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Value Array Of Type Java.lang.string Cannot Be Converted To Jsonarray

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android json parsing share|improve this question edited Dec 12 '14 at 17:27 SLee 1,4461617 asked Apr 22 '12 at 12:26 RCK69 2361819 add a comment| 12 Answers 12 active oldest votes there was some unwanted echo on php :( but i uped your answer cause it's wisefull ;) –Kenji Apr 29 '14 at 8:32 add a comment| up vote 7 down vote Our mission is to bring affordable, technology education to people everywhere, in order to help them achieve their dreams and change the world. Please accept the answer and upvote it. –Arshu Aug 14 '13 at 15:30 Am Facing the same problem am unable to solve it..Please help me out .. –Gowtham Kumar http://bovbjerg.net/cannot-be/type-java-lang-string-cannot-be-converted-to-jsonobject.php

here is error org.json.JSONException:Value

How To Convert Java.lang.string To Jsonobject

Ego Team 12,908 Points Ego Team Ego Team 12,908 Points 11mo ago Thanks so much mate. What is the most someone can lose the popular vote by but still win the electoral college? Dealing With Dragonslayers What is the meaning of ''cry oneself"?

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02-10 05:39:44.850: D/My String(889): 02-10 05:39:44.850: D/My String(889): ] and its length is :309 ' –user3279100 Feb 10 '14 at 11:05 Good, Memory and processing wise, it's pretty inefficient. Is it possible to check where an alias was defined? Error Parsing Data Org Json Jsonexception Value String Cannot Be Converted To Jsonobject Vent kitchen hood vent to roof turbine vent?

Help me finding it Please. 12-02 23:04:34.427: E/JSON Parser(1629): Error parsing data org.json.JSONException: Value  of type java.lang.String cannot be converted to JSONArray Code: public class Http { public static final Org.json.jsonexception: Value Of Type Java.lang.string Cannot Be Converted To Jsonobject If you make a connection with PDO, you can't use mysql_* queries to access the database. more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed more hot questions question feed lang-java about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation

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Org.json.jsonexception: Value Of Type Java.lang.string Cannot Be Converted To Jsonobject

Polyglot Anagrams Cops' Thread OBDII across the world? How can I claim compensation? How To Convert Java.lang.string To Jsonobject Sometimes you have to specifically tell your editor to display hidden / special characters. Value

 Of Type Java.lang.string Cannot Be Converted To Jsonobject The temp solution is return new JSONObject(json.substring(json.indexOf("{"), json.lastIndexOf("}") + 1)); But try to remove hidden characters on source String. 

To emailaddress: To name: From name: Extra information in the email body (optional): Email: I am sending you the codedump of JSONException: Value of type java.lang.String cannot be converted to JSONObject http://bovbjerg.net/cannot-be/type-org-json-jsonobject-cannot-be-converted-to-jsonarray.php But she command me to use something like dictionary Object or sth like that :( –user3032822 Dec 3 '13 at 4:56 Its the only possible with the looping only E/br.com.viprastreamento.rastreamento.ListaVeiculos: Look mistake here! java android json share|improve this question edited Aug 13 '13 at 10:20 asked Aug 11 '13 at 16:21 Anton Kashpor 1861319 your json is malformed. Value

I have no idea, you might try logging the output before you initialize the JSONArray to see the exact response. Lab colleague uses cracked software. For that, I would suggest you to handle your array as below: JSONArray jsonArr = null; String jsonObjRecv = JSONGet.getJSONfromURL(URL_LIST); if(!TextUtils.isEmpty(jsonObjRecv)){ jsonArr = new JSONArray(jsonObjRecv); } else{ Log.w("json", "jsonObjRecv is null"); check over here Polyglot Anagrams Robbers' Thread why does this error keep popping out?

Departing from airport before visa is valid, but arriving when it is Is it possible to sheathe a katana as a free action? Org.json.jsonexception Value Android What is the meaning of ''cry oneself"? I used this JSON-validator to ensure that the JSON-Syntax is correct.

share|improve this answer answered Mar 18 '13 at 4:45 khaintt 1,80311617 worked nice with me!

Bug? share|improve this answer answered Oct 24 at 18:15 Pablo Carbajal 434 add a comment| protected by Community♦ May 10 '15 at 5:56 Thank you for your interest in this question. Not the answer you're looking for? Java.lang.string Cannot Be Converted To Jsonobject Volley How can I claim compensation?

asked 2 years ago viewed 693 times Upcoming Events 2016 Community Moderator Election ends Nov 22 Related 0parssingorg.json.JSONException: Value of type java.lang.String cannot be converted to JSONArray2Android Value … of type Is it possible to hand start modern planes? What is the most someone can lose the popular vote by but still win the electoral college? this content I've just forgot to change sample of code, but it isn't a solution of my problem. –Anton Kashpor Aug 13 '13 at 10:34 @whisperofblood: See my update.

or how can I generate one with that log? –jason Nov 19 '15 at 9:41 1 "Additions":[....] is not a json array, more over, it is not valid json at How to capture disk usage percentage of a partition as an integer? Your Answer draft saved draft discarded Sign up or log in Sign up using Google Sign up using Facebook Sign up using Email and Password Post as a guest Name return jsonArray; } catch (JSONException e) { Log.e("JSON Parser", "Error parsing data " + e.toString()); } catch (Exception e) { Log.e("Buffer Error", "Error converting result " + e.toString()); } } catch

How to reduce the width of the equation in a text paragraph? jfernandojr commented Feb 16, 2016 I left my head as follows: @Override public Map getHeaders() throws AuthFailureError { Map headers = new HashMap(); headers.put("Content-Type", "application/json"); headers.put("imei", "459710040904196"); headers.put("acao", Isn't AES-NI useless because now the key length need to be longer? Terms Privacy Security Status Help You can't perform that action at this time.

Here is the code I wrote : String additionString = ""Additions":[{"AdditionID":0,"Type":-1,"Long":" ","Short":" "},{"AdditionID":8,"Type":6,"Long":"English,"Short":"Eng"},{"AdditionID":11,"Type":2,"Long":"French","Short":"Fr."},{"AdditionID":12,"Type":2,"Long":"German","Short":"Ger."}]"; JSONArray AdditionArray = new JSONArray(additionString); And I'm getting this error for that : Value Additions of type java.lang.String try { jObj = new JSONObject(json.substring(3)); } catch (JSONException e) { Log.e("JSON Parser", "Error parsing data [" + e.getMessage()+"] "+json); } share|improve this answer answered Jul 31 '12 at 6:51 MTurPash A route should consist of several sights where the user gets navigated to.