Convert byte array to JSON error invalid JSON

Question

I am trying to convert a byte[] to JSON that looks like: {"foo": [{...}, {...}, ...], "bar": []}

try {
  byte[] response = getExternalServiceResponse();
  JSONObject json = new JSONObject(new String(response));
  log.info(json.toString(4));
} catch (Exception e) {
  e.printStackTrace();
}

This works for most cases of response but there are some that throw the exception with org.json.JSONException: A JSONObject text must begin with '{' at 3 [character 2 line 2]. How can I find out what characters are causing the issue as I can't read a byte array and I am not sure what the input contains without first converting it to JSON which throws the error?

try to print the string of byte array which causes the exception. — Vishal Patel
– Vishal Patel, Commented Jul 5, 2019 at 18:09
"I can't read a byte array" You are converting the byte array to a String, which you then give to the JSON parser. You can read a String, can't you? Try printing it to see what it contains. See also What's the simplest way to print a Java array? — Andreas
– Andreas, Commented Jul 5, 2019 at 18:24

Behdad · Accepted Answer · 2019-07-05 18:26:22Z

1

Maybe this is because sometimes your service returns an error or something similar which is not json, like "Some Error has occurred here! ;-)"

It is better to log your response before converting it to json.It is even better to validate its json schema for production deployment.

String strResponse = new String(response).trim();
log.info( strResponse);

answered Jul 5, 2019 at 18:26

Behdad

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1 Comment

atkayla Over a year ago

Thanks, this was exactly it! There are are a few cases where it doesn't return JSON, but some garbled format

��q�s�A��Oքh��U�P�E>tT�jj�� ���9I<�f�)_��*=�[O��������: @ �쩭�Ka�

{��8�m��$g�h��k�B, which I did not expect.

Qingfei Yuan · Accepted Answer · 2019-07-05 18:25:12Z

1

I agree that use Exception to judge is bad idea. Maybe you can tell it invalidate by yourself easily.

byte[] response = getExternalServiceResponse();
String resStr = new String(response).trim();
if(!resStr.startWith("{")) throw Exception("invalid json input!");

edited Jul 5, 2019 at 18:25

answered Jul 5, 2019 at 18:11

Qingfei Yuan

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