required java.lang.String found org.json.JSONObject

Question

This is the code

@Override
protected String doInBackground(String... params) {
List<NameValuePair>param=new ArrayList<NameValuePair>();
param.add(new BasicNameValuePair("token_type",tToken));
param.add(new BasicNameValuePair("access_token",acToken));

String json=jsonParser.makeHttpRequest(URL_ORDERS,"GET",param);
Log.d("JSON ",json);

This is the tToken and acToken

JSONObject obj=new JSONObject(getIntent().getStringExtra("jsonToken"));
Log.d("Orders Response",obj.toString());

acToken= obj.getString("access_token");
tToken=obj.getString("token_type");

The problem is in the line String json=jsonParser.makeHttpRequest(URL_ORDERS,"GET",param);

It gives me this error required java.lang.String found org.json.JSONObject

i do understand the problem but i don't know how to fix it

maven thank you for the edit proposal , now i will know how to post my code on stackoverflow — Alex
– Alex, Commented Sep 19, 2014 at 11:48
use bold font only when they are needed .. else it will show arrogance ... :) — Maveňツ
– Maveňツ, Commented Sep 22, 2014 at 5:22

Blackbelt · Accepted Answer · 2014-09-19 11:15:09Z

1

it means that jsonParser.makeHttpRequest retuns a JSONObject and not a String. Change

 String json=jsonParser.makeHttpRequest(URL_ORDERS,"GET",param);

to

 JSONObject json=jsonParser.makeHttpRequest(URL_ORDERS,"GET",param);

answered Sep 19, 2014 at 11:15

Blackbelt

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1 Comment

Alex Over a year ago

Sit thank you a lot i will acept your'e answer later(now i can't before it's to early, or so stack overflow system says)

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