通过HttpUrlConnection发布在服务器上注册用户数据的请求

Question

我目前正在从事一个项目，在该项目中，我需要发送用户的用户名，密码和电子邮件以在服务器上注册该用户。 我为此使用了POST命令，并且由于必须发送3个值来注册用户，因此我使用了ContentValues。 这是我的代码：

@Override
    protected String doInBackground(String... params) {
        try {
            url = new URL(params[0]);
            httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setDoInput(true);
            httpURLConnection.setDoOutput(true);
            httpURLConnection.setRequestMethod("POST");

            ContentValues values = new ContentValues();
            values.put("email", "abc@xyz.com");
            values.put("password", "123");
            values.put("name","ABC");

            outputStream = httpURLConnection.getOutputStream();

            bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream));
            bufferedWriter.write(getQuery(values));
            bufferedWriter.flush();

            statusCode = httpURLConnection.getResponseCode();

            Log.i("Result",String.valueOf(statusCode));

            inputStream = new BufferedInputStream(httpURLConnection.getInputStream());

            InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
            int data = inputStreamReader.read();
            while (data != -1) {
                char current = (char) data;
                result += current;
                data = inputStreamReader.read();
            }

            JSONObject jsonObject = new JSONObject(result);

            Log.i("Result",String.valueOf(jsonObject));

            if (statusCode == 200) {
                inputStream = new BufferedInputStream(httpURLConnection.getInputStream());

                Log.i("Result","Correct Data Returned");

                JSONObject jsonObject = new JSONObject(response);

                return true;

            } else {
                Log.i("Result","Data not returned");
                return false;
            }

        } catch (Exception e) {

            e.printStackTrace();
        } finally {
            try {
                if (inputStream != null) {
                    inputStream.close();
                }
                if (outputStream != null) {
                    outputStream.close();
                }
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
        return null;
    }

这是我的getQuery方法：

private String getQuery(ContentValues values) throws UnsupportedEncodingException
    {
        StringBuilder result = new StringBuilder();
        boolean first = true;

        for (Map.Entry<String, Object> entry : values.valueSet())
        {
            if (first)
                first = false;
            else
                result.append("&");

            result.append(URLEncoder.encode(entry.getKey(), "UTF-8"));
            result.append("=");
            result.append(URLEncoder.encode(String.valueOf(entry.getValue()), "UTF-8"));
        }

        Log.i("Result",result.toString());

        return result.toString();
    }

但我得到以下回应：

Result: name=ABC&email=abc%40xyz.com&password=123
Result: 422
System.err: java.io.FileNotFoundException: http://docpanel.myworkdetails.com/api/auth/register

其中422是responseCode返回的状态码，表示：

"errors": {
     "email": [
       "The email field is required."
     ],
     "password": [
       "The password field is required."
     ],
     "name": [
       "The password field is required."
     ]
   },
   "status_code": 422

我没有得到如何通过POST方法传递争论，以使我的注册页面正常工作。 而且我有正确的网址。

这是服务器端故障还是在实施POST时出错？

请帮忙！ 提前致谢。

Answer 1

为什么不将url与参数连接起来呢？

该请求将如下所示：“ http://example.com/test?param1=a&param2=b&param3=c ”

Answer 2

您可以使用HttpPost代替HttpURLConnection

        final HttpParams httpParams = new BasicHttpParams();
        HttpConnectionParams.setConnectionTimeout(httpParams, Constants.CONNECTION_TIMEOUT);
        HttpClient httpClient = new DefaultHttpClient(httpParams);
        HttpPost httpPost = new HttpPost("yourBackendUrl");
        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
        nameValuePairs.add(new BasicNameValuePair("key","value"));
        httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
        HttpResponse response = httpClient.execute(httpPost);
        InputStream inStream = response.getEntity().getContent();
        builder = new StringBuilder();
        BufferedReader reader = new BufferedReader(new InputStreamReader(inStream));
        String line;
        while ((line = reader.readLine()) != null) {
            builder.append(line);
        }
        JSONObject json = new JSONObject(builder.toString());

Answer 3

解决了

我犯了一个错误，即我没有在正确的位置添加httpUrlConnection.connect（） ，也没有在正确的位置添加httpURLConnection.setRequestProperty（ “ Accept”，“ / ”）使其正常工作。 因此，这是使用HttpUrlConnection的POST请求用来使signUp页面成为可能的正确代码：

protected String doInBackground(String... params) {
        try {
            url = new URL(params[0]);
            httpURLConnection = (HttpURLConnection) url.openConnection();
            httpURLConnection.setDoInput(true);
            httpURLConnection.setDoOutput(true);
            httpURLConnection.setRequestMethod("POST");
            httpURLConnection.setRequestProperty("Accept","*/*");

            ContentValues values = new ContentValues();
            values.put("email", "abc@tjk.com");
            values.put("password", "hjh");
            values.put("name","hui");

            httpURLConnection.connect();

            outputStream = httpURLConnection.getOutputStream();

            bufferedWriter = new BufferedWriter(new OutputStreamWriter(outputStream));
            bufferedWriter.write(getQuery(values));
            bufferedWriter.flush();

            statusCode = httpURLConnection.getResponseCode();

            inputStream = httpURLConnection.getInputStream();

            if (statusCode == 200) {
                InputStreamReader inputStreamReader = new InputStreamReader(inputStream);
                int data = inputStreamReader.read();
                while (data != -1) {
                    char current = (char) data;
                    result += current;
                    data = inputStreamReader.read();
                }

                JSONObject jsonObject = new JSONObject(result);

                Log.i("Result",String.valueOf(jsonObject));

            } else {
                Log.i("Result","false");
                return false;
            }

        } catch (Exception e) {

            e.printStackTrace();
        } finally {
            try {
                if (inputStream != null) {
                    inputStream.close();
                }
                if (outputStream != null) {
                    outputStream.close();
                }
            } catch (Exception e) {
                e.printStackTrace();
            }
        }
        return null;
    }