我正在研究一些事情。我有一个json链接。http://api.thingspeak.com/channels/145827/feed/last.json我怎样才能为android解析这个。
答案 0 :(得分:0)
获取JSON的代码,这很好。
获取JSON的类
import android.util.Log;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.util.EntityUtils;
import org.json.JSONArray;
import org.json.JSONException;
import java.io.IOException;
public class ApiConnector { //Connector class
public JSONArray GetYourJson()
{
// URL for getting the json
String url = "http://api.thingspeak.com/channels/145827/feed/last.json";
// Get HttpResponse Object from url.
// Get HttpEntity from Http Response Object
HttpEntity httpEntity = null;
try
{
DefaultHttpClient httpClient = new DefaultHttpClient(); // Default HttpClient
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
httpEntity = httpResponse.getEntity();
} catch (ClientProtocolException e) {
// Signals error in http protocol
e.printStackTrace();
//Log Errors Here
} catch (IOException e) {
e.printStackTrace();
}
// Convert HttpEntity into JSON Array
JSONArray jsonArray = null;
if (httpEntity != null) {
try {
String entityResponse = EntityUtils.toString(httpEntity);
Log.e("Entity Response : ", entityResponse);
jsonArray = new JSONArray(entityResponse);
} catch (JSONException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
}
return jsonArray;
}
}
AsyckTaks运行ApiConnector,这样做你的UI不会冻结。
private class GetYourJsonTask extends AsyncTask<ApiConnector,Long,JSONArray>
{
@Override
protected JSONArray doInBackground(ApiConnector... params) {
// it is executed on Background thread
return params[0].GetYourJson();
}
@Override
protected void onPostExecute(JSONArray jsonArray) {
//TODO: Do what you want with your json here
}
}
然后从UI中调用它
new GetYourJsonTask().execute(new ApiConnector());