将java属性转换为json
将java属性转换为json
提问于 2014-05-26 21:50:02
有没有一种简单的方法可以将点符号的属性转换成json？
I.E
server.host=foo.bar
server.port=1234至
{
"server": {
"host": "foo.bar",
"port": 1234
}
} 回答 7
Stack Overflow用户
发布于 2014-08-07 16:54:37
这不是一种简单的方法，但我使用Gson库做到了这一点。结果将在jsonBundle字符串中。在本例中，我们获得了属性或包：
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);对于此实现，必须将ResourceBundle转换为包含String作为键和String作为值的Map。
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;
}我必须使用Gson为Map<String, String>创建自定义JSONSerializer
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;
}
}下面是创建JSON的主要逻辑：
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();
}
}最后，如果有一个值为John的键person.name.firstname，它会被转换成这样的JSON
{
"person" : {
"name" : {
"firstname" : "John"
}
}
}希望这能有所帮助:)
Stack Overflow用户
发布于 2019-03-02 01:40:00
使用lightbend配置java库(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());
}Stack Overflow用户
发布于 2014-05-26 21:58:01
下载并添加到您的库中非常简单：https://code.google.com/p/google-gson/
Gson gsonObj = new Gson();
String strJson = gsonObj.toJson(yourObject);页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/23871694
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