java - 如何使用 Jackson 对象映射器将 Java 对象中的并非所有属性转换为 json

Question

这个问题已经有答案了:

Jackson: how to prevent field serialization [duplicate] (9 个回答)

已关闭 7 年前。

我正在寻找一种通过 jackson 将部分对象(不是它的所有属性)转换为 json 的方法。目前我正在使用这个解决方案:

User userServed = User.find.byId(id);
ObjectMapper mapper = new ObjectMapper();
JsonNode node = null;
node = mapper.convertValue(userServed, JsonNode.class);

到目前为止，一切正常。

我的问题是，我不想将对象中的很多属性暴露给列表等 json 节点。有没有办法只选择所需的属性？

Answer 1

您可以使用 JsonIgnore 注释该字段吗？

来自http://wiki.fasterxml.com/JacksonAnnotations

@JsonIgnore (method/field): annotation used to completely disregard annotated method, regardless of auto-detection or other annotations

编辑:或者使用注释JsonIgnoreProperty。

(Jackson 1.4+) @JsonIgnoreProperties (class) can be used to indicate that certain properties are to be ignored for serialization and/or deserialization (handling differs a bit depending on which operation is affected):

◦String[] value() defines logical property names to ignore (names derived from getter/setter names, or by explicit annotations)

◦boolean ignoreUnknown() defines whether "unknown" JSON properties can be silently ignored during deserialization or not; does not affect serialization.