이 파서 클래스를 생성 한 다음
을 필요로하는 클래스의 객체를 생성하시기 바랍니다 당신의 도움이 필요합니다 응답 에서 문제가되었다
JSONParser 클래스.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UnsupportedEncodingException;
import java.util.List;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpGet;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.client.utils.URLEncodedUtils;
import org.apache.http.impl.client.DefaultHttpClient;
import org.json.JSONException;
import org.json.JSONObject;
import android.util.Log;
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List<NameValuePair> params) {
// Making HTTP request
try {
// check for request method
if (method == "POST") {
// request method is POST
// defaultHttpClient
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(params));
Log.d("url", url);
Log.d("params", params.toString());
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
} else if (method == "GET") {
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
Log.d("param", paramString);
if (!paramString.isEmpty()) {
url += "?" + paramString;
}
Log.d("url", url);
HttpGet httpGet = new HttpGet(url);
Log.d("httpGet", httpGet.toString());
HttpResponse httpResponse = httpClient.execute(httpGet);
Log.d("httpResponse", httpResponse.toString());
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
Log.d("is", is.toString());
}
} 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;
}
}
그리고이 클래스의 메소드를 호출하십시오. 매개 변수가없는 경우 null을 전달합니다. 아직 작업 공공 정적 무효의 다음 GetN (문자열 URL)
JSONParser jparser = new JSONParser();
JSONObject jobj = jparser.makeHttpRequest("url","GET",null);
logcat의 디버그 메시지는 무엇입니까? –
이 방법을 사용하려면 무엇이 필요합니까? 웹 응답을 다운로드 하시겠습니까? –
url "Http : //"+ ip + ":"+ port + "/? nekton ="+ 메시지 – abed