我在解析JSON时遇到问题。 http://www.json-generator.com/api/json/get/cdWQqIXQfC?indent=2 其实我想得到以下
public class MainActivity extends Activity {
String firstname = null;
String lastname = null;
String username = null;
String username1 = null;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
new AsyncTaskParseJson().execute();
}
public class AsyncTaskParseJson extends AsyncTask<String, String, String> {
final String TAG = "AsyncTaskParseJson.java";
String yourJsonStringUrl = "http://www.json-generator.com/api/json/get/cdWQqIXQfC?indent=2";
JSONArray dataJsonArr = null;
JSONObject dataJsonArry = null;
JSONObject dataJsonArry1 = null;
JSONObject dataJsonArry2 = null;
@Override
protected void onPreExecute() {
}
@Override
protected String doInBackground(String... arg0) {
try {
JsonParser jParser = new JsonParser();
JSONObject json = jParser.getJSONFromUrl(yourJsonStringUrl);
dataJsonArry = json.getJSONObject("online");
dataJsonArry1 = json.getJSONObject("men");
dataJsonArry2 = json.getJSONObject("clothes");
dataJsonArr = json.getJSONArray("jean");
for (int i = 0; i < dataJsonArr.length(); i++) {
JSONObject c = dataJsonArr.getJSONObject(i);
lastname = c.getString("cost");
username = c.getString("size");
Log.e("Hello", "firstname: " + firstname + ", lastname: "
+ lastname + ", username: " + username + "star:"
+ username1);
}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
@Override
protected void onPostExecute(String strFromDoInBg) {
}
}
}
从URL获取。然后需要设置listview。但首先我没有得到json数组本身。 以下代码抛出异常并且对男性没有任何价值。
我尝试使用
public class JsonParser {
final String TAG = "JsonParser.java";
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
public JSONObject getJSONFromUrl(String url) {
try {
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.d(TAG, "Error converting result " + e.toString());
}
// try parse the string to a JSON object
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e(TAG, "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
json解析器:
dependent: :destroy
我不知道如何实现这一点......任何人都可以帮助我......
答案 0 :(得分:0)
Jean是JSONObject
,您在代码中使用JSONArray
。此外,size
和cost
是数字,但您使用getString()
。所以这段代码应该可行(请使用变量的相关名称:不要在名为lastname的变量中存储成本)
JsonParser jParser = new JsonParser();
JSONObject json = jParser.getJSONFromUrl(yourJsonStringUrl);
JSONObject dataJsonOnline = json.getJSONObject("online");
JSONObject dataJsonMen = dataJsonOnline.getJSONObject("men");
JSONObject dataJsonCloth = dataJsonMen.getJSONObject("clothes");
JSONObject dataJsonJean = dataJsonCloth.getJSONObject("jean");
String color = dataJsonJean.getString("color");
int cost = dataJsonJean.getInt("cost");
int size = dataJsonJean.getString("size");
答案 1 :(得分:0)
试试这个:
void setup() {
Serial.begin(9600);
Serial.println("Start");
}
void loop() {
if(Serial.available()){
int a = Serial.read();
Serial.print(a);
}
}
你没有JSONArray。 JsonArray位于'['和']'括号之间。