I am getting a server side json response to load my menu, I tried twice and it gave this error message (the Error parsing data org.json.JSONException).
the reason for that is I'm getting the response partially, in both attempts i got different responses as shown in the images. i think I'm not getting the complete json response, getting only partial response. what should I do to get the complete response.
this is my code
#Override
protected JSONObject doInBackground(String... params) {
String path = null;
String response = null;
HashMap<String, String> request = null;
JSONObject requestJson = null;
DefaultHttpClient httpClient = null;
HttpPost httpPost = null;
StringEntity requestString = null;
ResponseHandler<String> responseHandler = null;
// get the email and password
try {
path = "http://xxxxxxxxxxxxxxxxxxx";
new URL(path);
} catch (MalformedURLException e) {
e.printStackTrace();
}
try {
// set the API request
request = new HashMap<String, String>();
request.put(new String("CetegoryCode"), "P");
request.entrySet().iterator();
// Store locations in JSON
requestJson = new JSONObject(request);
httpClient = new DefaultHttpClient();
httpPost = new HttpPost(path);
requestString = new StringEntity(requestJson.toString());
// sets the post request as the resulting string
httpPost.setEntity(requestString);
httpPost.setHeader("Content-type", "application/json");
// Handles the response
responseHandler = new BasicResponseHandler();
response = httpClient.execute(httpPost, responseHandler);
responseJson = new JSONObject(response);
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
try {
responseJson = new JSONObject(response);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
return responseJson;
}
this is the image
If your response is returning JsonArray thn need to set tht response string jsonarray. create instance of jsonarray and fill it up with the response.
if its normal get ws thn you can append parameters in url like query string
protected Void doInBackground(String... urls) {
/************ Make Post Call To Web Server ***********/
BufferedReader reader = null;
try {
// Append parameters with values eg ?CetegoryCode=p
String path = "http://xxxxxxxxxxxxxxxxxxx?CetegoryCode=p";
URL url = new URL(path);
URLConnection conn = url.openConnection();
conn.setDoOutput(true);
OutputStreamWriter wr = new OutputStreamWriter(
conn.getOutputStream());
wr.write(data);
wr.flush();
// Get the server response
reader = new BufferedReader(new InputStreamReader(
conn.getInputStream()));
StringBuilder sb = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null) {
sb.append(line + "");
}
Content = sb.toString();
JSONArray jArray = new JSONArray(Content);
if (jArray != null)
Log.e("Data", "" + jArray.length());
} catch (Exception ex) {
Error = ex.getMessage();
} finally {
try {
reader.close();
}
catch (Exception ex) {
}
}
/*****************************************************/
return null;
}
Try out below code to parse and get JSON response:
public static JSONObject getJSONFromUrl(String url) {
// Making HTTP request
try {
URL url1 = new URL(url);
HttpURLConnection conn = (HttpURLConnection) url1.openConnection();
conn.setReadTimeout(10000 /* milliseconds */);
conn.setConnectTimeout(15000 /* milliseconds */);
conn.setRequestMethod("POST");
conn.setDoInput(true);
// Starts the query
conn.connect();
InputStream stream = conn.getInputStream();
json = convertStreamToString(stream);
stream.close();
} catch (Exception e) {
e.printStackTrace();
}
// 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;
}
static String convertStreamToString(java.io.InputStream is) {
java.util.Scanner s = new java.util.Scanner(is).useDelimiter("\\A");
return s.hasNext() ? s.next() : "";
}
Use getJSONFromUrl method as below in your code:
#Override
protected JSONObject doInBackground(String... params) {
String path = null;
String response = null;
HashMap<String, String> request = null;
try {
responseJson = new JSONObject(response);
responseJson =getJSONFromUrl("http://xxxxxxxxxxxxxxxxxxx?CetegoryCode=p");
}catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
return responseJson;
}
Related
I am working on an application that interacts with a room security control device.
