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spring boot自带的json converter和fastjson对scala生成的对象,都不能很好的序列化成json。结果都会为空
解决办法:
[1] 需要序列化的类型,用kotlin定义.毕竟kotlin和scala的类定义语法基本相同
[2] 使用或者替换fastjson或者springboot的http message converter.如替换为json4s
[3] 附录:spring boot默认的converter如下:
- ByteArrayHttpMessageConverter – converts byte arrays
- StringHttpMessageConverter – converts Strings
- ResourceHttpMessageConverter – converts org.springframework.core.io.Resource for any type of octet stream
- SourceHttpMessageConverter – converts javax.xml.transform.Source
- FormHttpMessageConverter – converts form data to/from a MultiValueMap
. - Jaxb2RootElementHttpMessageConverter – converts Java objects to/from XML (added only if JAXB2 is present on the classpath)
- MappingJackson2HttpMessageConverter – converts JSON (added only if Jackson 2 is present on the classpath)
- MappingJacksonHttpMessageConverter – converts JSON (added only if Jackson is present on the classpath)
- AtomFeedHttpMessageConverter – converts Atom feeds (added only if Rome is present on the classpath)
- RssChannelHttpMessageConverter – converts RSS feeds (added only if Rome is present on the classpath)
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