How to validate a string is JSON or not in Java?

Question

I've already said, my question is different...question is already asked here, but i want a predefined method in java which checksgiven string is json format or not.

if there is no predefined method then at least tell a code which checks JSON Format or not, without using try catch block..

Thanks in advance...

Possible duplicate of How to check whether a given string is valid JSON in Java — Autar
– Autar, Commented Jan 25, 2017 at 11:24

Gowthaman M · Accepted Answer · 2018-02-23 05:41:25Z

3

Use JSONObject's constructor from a String

    try {
        JSONObject o = new JSONObject(yourString);
    } catch (JSONException e) {
        LOGGER.error("No valid json");
    }

edited Feb 23, 2018 at 5:41

Gowthaman M

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answered Jan 25, 2017 at 11:11

SCouto

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4 Comments

sai Over a year ago

is there any method ? which doesnt use try catch block?

SCouto Over a year ago

AFAIK this is the better way, you can build a method containing that block to avoid writing several catch's. Would that be acceptable according to your requirements?

sai Over a year ago

thanks for the reply. i just dont wanna use try catch blocks at all. like in javascript we hava a predefined function which validates string is json or not.. so i thought is there anything in java which automatically checks valid json or not.. i want boolean jsIsValid=method(string);

cassiomolin Over a year ago

@edsheeran First of all, Java is not JavaScript. Deal with it. And even in JavaScript you are supposed to use try-catch.

vaquar khan · Accepted Answer · 2019-07-03 17:59:11Z

import com.fasterxml.jackson.core.JsonFactory;
import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationFeature;
import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;    
/**
         * 
         * @param inputJosn
         * @return
         * @throws IOException 
         * @throws JsonParseException 
         * @throws JsonProcessingException
         */
        private static boolean isJsonValid(String inputJosn) throws JsonParseException, IOException  {
            ObjectMapper mapper = new ObjectMapper();
            mapper.enable(DeserializationFeature.FAIL_ON_READING_DUP_TREE_KEY);
            JsonFactory factory = mapper.getFactory();
            JsonParser parser = factory.createParser(inputJosn);
            JsonNode jsonObj = mapper.readTree(parser);
            System.out.println(jsonObj.toString());


            return true;
        }

Sanjay Kushwah · Accepted Answer · 2020-02-15 17:02:37Z

0

public boolean isValidJson(String jsonStr) {
    Object json = new JSONTokener(jsonStr)).nextValue();
    if (json instanceof JSONObject || json instanceof JSONArray) {
      return true;
    } else {
      return false;
    }
}

Just pass any string in this function.

edited Feb 15, 2020 at 17:02

answered May 29, 2019 at 9:56

Sanjay Kushwah

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3 Comments

simon_xia Over a year ago

var jsonStr doesn't appear in function body

Sanjay Kushwah Over a year ago

Just fixed the variable naming. Thanks for mentioning

Jangid Over a year ago

I think this one need some improvement. below example return true e.g. { "name" : mkj123 mk }

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