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JSON格式的数据如何提交到服务端

程序员文章站 2022-05-26 10:22:25
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这次给大家带来,JSON格式数据提交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

JSON格式的数据如何提交到服务端

在tomcat控制台看到传来的数据

JSON格式的数据如何提交到服务端

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