我有JSON解析器的以下代碼。其實我沒有自己寫過,我從各種stackoverflow問題收集它。但不知何故,它不起作用。我給使用互聯網的權限:異步JSON解析器不起作用
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
String url=null;
// constructor
public JSONParser() {
}
// function get JSON from URL
public JSONObject makeHttpRequest(String url) {
BackGroundTask Task= new BackGroundTask(url);
try {
return Task.execute().get();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
return null;
} catch (ExecutionException e) {
// TODO Auto-generated catch block
e.printStackTrace();
return null;
}
}
public class BackGroundTask extends AsyncTask<String, String, JSONObject>{
String URL=null;
public BackGroundTask(String url) {
URL = url;
}
@Override
protected JSONObject doInBackground(String... params) {
// TODO Auto-generated method stub
// Making HTTP request
try {
// Making HTTP request
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(
is, "iso-8859-1"), 8);
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "\n");
}
is.close();
json = sb.toString();
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
}
編輯:
你可以看到,可能會導致此類失敗的任何問題的一部分?我調試我的代碼,並發現該行
JSONParser jParser = new JSONParser();
JSONObject json = jParser.makeHttpRequest(URL);
的json
對象是null
後。
又是什麼「但不知何故,它不工作」是什麼意思?錯誤? – codeMagic 2013-05-07 13:45:41
我不小心按下了提問問題鏈接,您可以在編輯中找到更多信息。 – 2013-05-07 13:46:57
你不應該從'AsyncTask'使用'get()'...這幾乎是一樣的,就像你根本不會使用'AsyncTask'一樣... – Selvin 2013-05-07 13:48:49