android jsonObject throws exception

Question

My code:

string result = "[{"lat":"b", "lon":"d", "ulica":"ulica1"}]";
jArray = new JSONArray(result);
tab =new String[jArray.length()];
json = jArray.getJSONObject(0);

tab[0]=json.getString("lat");
tab[1]=json.getString("lon");  
tab[2]=json.getString("ulica");

it throws jsonException, no value for lon. Can anyone help me?

It was my mistake instead of lon I had lot in my result string... I was looking for this misspelling from early morning .... ;P

Can you post the message in the exception please?

Todd Davies
– Todd Davies

2012-01-03 09:03:03 +00:00
Commented Jan 3, 2012 at 9:03 — Todd Davies
– Todd Davies, Commented Jan 3, 2012 at 9:03
What exact error message you are getting?

MH16
– MH16

2012-01-03 09:05:53 +00:00
Commented Jan 3, 2012 at 9:05 — MH16
– MH16, Commented Jan 3, 2012 at 9:05

Dany's · Accepted Answer · 2012-01-03 09:05:02Z

3

you must escape " with \" so instead of "a" you need to have \"a\"

answered Jan 3, 2012 at 9:05

Dany's

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Todd Davies · Accepted Answer · 2012-01-03 09:05:56Z

0

I think these lines are the problem:

tab[0]=json.getString("lat"); tab[1]=json.getString("lon");
tab[2]=json.getString("ulica");

'lat', 'lon' and 'ulica' aren't mentioned in the JSON, so they won't be found. Try this:

tab[0]=json.getString("a"); tab[1]=json.getString("c");

Then you may get 'b' and 'd' respectively.

Note this is untested.

answered Jan 3, 2012 at 9:05

Todd Davies

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Comments

santBart · Accepted Answer · 2012-01-03 09:20:23Z

0

jArray.length() is actualy 1, so there was another(outOfRangeException) mistake.

answered Jan 3, 2012 at 9:20

santBart

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