Android JSONArray到ArrayList

Question

我试图在我的Android应用程序中解析JSONArray和ArrayList。 PHP脚本正确地重新调整了预期的结果，但是Java在resultsList.add(map)

处失败并出现空指针异常

public void agencySearch(String tsearch)    {
        // Setting the URL for the Search by Town
        String url_search_agency = "http://www.infinitycodeservices.com/get_agency_by_city.php";
        // Building parameters for the search
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("City", tsearch));

        // Getting JSON string from URL
        JSONArray json = jParser.getJSONFromUrl(url_search_agency, params);

        for (int i = 0; i < json.length(); i++) {
            HashMap<String, String> map = new HashMap<String, String>();

            try {
                JSONObject c = (JSONObject) json.get(i);
                //Fill map
                Iterator iter = c.keys();
                while(iter.hasNext())   {
                    String currentKey = (String) iter.next();
                    map.put(currentKey, c.getString(currentKey));
                }
                resultsList.add(map);

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

            }

        };

        MainActivity.setResultsList(resultsList);

    }

Answer 1

尝试这样可以帮到你，

public void agencySearch(String tsearch)    {
        // Setting the URL for the Search by Town
        String url_search_agency = "http://www.infinitycodeservices.com/get_agency_by_city.php";
        // Building parameters for the search
        List<NameValuePair> params = new ArrayList<NameValuePair>();
        params.add(new BasicNameValuePair("City", tsearch));

        // Getting JSON string from URL
        JSONArray json = jParser.getJSONFromUrl(url_search_agency, params);

       ArrayList<HashMap<String, String>> resultsList = new  ArrayList<HashMap<String, String>>();

        for (int i = 0; i < json.length(); i++) {
            HashMap<String, String> map = new HashMap<String, String>();

            try {
                JSONObject c = json.getJSONObject(position);
                //Fill map
               Iterator<String> iter = c.keys();
                while(iter.hasNext())   {
                    String currentKey = it.next();
                    map.put(currentKey, c.getString(currentKey));
                }
                resultsList.add(map);

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

            }

        };

        MainActivity.setResultsList(resultsList);

    }

Answer 2

使用自定义方法将JSONArray转换为List而不是迭代并构建List。

如何致电：

try {
     ArrayList<HashMap<String,String>> list = (ArrayList<HashMap<String,String>>) toList(json);
} catch (JSONException e) {
     e.printStackTrace();
}

将json数组转换为List：

private List toList(JSONArray array) throws JSONException {
    List list = new ArrayList();
    int size = array.length();
    for (int i = 0; i < size; i++) {
        list.add(fromJson(array.get(i)));
    }
    return list;
}

将json转换为Object：

private Object fromJson(Object json) throws JSONException {
    if (json == JSONObject.NULL) {
        return null;
    } else if (json instanceof JSONObject) {
        return jsonToMap((JSONObject) json);
    } else if (json instanceof JSONArray) {
        return toList((JSONArray) json);
    } else {
        return json;
    }
}

将json转换为地图：

public Map<String, String> jsonToMap(JSONObject object) throws JSONException {
    Map<String, String> map = new HashMap();
    Iterator keys = object.keys();
    while (keys.hasNext()) {
        String key = (String) keys.next();
        map.put(key, fromJson(object.get(key)).toString());
    }
    return map;
}