0

所以我想发送一个带参数的 GET 请求。但它似乎只对您发送请求的 url 有约定。与 POST 请求不同，我看不到在其中传递参数的方法。

我现在如何发送 GET 请求，不带参数（可能是错误的）：

String url = "http://api.netatmo.net/api/getuser";

            URL obj = new URL(url);
            HttpURLConnection con = (HttpURLConnection) obj.openConnection();

            // optional default is GET
            con.setRequestMethod("GET");

            //add request header
            con.setRequestProperty("User-Agent", USER_AGENT);

            int responseCode = con.getResponseCode();
            Log.v(TAG, ("\nSending 'GET' request to URL : " + url));
            Log.v(TAG, ("Response Code : " + responseCode));

            BufferedReader in = new BufferedReader(
                    new InputStreamReader(con.getInputStream()));
            String inputLine;
            StringBuffer response = new StringBuffer();

            while ((inputLine = in.readLine()) != null) {
                response.append(inputLine);
            }
            in.close();

            //print result
            Log.v(TAG, (response.toString()));

我如何发送带有参数的 POST 请求：

String url = "https://api.netatmo.net/oauth2/token";
            URL obj = new URL(url);
            HttpsURLConnection con = (HttpsURLConnection) obj.openConnection();

            //add request header
            con.setRequestMethod("POST");
            con.setRequestProperty("User-Agent", USER_AGENT);
            con.setRequestProperty("Accept-Language", "en-US,en;q=0.5");

            String urlParameters = "grant_type=password&client_id=myid&client_secret=mysecret&username=myusername&password=mypass";

            // Send post request
            con.setDoOutput(true);
            DataOutputStream wr = new DataOutputStream(con.getOutputStream());
            wr.writeBytes(urlParameters);
            wr.flush();
            wr.close();

            int responseCode = con.getResponseCode();
            Log.v(TAG, "\nSending 'POST' request to URL : " + url);
            Log.v(TAG, "Post parameters : " + urlParameters);
            Log.v(TAG, "Response Code : " + responseCode);

            BufferedReader in = new BufferedReader(
                    new InputStreamReader(con.getInputStream()));
            String inputLine;
            StringBuffer response = new StringBuffer();

            while ((inputLine = in.readLine()) != null) {
                response.append(inputLine);
            }
            in.close();

            //print result
            Log.v(TAG, response.toString());

            access_token = response.substring(17, 74);
            refresh_token = response.substring(93,150);
            getRequest = "/api/getuser?access_token=" + access_token + " HTTP/1.1";

            Log.v(TAG, access_token);
            Log.v(TAG, refresh_token);
            Log.v(TAG, getRequest);
4

4 回答 4

11

根据 HTTP 规范GET仅支持路径参数或 url 参数，因此您不能像在请求中那样将参数放在 HTTP 请求正文中POST。

正如 Sotirios 在评论中提到的那样，从技术上讲，您仍然可以在 GET 正文中推送参数，但如果 API 遵守规范，它们将不会为您提供执行此操作的方法。

于 2013-08-09T15:54:45.700 回答
2

您是否尝试将查询参数添加到请求 java.net.URL？

String url = "http://api.netatmo.net/api/getuser?access_token=" + access_token;
URL obj = new URL(url);
于 2013-08-09T16:50:45.030 回答
1

我遇到了同样的问题，试试这个：

String bla = "http://api.netatmo.net/api/devicelist?access_token=" + AUTH_TOKEN;
URL url = new URL(bla);

BufferedReader reader = new BufferedReader(new InputStreamReader(url.openStream()));

String line = "";
String message = "";

while ((line = reader.readLine()) != null)
{
    message += line;
}

我有一个例外，语法不正确。当我更改语法（例如使用 UTF 8 编码）时，API 只会返回错误（例如 404 not found ...）。

我终于让它工作了：

try
{

        System.out.println("Access Token: " + AUTH_TOKEN);

        String url = "http://api.netatmo.net/api/devicelist";
        String query = "access_token=" + URLEncoder.encode(AUTH_TOKEN, CHARSET);

        URLConnection connection = new URL(url + "?" + query).openConnection();
        connection.setRequestProperty("Accept-Charset", CHARSET);

        InputStream response = connection.getInputStream();

        BufferedReader reader = new BufferedReader(new InputStreamReader(response));

        String line = "";
        String message = "";

        while ((line = reader.readLine()) != null)
        {
            message += line;
        }

        return message;


    } catch (MalformedURLException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

注意：CHARSET = "UTF-8"

于 2013-09-12T19:26:48.083 回答
0

原来 API 提供的 url 让我很困惑。我修复了网址，现在可以使用了。

于 2013-08-14T12:59:32.963 回答