HTTP传输和HTTP请求

Question

我试图使用HTTP请求从API中获取JSON。但是，我仍然无法设置请求。这是我目前的代码

HttpTransport transport = new HttpTransport() {
    @Override
    protected LowLevelHttpRequest buildRequest(String s, String s1) throws IOException {
        return null;
    }
};

我被迫创建了一个，但是我不知道在其中放什么，文档也没有给我任何线索

@Override
public void action(String[] args, MessageReceivedEvent event) {
    HttpRequest request=new HttpRequest(transport,"GET");
    request.setUrl(url);
    request.setConnectTimeout(5000);
    try {
        request.execute();
    }
    catch (IOException e) {
        e.printStackTrace();
    }
}

url已经包含用于授权的Json Web令牌。

但是，在初始化HttpRequest时，它会给出以下错误：

'HttpRequest(com.google.api.client.http.HttpTransport, java.lang.String)' is not public
in 'com.google.api.client.http.HttpRequest'.
Cannot be accessed from outside package

当我清除括号时，甚至会出现此错误：

HttpRequest request=new HttpRequest();

有人知道我做错了什么吗？

更新:我已经修复了我的代码，但仍然有一个错误：

HttpTransport transport = new ApacheHttpTransport();
    HttpRequestFactory requestFactory = transport.createRequestFactory();
    try {
        HttpRequest request = requestFactory.buildGetRequest(url);
        request.execute();
        System.out.println(request.getContent());
    }
    catch (IOException e){
        e.printStackTrace();
    }

我看到的错误是：

avax.net.ssl.SSLHandshakeException: Received fatal alert: handshake_failure

编辑:即使我使用下面的示例，它也会返回此错误

Answer 1

这意味着HttpRequest构造函数受到保护以调用。您必须使用HttpRequestFactory工厂来创建HttpRequest实例。

要想得到HttpRequestFactory，你必须让HttpTransport见https://developers.google.com/api-client-library/java/google-http-java-client/reference/1.20.0/com/google/api/client/http/HttpTransport

HttpTransport transport = new ApacheHttpTransport();
HttpRequestFactory requestFactory = transport.createRequestFactory();
HttpRequest request = requestFactory.buildGetRequest(url);
HttpResponse response = request.execute();
     try {
       // process the HTTP response object
           InputStream is = response.getContent();
           try {
               // Process the input stream..
           } finally {
               is.close();
           }
     } finally {
       response.disconnect();
     }

从未试过这个例子，希望它能奏效。