Converting malformed json array string to Java object

Question

I have a malformed json array string which I get from an API call as follows:

[{\"ResponseCode\":1,\"ResponseMsg\":\"[{\"Code\":\"CA2305181\",\"Message\":\"Processed successfully\"}]\"}]

There is a double quote before open square bracket in the value of Response Msg property. Is there a way to convert this into Java object ?

What I have tried so far: I have used Jackson to parse it as follows but it gives error

ObjectMapper mapper = new ObjectMapper();
            mapper.setPropertyNamingStrategy(new ResponseNameStrategy());
            Response[] response = mapper.readValue(strOutput1, Response[].class);
Error: Can not deserialize instance of java.util.ArrayList out of VALUE_STRING token

I have also tried using Gson to parse it but it also gives error

Gson gson = new GsonBuilder()
                    .setFieldNamingPolicy(FieldNamingPolicy.UPPER_CAMEL_CASE)
                    .create();
            Response[] response = gson.fromJson(strOutput1, Response[].class);
Error: Expected BEGIN_ARRAY but was STRING at line 1 column 35 path $[0].ResponseMsg

I have gone through the following links on StackOverflow but none of them has addressed my issue:

How to Convert String Array JSON in a Java Object

Convert a JSON string to object in Java ME?

JSON Array to Java objects

Convert json String to array of Objects

converting 'malformed' java json object to javascript

I think it's time to go back to that API and figure out why it is sending out broken JSON. If it's their fault, then no one can consume their service. The alternative is that you did something to the data possibly when you received it. Maybe tell us exactly how you got this broken JSON data. — Tim Biegeleisen
– Tim Biegeleisen, Commented May 25, 2018 at 6:37
Your question does not make sense to me atleast. When something is not a valid JSON it cannot be called a JSON. It is just a string. — wandermonk
– wandermonk, Commented May 25, 2018 at 6:41
I got this json as part of Soap Web Service Response as follows: <Result>[{"ResponseCode":1,"ResponseMsg":"[{\"Code\":\"CA2305181\",\"Message\":\"Processed successfully.\"}]"}]</Result> — A_J
– A_J, Commented May 25, 2018 at 6:43
First of all, the service should be fixed right away. Also, i think there is another extra double quote at the end }]\"}]. That double quote is an error too? — rodridevops
– rodridevops, Commented May 25, 2018 at 6:45
That is the closing double quote of the one starting before the square bracket — A_J
– A_J, Commented May 25, 2018 at 6:50

MartinByers · Accepted Answer · 2018-05-25 07:24:11Z

2

I think the answer is in the comments, you appear to be trying to solve the issue on the wrong place.

You are receiving json which you wish to parse into java objects, unfortunately the json is malformed so will not parse.

As a general rule you should never be trying to solve the symptom, but should look for the root cause and fix that, it may sound trivial but fixing symptoms leads to messy, unpredictable, and unmaintainable systems.

So the answer is fix the json where it is being broken. If this is something or of your control, while you wait for the fix, you could put a hack in to fix the json before you parse it.

This way you won't compromise your parsing, and only have a small piece of string replacement to remove when the third party has fixed the issue. But do not go live with the hack, it should only be used during development.

answered May 25, 2018 at 7:24

MartinByers

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1 Comment

A_J Over a year ago

I don't have a control over what the API is returning. And as mentioned in the question I am unable to parse it. I have even tried to parse it using JSONArray jsonArray = new JSONArray(strOutput1); still getting the error Expected a ',' or '}' at 38 [character 39 line 1]

rodridevops · Accepted Answer · 2018-05-25 08:11:31Z

1

As i mentioned in the comment, you should prepare your service response in order to parse it.

I implemented an example:

public class JsonTest {
    public static void main(String args[]) throws JsonProcessingException, IOException{
        String rawJson = 
                "[{\"ResponseCode\":1,\"ResponseMsg\":\"[{\"Code\":\"CA2305181\",\"Message\":\"Processed successfully\"}]\"}]";
        String goodJson = "{"+rawJson.split("[{{.}]")[2]+"}";
        ObjectMapper mapper = new ObjectMapper();
        final ObjectNode node = mapper.readValue(goodJson, ObjectNode.class);
        System.out.println("Pretty Print: " + mapper.writerWithDefaultPrettyPrinter().writeValueAsString(node));

        System.out.println("Just code: " + node.get("Code"));   
    }

}

Which returns:

answered May 25, 2018 at 8:11

rodridevops

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1 Comment

A_J Over a year ago

In split method, single start curly brace in character class will also achieve the same result

A_J · Accepted Answer · 2018-05-25 11:23:42Z

This is how I finally solved my issue:

String inputJsonStr = "[{\"ResponseCode\":1,\"ResponseMsg\":\"[{\"Code\":\"CA2305181\",\"Message\":\"Claim has been added successfully.\"}"
                + "]\"}]";
int indexOfRes = inputJsonStr.indexOf("ResponseMsg");

        if(inputJsonStr.substring(indexOfRes+13,indexOfRes+14).equals("\""))
        {
            inputJsonStr = inputJsonStr.substring(0,indexOfRes+13) + inputJsonStr.substring(indexOfRes+14);
        }
        int indexOfFirstClosingSquare = inputJsonStr.indexOf("]");
        if(inputJsonStr.substring(indexOfFirstClosingSquare+1, indexOfFirstClosingSquare+2).equals("\"")) {
            inputJsonStr = inputJsonStr.substring(0, indexOfFirstClosingSquare+1)+inputJsonStr.substring(indexOfFirstClosingSquare+2);
        }

Now inputJsonStr contains a valid json array which can be parsed into Java custom object array easily with gson as given in this SO link:

Convert json String to array of Objects

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