Angular+servlet java实现前后端数据交互

来源：互联网发布：让ea客服转移游戏数据编辑：程序博客网时间：2024/06/06 01:49

每学习一个新知识我都会想到的是用它来实现对数据的CRUD，Angularjs也不例外，而实现CRUD的前提就是前后端数据能够进行交互，下面我就展示一个简易的前后端交互代码

Html+js代码:

<!DOCTYPE html><html lang="en" ><head><meta charset="utf-8"><title>servlet+angularjs数据交互</title> <script src="framework/angular-1.3.0.14/angular.js"></script>  <script src="framework/jquery-1.9.1.js"></script></head><body ng-app="myApp">    <div ng-controller="demo">            <input type="button" value="提交" ng-click="submit()"/>    </div><script>var app=angular.module("myApp", [])  app.config(['$httpProvider', function($httpProvider) { $httpProvider.defaults.headers.post["Content-Type"] =         "application/x-www-form-urlencoded";  $httpProvider.defaults.          transformRequest.unshift(function(data,headersGetter) {                var key, result = [];            for (key in data) {                 if (data.hasOwnProperty(key)) {                     result.push(encodeURIComponent(key) + "="                            + encodeURIComponent(data[key]));                 }             }             return result.join("&");        });}]); app.controller("demo", function($scope,$http) {            $scope.submit = function () {                var url = "demo";                var data = {"name": "你好"};                $http.post(url, data).success(function (data){                console.log(data);            });            };    });</script></body></html>

Java代码：

package com.whpu;import java.io.IOException;import java.io.PrintWriter;import javax.servlet.ServletException;import javax.servlet.http.HttpServlet;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;public class Demo extends HttpServlet{    private static final long serialVersionUID = 1L;    @Override    protected void doGet(HttpServletRequest req, HttpServletResponse resp)            throws ServletException, IOException {        doPost(req, resp);    }    @Override    protected void doPost(HttpServletRequest req, HttpServletResponse resp)            throws ServletException, IOException {        String name = req.getParameter("name");        System.out.println(name);        PrintWriter pw = resp.getWriter();        String jsonStr="{\"value\":\"hello\"}";        pw.println(jsonStr);    }}

Web.xml代码:

<?xml version="1.0" encoding="UTF-8"?><web-app id="WebApp_ID" version="2.4" xmlns="http://java.sun.com/xml/ns/j2ee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/j2ee http://java.sun.com/xml/ns/j2ee/web-app_2_4.xsd">    <display-name>AngularJs</display-name>    <servlet>        <servlet-name>demo</servlet-name>        <servlet-class>com.whpu.Demo</servlet-class>    </servlet></web-app>    <servlet-mapping>        <servlet-name>demo</servlet-name>        <url-pattern>/demo</url-pattern>    </servlet-mapping>

$http.post(url, data)data是一个json对象的参数，在后台通过request即可获取，然后你就可以进行相关处理，而后台传数据到前台

PrintWriterpw = resp.getWriter();

StringjsonStr="{\"value\":\"hello\"}";

pw.println(jsonStr);

通过这三行代码将{"value":"hello"}这个字符串形式的json放入流中在前台通过回调的

$http.post(url,data).success(function (data){

console.log(data);

});

即可访问data及上面{"value":"hello"}这个字符串，并且angularjs会自动解析成json对象

至于上面的$httpProvider配置则是设置了一些头部属性，以便正确获取参数并防止乱码。启动服务器后，后台java获取的值是你好而前台获得的是一个json对象至此前后台交互已经完成.

备注：1、servlet和Angularjs相关的包注意引入

2、小白写博客，不喜勿喷

3、有什么更好的建议希望提出来

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