[Update Sept 2020] Although my original answer here, from many years ago, seems to be helpful and is still getting upvotes, I now use the GSON library from Google, which I find to be more intuitive. I’ve got the following code: public void testJackson() throws IOException { ObjectMapper mapper = new ObjectMapper(); File from = … Read more
You can use Jackson’s class-level annotation: import com.fasterxml.jackson.annotation.JsonIgnoreProperties @JsonIgnoreProperties class { … } It will ignore every property you haven’t defined in your POJO. Very useful when you are just looking for a couple of properties in the JSON and don’t want to write the whole mapping. More info at Jackson’s website. If you want to … Read more
You need to create a TypeReference object for each generic type you use and use that for deserialization. For example – mapper.readValue(jsonString, new TypeReference<Data<String>>() {});
Here’s a good example. I use it to rename the variable because the JSON is coming from a .Net environment where properties start with an upper-case letter. public class Parameter { @JsonProperty(“Name”) public String name; @JsonProperty(“Value”) public String value; } This correctly parses to/from the JSON: “Parameter”:{ “Name”:”Parameter-Name”, “Value”:”Parameter-Value” }
Exactly how to do this depends on the version of Jackson that you’re using. This changed around version 1.9, before that, you could do this by adding @JsonIgnore to the getter. Which you’ve tried: Add @JsonIgnore on the getter method only Do this, and also add a specific @JsonProperty annotation for your JSON “password” field … Read more
Well, you can achieve that with Jackson, too. (and it seems to be more comfortable since you were considering using jackson). Use ObjectMapper‘s convertValue method: final ObjectMapper mapper = new ObjectMapper(); // jackson’s objectmapper final MyPojo pojo = mapper.convertValue(map, MyPojo.class); No need to convert into JSON string or something else; direct conversion does much faster.
Jackson provides an annotation that can be used on class level (JsonIgnoreProperties). Add the following to the top of your class (not to individual methods): @JsonIgnoreProperties(ignoreUnknown = true) public class Foo { … } Depending on the jackson version you are using you would have to use a different import in the current version it … Read more
You can configure individual ObjectMappers like this: ObjectMapper mapper = new ObjectMapper(); mapper.setVisibility(mapper.getSerializationConfig().getDefaultVisibilityChecker() .withFieldVisibility(JsonAutoDetect.Visibility.ANY) .withGetterVisibility(JsonAutoDetect.Visibility.NONE) .withSetterVisibility(JsonAutoDetect.Visibility.NONE) .withCreatorVisibility(JsonAutoDetect.Visibility.NONE)); If you want it set globally, I usually access a configured mapper through a wrapper class.
To convert your object in JSON with Jackson: import com.fasterxml.jackson.databind.ObjectMapper; import com.fasterxml.jackson.databind.ObjectWriter; ObjectWriter ow = new ObjectMapper().writer().withDefaultPrettyPrinter(); String json = ow.writeValueAsString(object);
You can put @JsonSerialize(using = CustomDateSerializer.class) over any date field of object to be serialized. public class CustomDateSerializer extends SerializerBase<Date> { public CustomDateSerializer() { super(Date.class, true); } @Override public void serialize(Date value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException { SimpleDateFormat formatter = new SimpleDateFormat(“EEE MMM dd yyyy HH:mm:ss ‘GMT’ZZZ (z)”); String format = formatter.format(value); … Read more