CODEWITHSUNDEEP
javajsonjacksonjsonnode
Senchu Thomas
Senchu Thomas

Reputation: 568

Get all the keys in a JSON string JsonNode in java

I have a json string which I need to validate and find any other keys other than in a list is there in the json string. The sample json string is

{
    "required" : true,
    "requiredMsg" : "Title needed",
    "choices" : [ "a", "b", "c", "d" ],
    "choiceSettings" : {
        "a" : {
            "exc" : true
        },
        "b" : { },
        "c" : { },
        "d" : {
            "textbox" : {
                "required" : true
            }
        }
    },
    "Settings" : {
        "type" : "none"
    }
}

To allow only predifined keys is exsist in the json string I want to get all the keys in the json string. How can I obtain all the keys in the json string. I am using jsonNode. My code till now is

        JsonNode rootNode = mapper.readTree(option);
        JsonNode reqiredMessage = rootNode.path("reqiredMessage");             
        System.out.println("msg   : "+  reqiredMessage.asText());            
        JsonNode drNode = rootNode.path("choices");
        Iterator<JsonNode> itr = drNode.iterator();
        System.out.println("\nchoices:");
        while (itr.hasNext()) {
            JsonNode temp = itr.next();
            System.out.println(temp.asText());
        }

How to get all the keys from the json string using JsonNode

Upvotes: 8

Views: 38116

Answers (4)

Eric
Eric

Reputation: 1

List in the json was not supported in the accepted solution. Here is what I propose:

public List<String> getAllNodeKeys(String json) throws JsonProcessingException {
    Map<String, Object> treeMap = objectMapper.readValue(json, new TypeReference<>() {
    });
    return findKeys(treeMap, new ArrayList<>());
}

private List<String> findKeys(Map<String, Object> treeMap, List<String> keys) {
    treeMap.forEach((key, value) -> {
        if (value instanceof LinkedHashMap) {
            LinkedHashMap map = (LinkedHashMap) value;
            findKeys(map, keys);
        } else if (value instanceof List) {
            ArrayList list = (ArrayList) value;
            list.forEach(map -> findKeys((LinkedHashMap) map, keys));

        }
        keys.add(key);
    });

    return keys;
}

Upvotes: 0

Dave
Dave

Reputation: 31

The accepted answer works out great but issues a warning, "Type safety: The expression of type Map needs unchecked conversion to conform to Map <String, Object>"

This answer led me to change that line to the following to eliminate the warning:

Map<String, Object> treeMap = mapper.readValue(json, new TypeReference<Map<String, Object>>() {});

Upvotes: 3

pvpkiran
pvpkiran

Reputation: 27048

This should do it. 

Map<String, Object> treeMap = mapper.readValue(json, Map.class);

List<String> keys  = Lists.newArrayList();
List<String> result = findKeys(treeMap, keys);
System.out.println(result);

private List<String> findKeys(Map<String, Object> treeMap , List<String> keys) {
    treeMap.forEach((key, value) -> {
      if (value instanceof LinkedHashMap) {
        Map<String, Object> map = (LinkedHashMap) value;
        findKeys(map, keys);
      }
      keys.add(key);
    });

    return keys;
  }

This will print out result as 

[required, requiredMsg, choices, exc, a, b, c, required, textbox, d, choiceSettings, type, Settings]

Upvotes: 7

Manos Nikolaidis
Manos Nikolaidis

Reputation: 22224

forEach will iterate over children of a JsonNode (converted to String when printed) and fieldNames() gets an Iterator<String> over keys. Here are some examples for printing elements of the example JSON:

JsonNode rootNode = mapper.readTree(option);

System.out.println("\nchoices:");
rootNode.path("choices").forEach(System.out::println);

System.out.println("\nAllKeys:");
rootNode.fieldNames().forEachRemaining(System.out::println);

System.out.println("\nChoiceSettings:");
rootNode.path("choiceSettings").fieldNames().forEachRemaining(System.out::println);

You'll probably need fields() at some point that returns an Iterator<Entry<String, JsonNode>> so you can iterate over key, value pairs.

Upvotes: 17

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