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

新建 Maven webapp 项目

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>

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;
    }
}

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);
    }
}

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>

编译并打包
```
mvn clean package
```
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

使用 Maven 创建 quickstart 程序

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>

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;
    }
}

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);        
    }
}

posted @ 2018-02-27 22:42 真栋哥阅读(303) 评论(0) 编辑收藏举报

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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

2 示例代码

2.1 Server 端

2.2 客户端 - Java

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