J2EE 练习题 - JSON HTTP Service

1 要求

2 示例代码

        2.1 Server 端

        2.2 客户端 - Java

1 要求

在 Tomcat 上布署一个 HTTP Service，使用 JSON 格式返回数据

2 示例代码

2.1 Server 端

基于 Maven 开发

1. 新建 Maven webapp 项目

2. 修改 pom.xml 如下

   pom.xml

   <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/maven-v4_0_0.xsd"> ???<modelVersion>4.0.0</modelVersion> ???<groupId>lld</groupId> ???<artifactId>http.json.test.server</artifactId> ???<packaging>war</packaging> ???<version>0.0.1-SNAPSHOT</version> ???<name>http.json.test.server Maven Webapp</name> ???<url>http://maven.apache.org</url> ???<properties> ???????<gson.version>2.8.2</gson.version> ???????<tomcat.version>6.0.53</tomcat.version> ???</properties> ???<dependencies> ???????<dependency> ???????????<groupId>com.google.code.gson</groupId> ???????????<artifactId>gson</artifactId> ???????????<version>${gson.version}</version> ???????</dependency> ???????<dependency> ???????????<groupId>org.apache.tomcat</groupId> ???????????<artifactId>servlet-api</artifactId> ???????????<version>${tomcat.version}</version> ???????????<scope>provided</scope> ???????</dependency> ???????<dependency> ???????????<groupId>junit</groupId> ???????????<artifactId>junit</artifactId> ???????????<version>3.8.1</version> ???????????<scope>test</scope> ???????</dependency> ???</dependencies> ???<build> ???????<finalName>http.json.test.server</finalName> ???</build></project>

3. 创建 POJO 类 User.java

   User.java

   package http.json.test.server.model;public class User { ???private String userId; ???private String userName; ???public String getUserId() { ???????return userId; ???} ???public void setUserId(String userId) { ???????this.userId = userId; ???} ???public String getUserName() { ???????return userName; ???} ???public void setUserName(String userName) { ???????this.userName = userName; ???}}

4. 创建 Servlet 类 UserServlet

   UserServlet.java

   package http.json.test.server.servlet;import java.io.IOException;import javax.servlet.ServletException;import javax.servlet.ServletOutputStream;import javax.servlet.http.HttpServlet;import javax.servlet.http.HttpServletRequest;import javax.servlet.http.HttpServletResponse;import com.google.gson.Gson;import http.json.test.server.model.User;public class UserServlet extends HttpServlet { ???private static final long serialVersionUID = -2118394734647389638L; ???@Override ???protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { ???????this.doPost(req, resp); ???} ???????@Override ???protected void doPost(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOException { ???????User user = new User(); ???????String userId = req.getParameter("userId"); ???????user.setUserId(userId); ???????user.setUserName("Lindong"); ???????String jsonString = new Gson().toJson(user); ???????ServletOutputStream outputStream = resp.getOutputStream(); ???????outputStream.print(jsonString); ???}}

5. 修改 web.xml 如下所示

   web.xml

   <!DOCTYPE web-app PUBLIC "-//Sun Microsystems, Inc.//DTD Web Application 2.3//EN" "http://java.sun.com/dtd/web-app_2_3.dtd" ><web-app> ???<display-name>HTTP JSON Service Demo</display-name> ???<servlet> ???????<servlet-name>getUser</servlet-name> ???????<servlet-class>http.json.test.server.servlet.UserServlet</servlet-class> ???</servlet> ???<servlet-mapping> ???????<servlet-name>getUser</servlet-name> ???????<url-pattern>/getUser.do</url-pattern> ???</servlet-mapping></web-app>

6. 编译并打包

   mvn clean package

7. 将生成的 war 包复制到 Tomcat webapps 目录并启动 Tomcat，打开浏览器输入 http://localhost:8080/http.json.test.server/getUser.do?userId=10001 后在浏览器中显示结果：

