我有一个函数可以将Kafka主题中的XML读取为字符串格式,然后再将其转换为JSON对象。
示例XML:
<Example>
<Object>
<Member1><![CDATA[]]</Member1>
<Member2><![CDATA[someText]]</Member2>
</Object>
</Example>然后我有了POJO类,比如:(使用lombok getter/setter/toString和jackson注解导入)
@Getter
@Setter
@ToString
@JacksonXMLRootElement(localName = "Example")
public class TXML {
@JacksonXmlProperty(localName = "Object")
private someObject object;
}@Getter
@Setter
@ToString
public class someObject {
@JacksonXmlProperty(localName = "Member1")
private String member1;
@JacksomXmlProperty(localName = "Member2")
private String member2;
}然后,我使用MappingJackson2XmlHttpMessageConverter获取对象映射器,并使用它将XML字符串映射到一个示例类。
@Autowired ObjectMapper xmlMapper;
@Autowired
private MappingJackson2XmlHttpMessageConverter xmlConverter;
...
xmlMapper = xmlConverter.getObjectMapper();
Example example = xmlMapper.readValue(xmlString, Example.class);这将产生一个具有以下内容的示例Class:
Example
-> object
-> member1 : ""
-> member2 : "someText"我希望member1为空,而不是空字符串。我如何才能做到这一点呢?
发布于 2020-10-17 06:59:56
您需要实现您自己的com.fasterxml.jackson.databind.util.Converter,它将在反序列化之后但在设置为所需的形式之前转换您的值。
示例实现:
class CDATAConverter implements Converter<String, String> {
@Override
public String convert(String value) {
return (value == null || value.length() == 0) ? null : value;
}
@Override
public JavaType getInputType(TypeFactory typeFactory) {
return typeFactory.constructType(String.class);
}
@Override
public JavaType getOutputType(TypeFactory typeFactory) {
return typeFactory.constructType(String.class);
}
}您需要指示Jackson使用它:
@Getter
@Setter
@ToString
class SomeObject {
@JsonDeserialize(converter = CDATAConverter.class)
@JacksonXmlProperty(localName = "Member1")
private String member1;
@JsonDeserialize(converter = CDATAConverter.class)
@JacksonXmlProperty(localName = "Member2")
private String member2;
}https://stackoverflow.com/questions/64394193
复制相似问题