java - 如何使用java复制curl命令？

Question

准确地说，我想执行下面的 curl 操作，它用 java 返回 json:

curl -H 'Client-ID: ahh_got_ya' -X GET 'https://api.twitch.tv/helix/streams'

这在 Linux shell 中工作得很好。

下面是我的脚本尝试使用 java json 执行上面的curl操作:

{String urly = "https://api.twitch.tv/helix/streams";
    URL obj = new URL(urly);
    HttpURLConnection con = (HttpURLConnection) obj.openConnection();

    con.setRequestMethod("GET");
    con.setRequestProperty("Content-Type","application/json");
    con.setRequestProperty("Client-ID","Ahh_got_ya");


    con.setDoOutput(true);
    DataOutputStream wr = new DataOutputStream(con.getOutputStream());
    wr.writeBytes("");
    wr.flush();
    wr.close();

    int responseCode = con.getResponseCode();
    System.out.println("Response Code : " + responseCode);

    BufferedReader iny = new BufferedReader(
    new InputStreamReader(con.getInputStream()));
      String output;
      StringBuffer jsonres = new StringBuffer();

      while ((output = iny.readLine()) != null) {
          jsonres.append(output);
      }
    iny.close();

    //printing result from response
    System.out.println(response.toString());
}

我得到:java.io.FileNotFoundException:https://api.twitch.tv/helix/streams 响应代码:404

非常感谢所有回复。

Answer 1

快到了!您正在进行 GET 调用，不需要使连接可写——因为您不打算发布。您需要删除该部分。另外 - 要准确了解您的curl 调用正在执行的操作，请删除Content-Type - 因为它不在curl 调用中使用。所以你的代码调整后应该是:

{
    String urly = "https://api.twitch.tv/helix/streams";
    URL obj = new URL(urly);
    HttpURLConnection con = (HttpURLConnection) obj.openConnection();

    //only 2 headers from cURL call
    con.setRequestMethod("GET");
    con.setRequestProperty("Client-ID","Ahh_got_ya");

    int responseCode = con.getResponseCode();
    System.out.println("Response Code : " + responseCode);

    BufferedReader iny = new BufferedReader(
    new InputStreamReader(con.getInputStream()));
      String output;
      StringBuffer jsonres = new StringBuffer();

      while ((output = iny.readLine()) != null) {
          jsonres.append(output);
      }
    iny.close();

    //printing result from response
    System.out.println(response.toString());
}

出现 404 错误的原因是您的请求与服务端点的期望不匹配。发送 POST 请求或其他类型的非预期内容将导致请求不匹配。删除多余的输出内容并尝试一下!