   {"userId":"10001","userName":"Lindong"}

2.2 客户端 - Java

使用 Java 程序获取 Service 响应

1. 使用 Maven 创建 quickstart 程序

2. 修改 pom.xml 如下

   pom.xml

   <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 http://maven.apache.org/xsd/maven-4.0.0.xsd"> ???<modelVersion>4.0.0</modelVersion> ???<groupId>lld</groupId> ???<artifactId>http.json.test.client</artifactId> ???<version>0.0.1-SNAPSHOT</version> ???<packaging>jar</packaging> ???<name>http.json.test.client</name> ???<url>http://maven.apache.org</url> ???<properties> ???????<project.build.sourceEncoding>UTF-8</project.build.sourceEncoding> ???</properties> ???<dependencies> ???????<dependency> ???????????<groupId>org.apache.httpcomponents</groupId> ???????????<artifactId>httpclient</artifactId> ???????????<version>4.5.2</version> ???????</dependency> ???</dependencies></project>

3. 创建工具类 HttpRequestUtil.java

   HttpRequestUtil.java

   package lld.http.json.test.client.util;import java.io.IOException;import java.util.ArrayList;import java.util.Iterator;import java.util.List;import java.util.Map;import java.util.Map.Entry;import org.apache.http.HttpEntity;import org.apache.http.HttpResponse;import org.apache.http.NameValuePair;import org.apache.http.client.ClientProtocolException;import org.apache.http.client.entity.UrlEncodedFormEntity;import org.apache.http.client.methods.HttpPost;import org.apache.http.impl.client.CloseableHttpClient;import org.apache.http.impl.client.HttpClientBuilder;import org.apache.http.message.BasicNameValuePair;import org.apache.http.util.EntityUtils;public class HttpRequestUtil { ???public static String callHttpService(String url, Map<String, String> parameters) throws ClientProtocolException, IOException { ???????return callHttpService(url, parameters, "utf-8"); ???} ???????public static String callHttpService(String url, Map<String, String> parameters, String charset) throws ClientProtocolException, IOException { ???????String result = null; ???????HttpClientBuilder httpClientBuilder = HttpClientBuilder.create(); ???????CloseableHttpClient httpClient = httpClientBuilder.build(); ???????HttpPost httpPost = new HttpPost(url); ???????List<NameValuePair> list = new ArrayList<NameValuePair>(); ???????if (parameters != null) { ???????????Iterator<Map.Entry<String, String>> iterator = parameters.entrySet().iterator(); ???????????while (iterator.hasNext()) { ???????????????Entry<String, String> entry = (Entry<String, String>) iterator.next(); ???????????????list.add(new BasicNameValuePair(entry.getKey(), entry.getValue())); ???????????} ???????} ???????if (list.size() > 0) { ???????????UrlEncodedFormEntity entity = new UrlEncodedFormEntity(list, charset); ???????????httpPost.setEntity(entity); ???????} ???????????????HttpResponse response = httpClient.execute(httpPost); ???????if (response != null) { ???????????HttpEntity responseEntity = response.getEntity(); ???????????????????????if (responseEntity != null) { ???????????????result = EntityUtils.toString(responseEntity); ???????????} ???????} ???????????????return result; ???}}

4. 调用 Http Service 例程

   App.java

   package lld.http.json.test.client;import java.io.IOException;import java.util.HashMap;import java.util.Map;import org.apache.http.client.ClientProtocolException;import lld.http.json.test.client.util.HttpRequestUtil;/** * Hello world! * */public class App { ???public static void main( String[] args ) throws ClientProtocolException, IOException ???{ ???????String url = "http://localhost:8080/http.json.test.server/getUser.do"; ???????Map<string, string=""> parameters = new HashMap<string, string="">(); ???????parameters.put("userId", "lld"); ???????String result = HttpRequestUtil.callHttpService(url, parameters); ???????System.out.println("Result is: "); ???????System.out.println(result); ???????????}}