I want to get devices information from API. I am using HttpUrlConnection and POST method. It hits the API and I get 200 OK response but I get the out
"{"json":{"control":{"cmd":"getdevice","uid":256}}} doesn't exist"
I have tried all the solutions from stackoverflow and other platforms but it's not giving the output.
Moreover I have tested this API on Postman and it's working there and giving the device information.
Here is the code:
public class HTTPRequestTask extends AsyncTask<Void, Void, Void> {
#Override
protected Void doInBackground(Void... params) {
String username = "admin";
String password = "888888";
URL url = null;
try {
url = new URL("http://192.168.100.25/network.cgi");
} catch (MalformedURLException e) {
e.printStackTrace();
}
assert url != null;
HttpURLConnection httpRequest = null;
try {
httpRequest = (HttpURLConnection) url.openConnection();
httpRequest.setRequestMethod("POST");
httpRequest.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
httpRequest.setDoInput(true);
httpRequest.setDoOutput(true);
android.util.Base64.encode(authString.getBytes(), android.util.Base64.DEFAULT);
httpRequest.addRequestProperty("Authorization", "Basic " + "YWRtaW46ODg4ODg4"); // This is auth bytecode
httpRequest.connect();
} catch (ProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
JSONObject json = new JSONObject();
JSONObject jsonObject = new JSONObject();
JSONObject jsonObjectControl = new JSONObject();
jsonObjectControl.put("cmd","getdevice");
jsonObjectControl.put("uid",256);
jsonObject.put("control",jsonObjectControl);
json.put("json", jsonObject);
String encodedData = URLEncoder.encode( json.toString(), "UTF-8" );
BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(httpRequest.getOutputStream()));
writer.write(encodedData);
writer.flush();
BufferedReader bufferedReader = null;
bufferedReader = new BufferedReader
(new InputStreamReader(httpRequest.getInputStream(), "UTF-8"));
String line = null;
StringBuilder sb = new StringBuilder();
do {
line = bufferedReader.readLine();
sb.append(line);
Log.i("Output line: ",sb.toString());
}
while(bufferedReader.readLine()!=null);
bufferedReader.close();
int responseCode = httpRequest.getResponseCode();
String resMsg = httpRequest.getResponseMessage();
String result = sb.toString();
Log.d("Output: ","--"+result);
Log.d("Response Code: "+responseCode, "!!");
Log.d("Response MSG ","--"+resMsg);
} catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
}
#Override
protected JSONObject doInBackground(String... params) {
String path = null;
String response = null;
HashMap<String, String> request = null;
JSONObject requestJson = null;
DefaultHttpClient httpClient = null;
HttpPost httpPost = null;
StringEntity requestString = null;
ResponseHandler<String> responseHandler = null;
// get the username and password
Log.i("Email", params[0]);
Log.i("Password", params[1]);
try {
path = "http://192.xxx.x.xxx/xxxxService/UsersService.svc/UserAuthentication";
new URL(path);
} catch (MalformedURLException e) {
e.printStackTrace();
}
try {
// set the API request
request = new HashMap<String, String>();
request.put(new String("Email"), params[0]);
request.put(new String("Password"), params[1]);
request.entrySet().iterator();
// Store locations in JSON
requestJson = new JSONObject(request);
httpClient = new DefaultHttpClient();
httpPost = new HttpPost(path);
requestString = new StringEntity(requestJson.toString());
// sets the post request as the resulting string
httpPost.setEntity(requestString);
httpPost.setHeader("Content-type", "application/json");
// Handles the response
responseHandler = new BasicResponseHandler();
response = httpClient.execute(httpPost, responseHandler);
responseJson = new JSONObject(response);
} catch (Exception e) {
Log.e("Buffer Error", "Error converting result " + e.toString());
}
try {
responseJson = new JSONObject(response);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
return responseJson;
}
I'm using this code to login to my app.
responseJson = new JSONObject(response);
I'm getting "success" for the response, but responseJson value is "null"
This is what I got:
Error converting result org.json.JSONException: Value Fail of type java.lang.String cannot be converted to JSONObject
I t will be good if i can get a salution
Thanks in advance.
