Question

我们正在尝试使用Google Places API查找附近的地方，并在仅作为Java项目运行时使其正常工作。但是，在Android应用程序内的按钮的onClick（）函数下运行相同的代码会导致从Google返回的JSON字符串为空。硬编码的GPS坐标用于两者。

为什么？

这是尝试获取JSON数据时显示的错误

05-14 00:54:12.828: E/PlacesService(8677): org.json.JSONException: End of input at character 0 of 
05-14 00:54:12.828: E/PlacesService(8677):  at org.json.JSONTokener.syntaxError(JSONTokener.java:450)
05-14 00:54:12.828: E/PlacesService(8677):  at org.json.JSONTokener.nextValue(JSONTokener.java:97)

这是进行API调用的地方：

private String makeUrl(double latitude, double longitude, String place) {
    StringBuilder urlString = new StringBuilder(
            "https://maps.googleapis.com/maps/api/place/search/json?");

    if (place.equals("")) {
        urlString.append("&location=");
           urlString.append(Double.toString(latitude));
           urlString.append(",");
           urlString.append(Double.toString(longitude));
           urlString.append("&radius=1000");
           // urlString.append("&types="+place);
           urlString.append("&sensor=false&key=" + API_KEY);

    } else {
        urlString.append("?types=" + "bar" + "|" + "night_club");
        urlString.append("&location=");
        urlString.append(Double.toString(latitude));
        urlString.append(",");
        urlString.append(Double.toString(longitude));
        urlString.append("&radius=1000");
        urlString.append("&sensor=false&key=" + API_KEY);
    }
    return urlString.toString();
}

protected String getJSON(String url) {
    return getUrlContents(url);
}

private String getUrlContents(String theUrl) {
    StringBuilder content = new StringBuilder();

    try {
        URL url = new URL(theUrl);
        URLConnection urlConnection = url.openConnection();
        BufferedReader bufferedReader = new BufferedReader(
                new InputStreamReader(urlConnection.getInputStream()), 8);
        String line;
        while ((line = bufferedReader.readLine()) != null) {
            content.append(line + "\n");
        }
        bufferedReader.close();
    }catch (Exception e) {
        e.printStackTrace();
    }
    return content.toString();
}

以下是我们如何调用上述方法来获取JSON字符串：

    String urlString = makeUrl(32.8400, -117.2769 placeSpacification);
    String json = getJSON(urlString);

这是我们的AndroidManifest.xml。我们已经包含了必要的位置和Internet权限。

<uses-sdk
    android:minSdkVersion="8"
    android:targetSdkVersion="19" />
<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION" />  
<uses-permission android:name="android.permission.INTERNET" />

<application
    android:allowBackup="true"
    android:icon="@drawable/ic_launcher"
    android:label="@string/app_name"
    android:theme="@style/AppTheme" >

    <activity
        android:name="com.example.testapp.ListViewAndroidExample"
        android:label="@string/app_name" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
</application>

谢谢！

Answer 1

嘿，将Aquery jar添加到您的项目中并尝试使用代码。

latitude =18 ;
longitude = 50 ;
 StringBuilder builder = new StringBuilder("https://maps.googleapis.com/maps/api/place/nearbysearch/json?");
    builder.append("location="+latitude+","+longitude);
    builder.append("&radius=100");
    builder.append("&sensor=true");
    builder.append("&types="+type);
    builder.append("&key=XXXXXXXXXXXXXXXXXXXXX");

    String url = builder.toString();

 aQuery.progress(progressDialog).ajax(url, JSONObject.class, new AjaxCallback<JSONObject>(){

        @Override
        public void callback(String url, JSONObject object,
                AjaxStatus status) {
            // TODO Auto-generated method stub
            super.callback(url, object, status);

            //System.out.println("objectttt ====  "+object);

            JSONArray jsonArray;

            try {

                jsonArray = object.getJSONArray("results");
                for(int j=0; j<jsonArray.length(); j++)
                {
                    JSONObject jsonObject = jsonArray.getJSONObject(j);
                    String placeName = jsonObject.getString("name");
                    String placeAddress= jsonObject.getString("vicinity");
                    String image=jsonObject.getString("icon");

                    JSONObject jsonObject2 = jsonObject.getJSONObject("geometry");
                    JSONObject jsonObject3 =jsonObject2.getJSONObject("location");

                    for(int i=0 ;i<jsonObject3.length();i++)
                    {
                        PlaceLatitude =jsonObject3.getDouble("lat");
                        PlaceLongitude =jsonObject3.getDouble("lng");

                    }

                    try 
                    {
                      imgURL= new URL(image);

                    } catch (MalformedURLException e) 
                    {
                        e.printStackTrace();
                    }

                    arrPlaces.add(placeName);
                    arrPlaceImage.add(imgURL);
                    arrPlaceAddress.add(placeAddress);
                    arrPlaceLatitude.add(PlaceLatitude);
                    arrPlaceLogitude.add(PlaceLongitude);


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

        }
    });

API调用在java项目中工作，但在Android应用程序中调用时则不行

1 个答案: