JAXB 或 Jackson 用于 xml 解组？

Question

Receiving an XML response, there are two main solutions decoding xml to a DTO object. Why should one prefer JAXB over jackson , or the other way around?

JAXB:

Unmarshaller unmarshaller = jaxbContext.createUnmarshaller();
XMLStreamReader reader = XMLInputFactory.newInstance().createXMLStreamReader(new StringReader(xmlString));
unmarshaller.unmarshal(reader, DtoObject.class);

Jackson:

mapper = new ObjectMapper() / new XmlMapper();
mapper.readValue(xmlString, DtoObjectc.class);

Answer 1

In my opinion, I will use JAXB, the following reason.

1. It is quite matured and part of JDK.
2. I do not need third party additional library to use
3. Jackson2 XML transformation is new as compared to JAXB which is in the industry for a quite long time and there are lot of community for it.

However I am not saying that which is good or bad. Again it is a choice to the developer to use.

Answer 2

From functionality, there is no difference. All is for java object <--> xml object. However, there is performance difference. I have tested with Jmeter between JAXB and Dozer Mapper. Result shows JAXB is about 50% efficient. I don't know what about Jackson. But from my personal experience. I prefer to JAXB. For your reference.

Answer 3

XmlMapper requires less configuration and on edge cases like PascalCase mapping it works when jaxb2 does not.

<dependency>
    <groupId>com.fasterxml.jackson.dataformat</groupId>
    <artifactId>jackson-dataformat-xml</artifactId>
    <version>2.9.8</version>
</dependency>

Use the same version as the other things you have with com.fasterxml.