("data");
 System.out.println("服务端接收到的数据是：" +data);
 JSONObject json=JSONObject.fromObject(data); 
 System.out.println("转换为JSON对象之后是："+ json);
 Hero hero = (Hero)JSONObject.toBean(json,Hero.class); 
 System.out.println("转换为Hero对象之后是："+hero);
 } 
}

1. 获取浏览器提交的字符串

2. 把字符串转换为JSON对象

3. 把JSON对象转换为Hero对象

最后配置web.xml

<?xml version="1.0" encoding="UTF-8"?> 
<web-app> 
 <servlet> 
 <servlet-name>SubmitServlet</servlet-name> 
 <servlet-class>SubmitServlet</servlet-class> 
 </servlet> 
 <servlet-mapping> 
 <servlet-name>SubmitServlet</servlet-name> 
 <url-pattern>/submitServlet</url-pattern> 
 </servlet-mapping> 
</web-app> 
<?xml version="1.0" encoding="UTF-8"?>
<web-app>
 <servlet>
 <servlet-name>SubmitServlet</servlet-name>
 <servlet-class>SubmitServlet</servlet-class>
 </servlet>
 <servlet-mapping>
 <servlet-name>SubmitServlet</servlet-name>
 <url-pattern>/submitServlet</url-pattern>
 </servlet-mapping>

</web-app>启动tomcat访问http://127.0.0.1:8080/项目名/submit.html

在tomcat控制台看到传来的数据

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如何上传json格式的数据到服务器?