Try below code:
Object json = new JSONTokener(response).nextValue();
if (json instanceof JSONArray) {
JSONArray jsonArray = (JSONArray) json;
// your logic
}else if(json instanceof JSONObject){
//your logic
}
Im trying to parse the json returned from graph facebook, a feed from a page.
The url is something like this: https://graph.facebook.com/pageId/feed?access_token=MyAppId|MySecretKey&limit=10.
But I'm getting the error: Illegal character in query at index 7. The character at index 77 it's the "|". My code:
private JSONObject getJSONFromUrl(String url) throws UnsupportedEncodingException {
// Making HTTP request
InputStream is = null;
JSONObject jObj = null;
String json = "";
try {
// 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;
}
If I replace the URL, using the encoder, like this:
String url = "https://graph.facebook.com/pageId/feed?access_token=MyAppId" + URLEncoder.encode("|", "UTF-8") + "mySecretKey&limit=10";
I receive the error:
{"error":{"message":"(#200) The user hasn't authorized the application to perform this action","type":"OAuthException","code":200}}
But If I copy and paste the original URL, it's returning successfully the JSON
How can I send correctly the url on this case?
I have two text boxes, 1 for username and the other for password.
I wanted to pass what the user enters into the edit texts with the post method
String request = "https://beta135.hamarisuraksha.com/web/webservice/HamariSurakshaMobile.asmx/getIMSafeAccountInfoOnLogon";
URL url;
try {
url = new URL(request);
HttpURLConnection connection = (HttpURLConnection) url
.openConnection();
connection.setDoOutput(true);
connection.setDoInput(true);
connection.setInstanceFollowRedirects(false);
connection.setRequestMethod("POST");
connection.setRequestProperty("Connection", "Keep-Alive");
connection.setRequestProperty("Content-Type",
"application/x-www-form-urlencoded;");// boundary="+CommonFunctions.boundary
connection.setUseCaches(false);
DataOutputStream wr = new DataOutputStream(
connection.getOutputStream());
wr.writeBytes(urlParameters);
wr.flush();
wr.close();
int responseCode = connection.getResponseCode();
/*
* System.out.println("\nSending 'POST' request to URL : " +
* url); System.out.println("Post parameters : " +
* urlParameters);
*/
System.out.println("Response Code : " + responseCode);
InputStream errorstream = connection.getErrorStream();
BufferedReader br = null;
if (errorstream == null) {
InputStream inputstream = connection.getInputStream();
br = new BufferedReader(new InputStreamReader(inputstream));
} else {
br = new BufferedReader(new InputStreamReader(errorstream));
}
String response = "";
String nachricht;
while ((nachricht = br.readLine()) != null) {
response += nachricht;
}
// print result
// System.out.println(response.toString());
return response.toString();
} catch (MalformedURLException e) {
// TODO Auto-generated catch block
e.printStackTrace();
return null;
} catch (ProtocolException e) {
// TODO Auto-generated catch block
e.printStackTrace();
return null;
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
return null;
}
}
if i am getting your question correctly , you need to pass your parameters to a web service. in my case i have implemented a method to get the web service response by giving the url and the values as parameters. i think this will help you.
