Question

我正在后端构建一个带有关联REST API的webapp。

使用django-rest-framework创建的API工作正常，我可以从命令行执行以下操作：

$ http http://localhost:8000/API/api-token-auth/ username=ortho1 password=123456

HTTP/1.0 200 OK
Allow: POST, OPTIONS
Content-Type: application/json
Date: Wed, 11 May 2016 12:43:05 GMT
Server: WSGIServer/0.2 CPython/3.5.1+
Vary: Cookie
X-Frame-Options: SAMEORIGIN

{
    "token": "4d6e32726e321448fc212242fc03a3fa84c80510"
}

但是从我的Android应用程序中我尝试执行以下操作：

protected String doInBackground(String... urls) {
    // Starting the request
    URL url;
    HttpURLConnection urlConnection = null;
    try {

        url = new URL("http://10.0.2.2:8000/API/api-token-auth/");
        urlConnection = (HttpURLConnection) url.openConnection();

             /* optional request header */
        urlConnection.setRequestProperty("Content-Type", "application/json");
        urlConnection.setRequestProperty("Accept", "application/json");

            /* for GET request */
        urlConnection.setRequestMethod("POST");

        String urlParameters =
                "username=" + URLEncoder.encode("ortho1", "UTF-8") +
                        "&password=" + URLEncoder.encode("123456", "UTF-8");

        urlConnection.setRequestProperty("Content-Length", "" +
                Integer.toString(urlParameters.getBytes().length));

        urlConnection.setDoOutput(true);
        urlConnection.setDoInput(true);

        DataOutputStream wr = new DataOutputStream (
                urlConnection.getOutputStream ());
        wr.writeBytes (urlParameters);
        wr.flush ();
        wr.close ();

        InputStream inputStream;
        if (urlConnection.getResponseCode() == 200) { // 2xx code means success
            inputStream = urlConnection.getInputStream();
        } else {

            inputStream = urlConnection.getErrorStream();

        }

        inputStream = new BufferedInputStream(inputStream);


        String response = convertInputStreamToString(inputStream);

        Log.i("Request", response);

        return response;

    } catch (Exception e) {

        e.printStackTrace();
        return null;

    } finally {
        if(urlConnection != null) {
            urlConnection.disconnect();
        }
    }

}

private String convertInputStreamToString(InputStream inputStream) throws IOException {

    BufferedReader bufferedReader = new BufferedReader( new InputStreamReader(inputStream));

    String line = "";
    String result = "";

    while((line = bufferedReader.readLine()) != null){
        result += line;
    }

        /* Close Stream */
    if(null!=inputStream){
        inputStream.close();
    }

    return result;
}

我作为来自服务器的响应进入控制台：

{"detail":"JSON parse error - Expecting value: line 1 column 1 (char 0)"}

结果。

我可能做错了什么？

Answer 1

尝试删除此行，

urlConnection.setRequestProperty（“Content-Type”，“application / json”）;

对于您的信息，处理任何 REST API 的最佳库是 RETROFIT http://square.github.io/retrofit/

Answer 2

（狂野猜测）

我看到您的服务器在localhost中运行，并且您正在尝试使用无法正常工作的IP来访问它。如果您在本地测试，请尝试在广播IP中运行服务器： $ python manage.py runserver 0.0.0.0:8000 或者如果您需要公共地址，您可以使用ngrok之类的服务将您的请求路由到本地服务器

Answer 3

urlConnection.setRequestProperty("Content-Type", "application/json");.

你在控制台命令中的哪个位置说明必须发送json？ django期待json吗？ Django发送json是的。更改为表单url编码而不是

对Android上的REST API的POST请求没有获取参数

3 个答案: