How to convert List <JSONObject> to Json String in java (com.amazonaws.util.json.JSONObject)

Question

com.amazonaws.util.json.JSONObject Below is the List, i want to convert it into json string.

List<JSONObject> jsonObjlist is

[{"Attribute":"EmailAddress","Value":"[email protected]"}, {"Attribute":"Source","Value":"Missing_Fixed"}, {"Attribute":"mx_Lead_Status","Value":"Registered User"}, {"Attribute":"mx_Age","Value":""}, {"Attribute":"mx_LoginID","Value":"[email protected]"}, {"Attribute":"mx_Registration_Source","Value":"EMAIL"}]

ObjectMapper objectMapper = new ObjectMapper();
        objectMapper.setVisibility(PropertyAccessor.FIELD, Visibility.ANY);
        objectMapper.enable(SerializationFeature.INDENT_OUTPUT);
        String arrayToJson = objectMapper.writeValueAsString(jsonObjlist);

Output i get is

[{"map":{"Attribute":"EmailAddress","Value":"[email protected]"}},{"map":{"Attribute":"Source","Value":"Missing_Fixed"}},{"map":{"Attribute":"mx_Lead_Status","Value":"Registered User"}},{"map":{"Attribute":"mx_Age","Value":""}},{"map":{"Attribute":"mx_LoginID","Value":"[email protected]"}},{"map":{"Attribute":"mx_Registration_Source","Value":"EMAIL"}}]

Desired out put is

"[{"Attribute":"EmailAddress","Value":"[email protected]"}, {"Attribute":"Source","Value":"Missing_Fixed"}, {"Attribute":"mx_Lead_Status","Value":"Registered User"}, {"Attribute":"mx_Age","Value":""}, {"Attribute":"mx_LoginID","Value":"[email protected]"}, {"Attribute":"mx_Registration_Source","Value":"EMAIL"}]"

You should also add your desired output, not just the one you get now. — Florian Albrecht
– Florian Albrecht, Commented Apr 4, 2018 at 11:57
If you are using Amazon AWS's JSONObject, you should also include relevant import statements, as most readers (including me) will otherwise assume that you are using org.json classes. — Florian Albrecht
– Florian Albrecht, Commented Apr 4, 2018 at 12:25

Florian Albrecht · Accepted Answer · 2018-04-04 12:17:42Z

6

You should convert your list to a JSON Array, and just use its toString() function:

JSONArray myArray = new JSONArray(jsonObjlist);

// ...
String arrayToJson = myArray.toString(2);

The int parameter specifies the indent factor to use for formatting.

edited Apr 4, 2018 at 12:17

answered Apr 4, 2018 at 12:03

Florian Albrecht

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2 Comments

Vinay Sawant Over a year ago

I am getting this {"myArrayList":[{"map":{"Attribute":"EmailAddress","Value":"[email protected]"}},{"map":{"Attribute":"Source","Value":"Missing_Fixed"}},{"map":{"Attribute":"mx_Lead_Status","Value":"Registered User"}},{"map":{"Attribute":"mx_Age","Value":""}},{"map":{"Attribute":"mx_LoginID","Value":"[email protected]"}},{"map":{"Attribute":"mx_Registration_Source","Value":"EMAIL"}}]}

Florian Albrecht Over a year ago

@VinaySawant Right, ObjectMapper is for Java -> JSON objects. You can just use toString(), changed my answer.

Deb · Accepted Answer · 2018-04-04 12:22:00Z

-1

You can also direct use

 String jsonString = new ObjectMapper().writeValueAsString(jsonObjectList)

To get the desired ouptput

Here is the small example

List<JSONObject> jsons = new ArrayList<>();
jsons.add(new JSONObject(ImmutableMap.of("Attribute", "EmailAddress", "Value", "[email protected]")));
jsons.add(new JSONObject(ImmutableMap.of("Attribute1", "EmailAddress3", "Value1", "[email protected]")));
System.out.println(new ObjectMapper().writeValueAsString(jsons));

The output is

[{"Value":"[email protected]","Attribute":"EmailAddress"},{"Attribute1":"EmailAddress3","Value1":"[email protected]"}]

I suspect there are some toString() method in any object that you have override?

edited Apr 4, 2018 at 12:22

answered Apr 4, 2018 at 12:12

Deb

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5 Comments

Florian Albrecht Over a year ago

Isn't this (almost) exactly what the OP tries, and what does not have desired effect?

Vinay Sawant Over a year ago

com.fasterxml.jackson.databind.JsonMappingException: No serializer found for class com.amazonaws.util.json.JSONObject and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: java.util.ArrayList[0])

Vinay Sawant Over a year ago

com.fasterxml.jackson.databind.JsonMappingException: No serializer found for class com.amazonaws.util.json.JSONObject and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationFeature.FAIL_ON_EMPTY_BEANS) ) (through reference chain: java.util.ArrayList[0])

Deb Over a year ago

Did you check if there are any toString method in your bean?

Deb Over a year ago

Just try out what you have tried in the question, but only this time without the toString method. I assume I did not use any kind of bean that's why I cannot reproduce it.

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