public JSONObject getJSONFromUrl(JSONObject parm,String url) throws JSONException {
InputStream is = null;
JSONObject jObj = null;
String json = "";
// Making HTTP request
try {
// defaultHttpClient
/*JSONObject parm = new JSONObject();
parm.put("agencyId", 27);
parm.put("caregiverPersonId", 47);*/
/* if(!(jObj.isNull("d"))){
jObj=null;
}
*/
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.addHeader("Content-Type", "application/json; charset=utf-8");
HttpEntity body = new StringEntity(parm.toString(), "utf8");
httpPost.setEntity(body);
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
is = httpEntity.getContent();
/* String response = EntityUtils.toString(httpEntity);
Log.w("myApp", response);*/
} 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());
}
// JSONObject jObj2 = new JSONObject(json);
// 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;
}
this method take two parameters. one is the url, other one is the values that we should send to the web service. and simply returns the json object. hope this will help you
EDIT
to pass your username and password just use below code
JsonParser jp = new JsonParser(); // create instance for the jsonparse class
String caregiverID = MainActivity.confirm.toString();
JSONObject param = new JSONObject();
JSONObject job = new JSONObject();
try {
param.put("username", yourUserNAme);
job = jp.getJSONFromUrl(param, yourURL);
I am not sure why this url is throwing a MalformedURL exception: http%3A%2F%2Fapi.themoviedb.org%2F3%2Fsearch%2Fperson%3Fapi_key%3secret%26query%3Dchristopher_guest
This is the url required by the api that I need to use. http://api.themoviedb.org/3/search/person?api_key=secret&query=christopher_guest
I have been getting target host must not be null errors using this url then I changed my coded to what you are seeing below. Not sure whats going on here although I have heard urls that contain underscores dont validate outside of web browsers and cause these types of situations.
Any ideas around this?
This is where I build the url
package com.tot.tipofthetongue;
import android.widget.EditText;
public class getName {
static String nameOne = null;
static String nameTwo = null;
static StringBuilder personURLOne = new StringBuilder();
static StringBuilder personURLTwo = new StringBuilder();
public static String personURL = "http://api.themoviedb.org/3/search/person?api_key=secret&query=";
public static StringBuilder getName1(EditText searchOne){
nameOne = searchOne.getText().toString();
nameOne = nameOne.replace(" ", "_");
personURLOne.append(personURL);
personURLOne = personURLOne.append(nameOne);
return personURLOne;
}
And this is my jsonparser that I pass that url to.
public class JSONParser extends AsyncTask<String, Void, JSONObject> {
static InputStream inputStream = null;
static JSONObject jObject = null;
static String jSon = "";
public String myURL;
String host;
HttpRequest request;
protected JSONObject doInBackground(String... url) {
// TODO Auto-generated method stub
//Make HTTP Request
try {
//defaultHttpClient
for(int i = 0; i < url.length; i++){
myURL = url[0];
myURL = URLEncoder.encode(myURL, "utf-8");
}
HttpGet httpGet = new HttpGet(myURL);
//header
httpGet.setHeader("Accept", "application/json");
HttpResponse httpResponse = new DefaultHttpClient().execute(new HttpHost(new URL(myURL).getHost()), request);
HttpEntity httpEntity = httpResponse.getEntity();
inputStream = httpEntity.getContent();
} catch (UnsupportedEncodingException e){
e.printStackTrace();
} catch (ClientProtocolException e){
e.printStackTrace();
}catch (IOException e){
e.printStackTrace();
}
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(inputStream, "UTF-8"), 8);
StringBuilder stringBuilder = new StringBuilder();
String line = null;
while ((line = reader.readLine()) != null){
stringBuilder.append(line + "\n");
}
Log.d("JSON Contents", stringBuilder.toString());
inputStream.close();
jSon = stringBuilder.toString();
} catch (Exception e){
Log.e("Buffer Error", "Error converting result " + e.toString());
}
//try to parse the string to JSON Object
try {
jObject = new JSONObject(jSon);
} catch (JSONException e){
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
//return JSON String
return jObject;
}
}
Print the String you formed before final submission to form Uri. And attach this to your question. It would be much easier to answer.
Try using HttpGet(URI uri) instead of HttpGet(String uri)
The reason is pretty simple. If you are using Uri, you will get immediately the Exception.
Hope this will help you to debug quickly.