Question

我对帖子请求有疑问。服务器给我回复消息确定，但没有识别json我发送的帖子是什么。我几乎尝试了一切。 Php属性$ _POST始终为空。谢谢你的回答。

Json有结构：

sizeof

Android代码：

{
  "data": {
       "email": "something@something.com",
       "password": "tralala"
   }
}

服务器响应：

public static Pair<Integer, String> signUpByEmailPost(String username, String passwd) {
    try {
        URL url = new URL(SERVER_URL + "create/createUser.php");
        HttpURLConnection urlConn = (HttpURLConnection) url.openConnection();
        urlConn.setRequestMethod("POST");
        urlConn.setRequestProperty("Content-Type", "application/json");
        urlConn.setDoOutput(true);
        urlConn.setDoInput(true);
        urlConn.setUseCaches(false);
        urlConn.connect();

        JSONObject json = createJsonCredentials(username, passwd);
        String dataString = json.toString();
        Log.i(TAG, dataString);

        OutputStreamWriter out = new OutputStreamWriter(urlConn.getOutputStream());
        out.write(dataString);
        out.flush();

        int httpResult = resolveHttpResponseCode(urlConn);
        if(httpResult > 0) {
            return new Pair<>(httpResult, null);
        }

        String receivedDataString = getStringContentFromConnection(urlConn);
        return new Pair<>(0, receivedDataString);

    } catch (Exception e) {
        e.printStackTrace();
    }
    return new Pair<>(3, null);
}

private static JSONObject createJsonCredentials(String username, String password) {
    try {
        JSONObject jsonInner = new JSONObject();
        jsonInner.put("email", "something@something.com");
        jsonInner.put("password", "tralala");

        JSONObject jsonOuter = new JSONObject();
        jsonOuter.put("data", jsonInner);

        return jsonOuter;
    } catch(JSONException ex) {
        ex.printStackTrace();
    }
    return null;
}

服务器端：

{
    success: false,
    errors: {
        email: "blank_email",
        password: "blank_password"
    }
}

Answer 1

您好在这里从方法中返回null值。试试这个代码。它会起作用。

private static JSONObject createJsonCredentials(String username, String password) {

        JSONObject jsonOuter = new JSONObject();

        try {
            JSONObject jsonInner = new JSONObject();
            jsonInner.put("email", "something@something.com");
            jsonInner.put("password", "tralala");


            jsonOuter.put("data", jsonInner);


        } catch(JSONException ex) {
            ex.printStackTrace();
        }
        return jsonOuter;
    }

HttpUrlConnection发布请求问题

1 个答案: