A JSONObject text must begin with '{' at 1 [character 2 line 1] with '{' error

Question

String JSON = "http://www.json-generator.com/j/cglqaRcMSW?indent=4";

JSONObject jsonObject = new JSONObject(JSON);
JSONObject getSth = jsonObject.getJSONObject("get");
Object level = getSth.get("2");

System.out.println(level);

I referred many solutions for parsing this link, still getting the same error in question. Can any give me a simple solution for parsing it.

Solution 1

Your problem is that String JSON = "http://www.json-generator.com/j/cglqaRcMSW?indent=4"; is not JSON.
What you want to do is open an HTTP connection to "http://www.json-generator.com/j/cglqaRcMSW?indent=4" and parse the JSON response.

String JSON = "http://www.json-generator.com/j/cglqaRcMSW?indent=4";
JSONObject jsonObject = new JSONObject(JSON); // <-- Problem here!

Will not open a connection to the site and retrieve the content.

Solution 2

While the json begins with "[" and ends with "]" that means this is the Json Array, use JSONArray instead:

JSONArray jsonArray = new JSONArray(JSON);

And then you can map it with the List Test Object if you need:

ObjectMapper mapper = new ObjectMapper();
List<TestExample> listTest = mapper.readValue(String.valueOf(jsonArray), List.class);

Solution 3

I had same issue. My Json response from the server was having [, and, ]:

[{"DATE_HIRED":852344800000,"FRNG_SUB_ACCT":0,"MOVING_EXP":0,"CURRENCY_CODE":"CAD  ","PIN":"          ","EST_REMUN":0,"HM_DIST_CO":1,"SICK_PAY":0,"STAND_AMT":0,"BSI_GROUP":"           ","LAST_DED_SEQ":36}]

http://jsonlint.com/ says valid json. you can copy and verify it.

I have fixed with below code as temporary solution:

BufferedReader br = new BufferedReader(new InputStreamReader((response.getEntity().getContent())));
String result ="";
String output = null;
while ((result = br.readLine()) != null) {
    output = result.replace("[", "").replace("]", "");
    JSONObject jsonObject = new JSONObject(output); 
    JSONArray jsonArray = new JSONArray(output); 
    .....   
}

Solution 4

I had the same, there was an empty new line character at the beginning. That solved it:

int i = result.indexOf("{");
result = result.substring(i);
JSONObject json = new JSONObject(result.trim()); 
System.out.println(json.toString(4));  

Solution 5

I had the same error and struggled to fix it, then answer above by Nagaraja JB helped me to fix it. In my case:

Was before: JSONObject response_json = new JSONObject(response_data);

Changed it to: JSONArray response_json = new JSONArray(response_data);

This fixed it.

View more solutions

Comments

[Scary Wombat]

this is valid json see jsonlint.com maybe it is your code?

[Jitesh Upadhyay]

This given json lik is correct..your code which you tried and also please log cat

[fge]

The bug is somewhere in your code; it looks like you are trying to parse a JSON value which is not an object using JSONObject. As an alternative, use a better JSON library, such as Jackson.

[ug_]

Try printing str.charAt(0) and see what the first char is. It could be a [ in that case its a json array. Or you might have a hidden char of some sort.

[OrhanC1]

@ns47731, he has posted a link to the JSON he's parsing and there is no [. str.charAt(0) is still a good debug step, though.

[Assaf Gamliel]

I'm pretty sure this is incorrect. Can you explain please how exactly did this fix your problem? It's still not opening a connection to retrieve the JSON from the server and it's not a valid JSON, it's a URL. Moreover, new StringBuilder.append is not valid Java syntax. As you unaccepted my answer I want to know why this is better, thanks.

[Itaypk]

The code here is incorrect, why not use Assaf's answer?

[Léo Natan]

How exactly does using a string builder solve the problem? You are still passing a URL and trying to create a JSON object from it, which is incorrect.

[Tim Autin]

Or you may use JSONArray jsonArray = new JSONArray(output);

[ucMedia]

Works like charm