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Converting malformed json array string to Java object

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 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.

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:

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

Convert XML to JSON, Issue with jaxb class being used with FasterXML

I need to convert XML to JSON at runtime, using FasterXML and generated beans from JAXB tool xjc using the source XSD.
I am using follwing code
public static void main(String[] args) {
try {
Testing tObj=new Testing();
ObjectMapper tester=tObj.createJaxbObjectMapper();
CustomerOrderType data=tester.readValue(TEST_XML_STRING, CustomerOrderType.class);
//ObjectMapper serializr=new ObjectMapper();
//serializr.writeValue(System.out, data);
} catch (Exception je) {
System.out.println(je.toString());
}
}
public ObjectMapper createJaxbObjectMapper()
{
final ObjectMapper mapper = new ObjectMapper();
final TypeFactory typeFactory = TypeFactory.defaultInstance();
final AnnotationIntrospector introspector = new JaxbAnnotationIntrospector(typeFactory);
// make deserializer use JAXB annotations (only)
mapper.getDeserializationConfig().with(introspector);
// make serializer use JAXB annotations (only)
mapper.getSerializationConfig().with(introspector);
return mapper;
}
But I am getting an error as follows.
com.fasterxml.jackson.core.JsonParseException: Unexpected character ('<' (code 60)): expected a valid value (number, String, array, object, 'true', 'false' or 'null')
at [Source: (String)"<ns1:orderDetail xmlns:ns1="http://www.colt.net/xml/ns/webservice/manord/v1.0"><ns7:Customer_Order_Status xmlns:ns7="http://www.colt.net/xml/ns/cbe/ord/v1.0">New</ns7:Customer_Order_Status><ns8:Order_Creation_Date xmlns:ns8="http://www.colt.net/xml/ns/cbe/ord/v1.0">2018-03-06T09:03:25</ns8:Order_Creation_Date><ns9:Opportunity_Number xmlns:ns9="http://www.colt.net/xml/ns/cbe/ord/v1.0">8798</ns9:Opportunity_Number><ns10:Order_Source xmlns:ns10="http://www.colt.net/xml/ns/cbe/ord/v1.0">eOrder Lite<"[truncated 9232 chars]; line: 1, column: 2]

It looks that you are using JSON-java lib, which is extremely lightweight and does not provide a possibility to skip namespaces from parsed XML tag names.
But this can be achieved using a more advanced XML/JSON processing lib, for example, FasterXML/Jackson.
Update
To convert String xml to Json:
1) Use com.fasterxml.jackson.dataformat:jackson-dataformat-xml lib dependency.
2) Conversion sequence is similar to:
XmlMapper xmlMapper = new XmlMapper();
MyObject myObj = xmlMapper.readValue(xml, MyObject.class);
ObjectMapper jsonMapper = new ObjectMapper();
String json = jsonMapper.writeValueAsString(myObj);
System.out.println(json);

An off-the-shelf XML-to-JSON converter is never going to give you optimal output for your particular application. You generally need to apply either some preprocessing (an XML transformation using XSLT) or some post-processing of the JSON.
In this case I would suggest doing an XSLT transformation to drop the namespaces.

Convert nested JSON string to list of java object

I have this json String
{"data":"[Level [key=LevelKey [keyEnd=0], Description=abc], Level [key=levelKey [keyEnd=1], Description=xyz]", "id":"123"}
And corresponding java classes are
public class Level {
public LevelKey key;
public String id;
}
public class LevelKey{
public String keyEnd;
}
I want to convert this data json string to list of Level object using Jackson
ObjectMapper mapper = new ObjectMapper();
List<Level> arr = mapper.readValue(data, new TypeReference<List<Level>>(){});
But I am getting below error
com.fasterxml.jackson.core.JsonParseException: Unrecognized token 'Level': was expecting ('true', 'false' or 'null')
Is there any other method to parse it?

The below does not look like a proper JSON for the purpose (except for a standard fixed string)
"[Level [key=LevelKey [keyEnd=0], Description=abc], Level [key=levelKey [keyEnd=1], Description=xyz]"
You could correct the data part of your JSON to something like below (Closest to your JSON in question) :
[\"Level [key=LevelKey [keyEnd=0], Description=abc]\",\" Level [key=levelKey [keyEnd=1], Description=xyz]\"]
Is there any other method to parse it?
You could use a direct class reference of ArrayList instead of having to instantiate TypeReference like below to parse the above (corrected) json string :
List<Level> arr = mapper.readValue(data, (new ArrayList<Level>()).getClass());

This was an interesting one I must say. Take a look at code snippet I think I got it correct :
String data ="{\"data\":\"[Level [key=LevelKey [keyEnd=0], Description=abc], Level [key=levelKey [keyEnd=1], Description=xyz]\", \"id\":\"123\"}";
ObjectMapper mapper = new ObjectMapper();
mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
List<Level> arr = mapper.readValue(data, (new ArrayList<Level>()).getClass());
System.out.println(arr);
I got the following output :
[{data=[Level [key=LevelKey [keyEnd=0], Description=abc], Level [key=levelKey [keyEnd=1], Description=xyz], id=123}]
Also if you encountered any JsonParseException which according to documentation means :
Exception type for parsing problems, used when non-well-formed content
(content that does not conform to JSON syntax as per specification) is
encountered.
So while hacking the JSON you can update the ObjectMapper object like this :
mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
Also as mentioned by Exception_al using a direct class reference of ArrayList instead of having to instantiate TypeReference like below to parse the above (corrected) json string.
List<Level> arr = mapper.readValue(data, (new ArrayList<Level>()).getClass());
Hope this helped.

The String is not the JSON representation of what you expect you get deserialized into JAVA.
This is a JSON String:
"{"data":[{"key":{"keyEnd":0},"Description":"abc"},{"key":{"keyEnd":1},"Description":"abc"}],"id":"123"}"
So, there is either a problem with the String or you need to do the parsing yourself.

How to disable character escaping by jackson

I want to include html in my JSON response.
MyClass obj= new MyCLass();
obj.setHTML("<div style='display:none'>4</div>");
ObjectMapper mapper=new ObjectMapper();
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
String jsonResponse=mapper.writeValueAsString(obj);
System.out.println(jsonResponse);
O/P i get
{"html":"<div style=\"display:none\">4</div>"}
Required O/P
{"html":"<div style='display:none'>4</div>"}
Since I want to use the json response directly. Can i disable the escaping of qoutes by object mapper.

You can annotate your getHtml method with
#JsonRawValue

Add a property to a json string with jackson json

I am storing a json string into a text field in mysql.
After the insertion, i want to update my json string and add the mysql line id into it with jackson json.
I have a java String which is in Json format
{
"thing":"val"
}
I'm looking to add another K/V without writing lines of codes.
to finally have this :
{
"thing":"val"
"mysqlId":10
}
I can convert my String to a JsonNode :
ObjectMapper mapper = new ObjectMapper();
JsonNode json = mapper.readTree( jsonStr);
Looking to do something like this
json.put("mysqlId",10);
json.toString();
then update in my text field with new json string in mysql
I can't make it.
I don't want use many class is there a simple way to do so with jackson?

Try casting your JsonNode to an com.fasterxml.jackson.databind.node.ObjectNode and then calling put set (or replace) on it.