Apache HttpClient之fluent API的使用

Posted 2020-10-15 v小周与海神

该方法为Apache HttpClient 4.5以上的版本支持，在官网有明确的说明。
对比以前的方式，其优点是代码更简洁，同时为线程安全的。仅举一个最简单的post栗子

JAR包信息：

        <dependency>
            <groupId>org.apache.httpcomponents</groupId>
            <artifactId>httpclient</artifactId>
            <version>4.5.1</version>
        </dependency>

package utils;

import com.alibaba.fastjson.JSONObject;
import org.apache.http.client.fluent.Request;
import org.apache.http.entity.ContentType;

import java.io.IOException;

/**
 * Created with IntelliJ IDEA.
 * Description:
 *         Http请求工具类
 *
 * @author
 *         Date: 2017-11-01
 *         Time: 10:24
 */
public class HttpClientUtil {

    /**
     * @return String
     * 响应体
     */
    public String jsonPostHttp(JSONObject jsonObject, String url) throws IOException {

        String response = Request.Post(url).bodyString(jsonObject.toString(), ContentType.APPLICATION_JSON).execute().returnContent().asString();

        System.out.println(response);   // 将返回的内容打印出来便于调试
        return response;
    }
    
}

测试代码：

package utilsTest;

import com.alibaba.fastjson.JSONObject;
import org.junit.Test;
import utils.HttpClientUtil;

import java.io.File;
import java.io.IOException;

public class HttpClientUtilTest {

    @Test
    public void httpClientUtilTest(){

        JSONObject jsonObject = new JSONObject();
        // 根据业务需要，定义JSONObject所需的参数
        jsonObject.put("appId", "1");
        jsonObject.put("userName", "admin");
        jsonObject.put("password", "111111");
        
        HttpClientUtil httpClientUtil = new HttpClientUtil();

        try {
            httpClientUtil.jsonPostHttp(jsonObject,"http://localhost/manage/auth/login.do");
        } catch (IOException e) {
            e.printStackTrace();
        }
    }