java - 无法解析 google place JSON 响应 - org.json.JSONException

Question

我收到了来自 Google Places API 的 JSON 响应，我正在尝试解析该响应。但我收到了 org.json.JSONException。

这是我的 JSON 响应。

https://maps.googleapis.com/maps/api/place/details/json?placeid=ChIJbf6hrtZ2AjoRwUPd_nrhVjM&key=AIzaSyBWKQHS39-SYUNxEEAry1FxrMET2NwhqxE

我正在使用以下代码来检索格式化的地址。

  try {

        Log.e("test-ttt", jsonResults.toString());

        // Create a JSON object hierarchy from the results
        JSONObject jsonObj = new JSONObject(jsonResults.toString());

        JSONObject placeDetailsJsonArray = jsonObj.getJSONObject("result");

        // Extract the Place descriptions from the results
        placeDetails = "NAME: " + placeDetailsJsonArray.getJSONObject("name").toString();
        placeDetails += "ADDRESS: " + placeDetailsJsonArray.getJSONObject("formatted_address").toString();


    } catch (JSONException e) {
        Log.e(TAG, "Cannot process JSON results", e);
    }

这是我收到的 logcat 异常:

org.json.JSONException: Value South Point School at name of type java.lang.String cannot be converted to JSONObject
        at org.json.JSON.typeMismatch(JSON.java:100)
        at org.json.JSONObject.getJSONObject(JSONObject.java:578)
        at manasthemarvel.triptimeline.PlaceAPI.getPlaceDetails(PlaceAPI.java:78)
        at manasthemarvel.triptimeline.placeInfo.doInBackground(placeInfo.java:14)
        at manasthemarvel.triptimeline.placeInfo.doInBackground(placeInfo.java:10)
        at android.os.AsyncTask$2.call(AsyncTask.java:288)
        at java.util.concurrent.FutureTask.run(FutureTask.java:237)
        at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:231)
        at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1112)
        at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:587)
        at java.lang.Thread.run(Thread.java:841)


        // Extract the Place descriptions from the results
        placeDetails = "NAME: " + placeDetailsJsonArray.getJSONObject("name").toString();
        placeDetails += "ADDRESS: " + placeDetailsJsonArray.getJSONObject("formatted_address").toString();

我在这里做错了什么？

Answer 1

您将“名称”和“formatted_address”视为 JSONObject，而不是普通的键/值对。

试试这个:

JSONObject placeDetailsJsonArray = jsonObj.getJSONObject("result");
String name = placeDetailsJsonArray.getString("name");