Java Http 请求 JSON 和响应处理

标签 java json http apache-httpclient-4.x

我已经看过其他几个问题,但我仍然没有完全理解这一点。 我想将 JSON 字符串发布到远程地址,然后从 JSON 响应中检索值。 我正在使用适用于 Java 的 Apache 库。

public HttpResponse http(String url, String body) throws IOException {

    try (CloseableHttpClient httpClient = HttpClientBuilder.create().build()) {
        HttpPost request = new HttpPost(url);
        StringEntity params = new StringEntity(body);
        request.addHeader("content-type", "application/json");
        request.setEntity(params);
        //httpClient.execute(request);
        HttpResponse result = httpClient.execute(request);

    } catch (IOException ex) {
    }
    return null;
}

作为正文,我将传递以下内容(示例):

{"example":1,"fr":"lol"}

我也完全不知道如何从响应中检索 JSON 值。

最佳答案

最简单的方法是使用像 google-http-java-client 这样的库但是如果你想自己解析 JSON 响应,你可以通过多种方式来解析,你可以使用 org.json , json-simple , Gson , minimal-json , jackson-mapper-asl (从 1.x 开始)...等等

一组简单的例子:

使用 Gson:

import java.io.IOException;

import org.apache.http.HttpResponse;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.CloseableHttpClient;
import org.apache.http.impl.client.HttpClientBuilder;
import org.apache.http.util.EntityUtils;

public class Gson {

    public static void main(String[] args) {
    }

    public HttpResponse http(String url, String body) {

        try (CloseableHttpClient httpClient = HttpClientBuilder.create().build()) {
            HttpPost request = new HttpPost(url);
            StringEntity params = new StringEntity(body);
            request.addHeader("content-type", "application/json");
            request.setEntity(params);
            HttpResponse result = httpClient.execute(request);
            String json = EntityUtils.toString(result.getEntity(), "UTF-8");

            com.google.gson.Gson gson = new com.google.gson.Gson();
            Response respuesta = gson.fromJson(json, Response.class);

            System.out.println(respuesta.getExample());
            System.out.println(respuesta.getFr());

        } catch (IOException ex) {
        }
        return null;
    }

    public class Response{

        private String example;
        private String fr;

        public String getExample() {
            return example;
        }
        public void setExample(String example) {
            this.example = example;
        }
        public String getFr() {
            return fr;
        }
        public void setFr(String fr) {
            this.fr = fr;
        }
    }
}

使用 json-simple:

import java.io.IOException;

import org.apache.http.HttpResponse;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.entity.StringEntity;
import org.apache.http.impl.client.CloseableHttpClient;
import org.apache.http.impl.client.HttpClientBuilder;
import org.apache.http.util.EntityUtils;
import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;

public class JsonSimple {

    public static void main(String[] args) {

    }

    public HttpResponse http(String url, String body) {

        try (CloseableHttpClient httpClient = HttpClientBuilder.create().build()) {
            HttpPost request = new HttpPost(url);
            StringEntity params = new StringEntity(body);
            request.addHeader("content-type", "application/json");
            request.setEntity(params);
            HttpResponse result = httpClient.execute(request);

            String json = EntityUtils.toString(result.getEntity(), "UTF-8");
            try {
                JSONParser parser = new JSONParser();
                Object resultObject = parser.parse(json);

                if (resultObject instanceof JSONArray) {
                    JSONArray array=(JSONArray)resultObject;
                    for (Object object : array) {
                        JSONObject obj =(JSONObject)object;
                        System.out.println(obj.get("example"));
                        System.out.println(obj.get("fr"));
                    }

                }else if (resultObject instanceof JSONObject) {
                    JSONObject obj =(JSONObject)resultObject;
                    System.out.println(obj.get("example"));
                    System.out.println(obj.get("fr"));
                }

            } catch (Exception e) {
                // TODO: handle exception
            }

        } catch (IOException ex) {
        }
        return null;
    }
}

等...

关于Java Http 请求 JSON 和响应处理,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22816335/

相关文章:

java - XQuery 错误处理

java - 如何将异步 HTTP 请求响应合并为一个?

java - 变量未发送到服务器

.net - WCF 返回未实时找到端点

http - 支持使用 HTML 图片或服务器端通过 http header 的 WebP?

Java-使用 float -构造函数错误

iOS 解析内部 Json

json - DocumentDB,对 JSON 文档使用小字段名称以最小化长度

javascript - 从回调中创建 XMLHttpRequest 如何影响缓存?

http - openssl s_client 通过代理与 clientAuth 建立 TLS 连接