请求使用Google Places API for Android Web应用程序中使用的服务器Web拒绝

时间:2015-06-23 18:58:10

标签: java android google-places-api

我正在尝试从Android应用的Google商家信息API获取信息。为此,我首先在Google帐户中启用了此API。

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其次,我已经为浏览器创建了一个API KEY。由于其他API,我已经有了一个API KEY服务器。

enter image description here

所以,在我的代码中,我已经使用这两个密钥进行了测试,并且两者都得到了相同的结果!

  

{

     

“error_message”:“此服务需要API密钥。”,

     

“html_attributions”:[],

     

“结果”:[],

     

“status”:“REQUEST_DENIED”

     

}

我用来拨打电话的代码是......

        @Override
    protected String doInBackground(LocationService... ls) {
        JSONObject result = new JSONObject();
        URL url;
        HttpsURLConnection urlConnection;

        // Making HTTP request
        try {

            //Define connection
            url = new URL("https://maps.googleapis.com/maps/api/place/nearbysearch/json");
            urlConnection = (HttpsURLConnection)url.openConnection();

            urlConnection.setRequestMethod("POST");
            urlConnection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
            urlConnection.setRequestProperty("charset", "utf-8");
            urlConnection.setRequestProperty("Accept", "application/json");
            urlConnection.setDoOutput(true);
            urlConnection.setDoInput(true);
            urlConnection.setUseCaches(false);

            //Send data
            String parameters = "?location=" + String.valueOf(ls[0].getLocation().getLatitude()) + "," + String.valueOf(ls[0].getLocation().getLongitude());
            parameters+="&radius=5000";
            parameters+="&types=restaurant|health|city_hall|gas_station|shopping_mall|grocery_or_supermarket";
            parameters+="&sensor=false";
            parameters+="&key=" + Constants.API_KEY_BROWSER_APPLICATIONS;
            byte[] postData = parameters.getBytes(Charset.forName("UTF-8"));
            int postDataLength = postData.length;
            urlConnection.setRequestProperty("Content-Length", Integer.toString(postDataLength));
            DataOutputStream data = new DataOutputStream(urlConnection.getOutputStream());
            data.write(postData);
            data.flush();
            data.close();

            Log.d(TAG, "Datos enviados");
            Log.d(TAG, "ResponseCode: " + String.valueOf(urlConnection.getResponseCode()));

            //Display what returns POST request

            StringBuilder sb = new StringBuilder();

            int HttpResult = urlConnection.getResponseCode();

            if(HttpResult == HttpURLConnection.HTTP_OK){
                String json;

                BufferedReader br = new BufferedReader(new InputStreamReader(urlConnection.getInputStream(),"utf-8"));

                String line;

                while ((line = br.readLine()) != null) {
                    sb.append(line + "\n");
                }

                br.close();

                //System.out.println(""+sb.toString());
                Log.d(TAG, "json: " + sb.toString());

                FileService file = new FileService();
                file.writeLog(POIActivity.TAG, getClass().getName(), POIActivity.urlConnection + parameters);
                file.writeLog(POIActivity.TAG, "doInBackground", sb.toString());

                // Parse the String to a JSON Object
                result = new JSONObject(sb.toString());

            }else{
                //System.out.println(urlConnection.getResponseMessage());
                Log.d(TAG, "urlConnection.getResponseMessage(): " + urlConnection.getResponseMessage());
                result = null;
            }

        } catch (UnsupportedEncodingException e) {
            e.printStackTrace();
            Log.d(TAG, "UnsuppoertedEncodingException: " + e.toString());
        } catch (JSONException e) {
            e.printStackTrace();
            Log.d(TAG, "Error JSONException: " + e.toString());
        } catch (IOException e) {
            e.printStackTrace();
            Log.d(TAG, "IOException: " + e.toString());
        }

        // Return JSON Object
        return result.toString();


    }

当我调用API时,我得到了ResponseCode = 200,而我构建的调用最终就是那样......

https://maps.googleapis.com/maps/api/place/nearbysearch/json?location=38.26790166666667,-0.7052183333333333&radius=5000&types=restaurant|health|city_hall|gas_station|shopping_mall|grocery_or_supermarket&sensor=false&key=API_KEY

请记住,像API_KEY一样,我已经同时使用了服务器应用程序的Api Key和浏览器应用程序的Api Key,而且两者都得到了相同的结果。

真诚地,我对这个问题感到绝望,因为我不知道我做错了什么!!!

3 个答案:

答案 0 :(得分:6)

问题是您没有使用Google Places API for Android,而是使用Google Places API Web Service

Here是使用Google Places API for Android的示例,而here是使用Google Places API Web Service的示例。你肯定是在使用后者。

启用Google Places API Web Service,它将起作用:

如果您在登录Google云端控制台帐户时转到此链接: https://console.cloud.google.com/apis/library?filter=category:maps

这是应该启用的API:

enter image description here

答案 1 :(得分:1)

来自文档:

  

注意:您需要Android API密钥,而不是浏览器密钥。您可以为Google Maps Android API v2应用和适用于Android应用的Google Places API使用相同的API密钥。

检查this以获取更多帮助。

答案 2 :(得分:0)

更简单的方法是尝试最新的GCM配置文件实现,并使用他们的开发人员界面轻松创建项目。

Enable Google services for your app