How to deserialize Spring Boot actuator environment

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How to deserialize Spring Boot actuator environment



I would like to deserialize the Spring Boot Environment object returned by:



http://hostname:port/actuator/env



I'm using Jackson's ObjectMapper:


ObjectMapper


private static final ObjectMapper mapper = new ObjectMapper().configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);
...
ClientResponse clientResponse = resource.type(MediaType.APPLICATION_JSON).get(ClientResponse.class);
InputStream is = clientResponse.getEntityInputStream();

org.springframework.core.env.Environment e = mapper.readValue(is, org.springframework.core.env.Environment.class);



The code above fails with the following error, which makes sense:


com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of org.springframework.core.env.Environment, problem: abstract types either need to be mapped to concrete types, have custom deserializer, or be instantiated with additional type information



But I've tried all the implementations of the Environment class (AbstractEnvironment, MockEnvironment, StandardEnvironment, StandardServletEnvironment) and they all fail as well.


Environment


AbstractEnvironment


MockEnvironment


StandardEnvironment


StandardServletEnvironment



Which class should I use?





That isn't the environment object but a representation of what Spring Boot calls the environment. Trying to deserialize that into one of the Environment classes simply won't work.
– M. Deinum
Aug 8 at 11:19


Environment





That's interesting. So I'm basically wasting my time.
– Gep
Aug 8 at 11:34





The Environment is a heavy and custom object to instantiate, especially the internal PropertySource instances. You cannot simply deserialize them.
– M. Deinum
Aug 8 at 11:35


Environment


PropertySource




1 Answer
1



org.springframework.core.env.Environment is an interface. So ObjectMapper can not guess what concrete class to instantiate. You need to tell your ObjectMapper which class to use. So in your line
org.springframework.core.env.Environment e = mapper.readValue(is,org.springframework.core.env.Environment.class); You need to replace org.springframework.core.env.Environment.class with some concrete class. For example org.springframework.core.env.StandardEnvironment (depending on what kind of environment actually being returned). Otherwise de-serialize it to map:


org.springframework.core.env.Environment


ObjectMapper


ObjectMapper


org.springframework.core.env.Environment e = mapper.readValue(is,org.springframework.core.env.Environment.class);


org.springframework.core.env.Environment.class


org.springframework.core.env.StandardEnvironment


Map<String, Object> map = mapper.readValue(is,HashMap<String, Object>.class);



And then go from there





Thank you guys. I don't have the source code of the service I'm trying to inspect, so I don't know what kind of environment is being returned. Looking at the json I would guess a StandardServletEnvironment: {"activeProfiles":["aaaaa","bbbbbbbbbbb"],"propertySources":["name":"server.ports","properties":"local.server.port":"value":50053,"name":"servletContextInitParams","properties":,{"name":"systemProperties","properties":{"java.runtime.name":{"value":"Java(TM) SE Runtime... Anyway, whether I use StandardServletEnvironment or StandardEnvironment I get the same error:
– Gep
Aug 8 at 11:28






Can not deserialize instance of org.springframework.core.env.MutablePropertySources out of START_ARRAY token at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@67f9fca4; line: 1, column: 42] (through reference chain: org.springframework.web.context.support.StandardServletEnvironment["propertySources"])
– Gep
Aug 8 at 11:30






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