【JavaWeb】jQuery对Ajax的支持

jQuery对Ajax的支持

jQuery对Ajax进行封装，提供了$.ajax()方法
语法：$.ajax(options)

常用设置项	说明
url	发送请求地址
type	请求类型get\|post
data	向服务器传递的参数
dataType	服务器响应的数据类型 text\|json\|xml\|html\|jsonp\|script
success	接收响应时的处理函数
error	请求失败时的处理函数

实例代码

MusicServlet.java

package demo;

import java.io.IOException;

import java.util.ArrayList;

import java.util.List;

import javax.servlet.ServletException;

import javax.servlet.annotation.WebServlet;

import javax.servlet.http.HttpServlet;

import javax.servlet.http.HttpServletRequest;

import javax.servlet.http.HttpServletResponse;

import com.alibaba.fastjson.JSON;

/**

 * Servlet implementation class MusicSetvlet

 */

@WebServlet("/music")

public class MusicServlet extends HttpServlet {

	private static final long serialVersionUID = 1L;

    /**

     * @see HttpServlet#HttpServlet()

     */

    public MusicServlet() {

        super();

        // TODO Auto-generated constructor stub

    }

	/**

	 * @see HttpServlet#doGet(HttpServletRequest request, HttpServletResponse response)

	 */

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

		String songType = request.getParameter("songType");

		System.out.println(songType);

		List<String> song = new ArrayList<>();

		if(songType.equals("流行歌曲")) {

			song.add("稻香");

			song.add("晴天");

			song.add("告白气球");

		}else if(songType.equals("经典歌曲")) {

			song.add("千千阙歌");

			song.add("傻女");

			song.add("七友");

		}else if(songType.equals("摇滚歌曲")) {

			song.add("一块红布");

			song.add("假行僧");

			song.add("星长征路上的摇滚");

		}

		String json = JSON.toJSONString(song);

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

		response.getWriter().println(json);

	}

}

musicList.jsp

<%@ page language="java" contentType="text/html; charset=utf-8"

	pageEncoding="utf-8"%>

<!DOCTYPE html>

<html>

<head>

<meta charset="utf-8">

<title>Insert title here</title>

<style>

div {

	text-align: center;

}

.mystyle {

	width: 30%;

	cursor: pointer;

}

</style>

</head>

<body>

	<div>

		<input class="mystyle" type="button" value="流行歌曲"> <input

			class="mystyle" type="button" value="经典歌曲"> <input

			class="mystyle" type="button" value="摇滚歌曲">

	</div>

	<div id="divContent"></div>

	<script type="text/javascript" src="js/jquery-3.4.1.js"></script>

	<script type="text/javascript">

		$(":button").click(function() {

			var songType = this.value;

			$(function() {

				$.ajax({

					"url" : "/ajax/music",

					"type" : "get",

					"data" : {

						"songType" : songType

					},

					"dataType" : "json",

					"success" : function(json) {

						$("#divContent>span").remove();

						$("#divContent>br").remove();

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

							var html = "<span>" + json[i] + "</span><br>";

							$("#divContent").append(html);

						}

					},

					"error" : function(xmlhttp, errorText){

                        console.log(xmlhttp);

                        console.log(errorText);

                        if(xmlhttp.status == "405"){

                            alert("无效的请求方式");

                        }else if(xmlhttp.status == "404"){

                        	alert("未找到URL资源");

                        }else if(xmlhttp.status == "500"){

                            alert("服务器内部错误，请联系管理员");

                        }else{

                            alert("产生异常，请联系管理员");

                        }

                    }

				})

			})

		})

	</script>

</body>

</html>

巴特西

【JavaWeb】jQuery对Ajax的支持

jQuery对Ajax的支持

实例代码

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