Alt*_*s34 6 spring json jersey spring-boot
我在Spring启动项目中有一个与Jackson配置相关的问题
我尝试自定义我的对象序列化.
在我的配置中添加了一个新的配置bean
@Bean
public Jackson2ObjectMapperBuilder jacksonBuilder() {
Jackson2ObjectMapperBuilder builder = new Jackson2ObjectMapperBuilder();
builder.propertyNamingStrategy(PropertyNamingStrategy.CAMEL_CASE_TO_LOWER_CASE_WITH_UNDERSCORES);
return builder;
}
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当我尝试输出我的类User的实例时,json结果不在CAMEL_CASE_TO_LOWER_CASE_WITH_UNDERSCORES中
Class User {
private String firstName = "Joe Blow";
public String getFirstName() {
return firstName;
}
}
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json输出是:
{
"firstName": "Joe Blow"
}
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并不是
{
"first_name": "Joe Blow"
}
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也许我需要在Jersey配置中注册一些东西来激活我的自定义obejctMapper配置
@Configuration
public class JerseyConfig extends ResourceConfig {
public JerseyConfig() {
packages("my.package);
}
}
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谢谢
Pau*_*tha 12
配置ObjectMapperfor JAX-RS/Jersey应用程序的一般方法是使用a ContextResolver.例如
@Provider
public class ObjectMapperContextResolver implements ContextResolver<ObjectMapper> {
private final ObjectMapper mapper;
public ObjectMapperContextResolver() {
mapper = new ObjectMapper();
mapper.setPropertyNamingStrategy(
PropertyNamingStrategy.CAMEL_CASE_TO_LOWER_CASE_WITH_UNDERSCORES
);
}
@Override
public ObjectMapper getContext(Class<?> type) {
return mapper;
}
}
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它应该通过包扫描获取,或者如果它不在包范围内,您可以显式注册它
public JerseyConfig() {
register(new ObjectMapperContextResolver());
// Or if there's is an injection required
// register it as a .class instead of instance
}
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在ContextResolver打包和解包时被调用.被序列化或反序列化的类/类型将传递给该getContext方法.因此,您甚至可以为不同类型甚至更多用例使用多个映射器.
从Spring Boot 1.4开始,您可以创建一个ObjectMapperSpring bean,Spring Boot将为ContextResolver您创建,并使用您的ObjectMapper
// in your `@Configuration` file.
@Bean
public ObjectMapper mapper() {}
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