java properties to json

Question

Is there an easy way to convert properties with dot notation to json

I.E

server.host=foo.bar server.port=1234 

TO

{  "server": {     "host": "foo.bar",     "port": 1234   } }  

Answer 1

Using lightbend config java library (https://github.com/lightbend/config)

String toHierarchicalJsonString(Properties props) {   com.typesafe.config.Config config = com.typesafe.config.ConfigFactory.parseProperties(props);   return config.root().render(com.typesafe.config.ConfigRenderOptions.concise()); } 

Answer 2

Not the easy way, but I managed to do that using Gson library. The result will be in the jsonBundle String. Here we getting the properties or bundles in this case:

final ResourceBundle bundle = ResourceBundle.getBundle("messages"); final Map<String, String> bundleMap = resourceBundleToMap(bundle);  final Type mapType = new TypeToken<Map<String, String>>(){}.getType();  final String jsonBundle = new GsonBuilder()         .registerTypeAdapter(mapType, new BundleMapSerializer())         .create()         .toJson(bundleMap, mapType); 

For this implementation ResourceBundle have to be converted to Map containing String as a key and String as a value.

private static Map<String, String> resourceBundleToMap(final ResourceBundle bundle) {     final Map<String, String> bundleMap = new HashMap<>();      for (String key: bundle.keySet()) {         final String value = bundle.getString(key);          bundleMap.put(key, value);     }      return bundleMap; } 

I had to create custom JSONSerializer using Gson for Map<String, String>:

public class BundleMapSerializer implements JsonSerializer<Map<String, String>> {      private static final Logger LOGGER = LoggerFactory.getLogger(BundleMapSerializer.class);      @Override     public JsonElement serialize(final Map<String, String> bundleMap, final Type typeOfSrc, final JsonSerializationContext context) {         final JsonObject resultJson =  new JsonObject();          for (final String key: bundleMap.keySet()) {             try {                 createFromBundleKey(resultJson, key, bundleMap.get(key));             } catch (final IOException e) {                 LOGGER.error("Bundle map serialization exception: ", e);             }         }          return resultJson;     } } 

And here is the main logic of creating JSON:

public static JsonObject createFromBundleKey(final JsonObject resultJson, final String key, final String value) throws IOException {     if (!key.contains(".")) {         resultJson.addProperty(key, value);          return resultJson;     }      final String currentKey = firstKey(key);     if (currentKey != null) {         final String subRightKey = key.substring(currentKey.length() + 1, key.length());         final JsonObject childJson = getJsonIfExists(resultJson, currentKey);          resultJson.add(currentKey, createFromBundleKey(childJson, subRightKey, value));     }      return resultJson; }      private static String firstKey(final String fullKey) {         final String[] splittedKey = fullKey.split("\\.");          return (splittedKey.length != 0) ? splittedKey[0] : fullKey;     }      private static JsonObject getJsonIfExists(final JsonObject parent, final String key) {         if (parent == null) {             LOGGER.warn("Parent json parameter is null!");             return null;         }          if (parent.get(key) != null && !(parent.get(key) instanceof JsonObject)) {             throw new IllegalArgumentException("Invalid key \'" + key + "\' for parent: " + parent + "\nKey can not be JSON object and property or array in one time");         }          if (parent.getAsJsonObject(key) != null) {             return parent.getAsJsonObject(key);         } else {             return new JsonObject();         }    } 

In the end, if there were a key person.name.firstname with value John, it will be converted to such JSON:

{      "person" : {          "name" : {              "firstname" : "John"          }      } } 

Hope this will help :)