来自 Jackson 的 kotlin 的自定义 json 序列化 java 原语
Custom json serialization java primitives from kotlin by Jackson
在我的项目中有 java 和 kotlin 模块。在某些 kotlin 模块中,我需要自定义序列化。所有长值都应序列化为字符串。我用杰克逊。
示例:
科特林
data class KotlinRecord(val id: Long)
java
public class JavaRecord {
private Long id;
public Long getId() {
return id;
}
public JavaRecord setId(Long id) {
this.id = id;
return this;
}
}
当 ObjectMapper 在 java 模块中配置时,我像这样序列化值:
ObjectMapper mapper = new ObjectMapper()
.registerModule(new KotlinModule())
.registerModule(new JavaTimeModule())
.registerModule(
new SimpleModule()
.addSerializer(Long.class, LongToStringJSONSerializer.ofObject())
.addSerializer(long.class, LongToStringJSONSerializer.ofPrimitive())
);
JavaRecord javaRecord = new JavaRecord().setId(Long.MAX_VALUE);
KotlinRecord kotlinRecord = new KotlinRecord(Long.MAX_VALUE);
System.out.println("Java record: " + mapper.writeValueAsString(javaRecord));
System.out.println("Kotlin record: " + mapper.writeValueAsString(kotlinRecord));
我得到以下结果:
Java record: {"id":"9223372036854775807"}
Kotlin record: {"id":"9223372036854775807"}
没关系。这就是我需要的。
但是如果我像这样在 kotlin 模块中做同样的事情:
val mapper = ObjectMapper()
.registerModule(KotlinModule())
.registerModule(JavaTimeModule())
.registerModule(
SimpleModule()
.addSerializer(Long::class.java, LongToStringJSONSerializer.ofObject())
.addSerializer(Long::class.javaPrimitiveType, LongToStringJSONSerializer.ofPrimitive())
)
val javaRecord = JavaRecord().setId(Long.MAX_VALUE)
val kotlinRecord = KotlinRecord(Long.MAX_VALUE)
println("Java record: ${mapper.writeValueAsString(javaRecord)}")
println("Kotlin record: ${mapper.writeValueAsString(kotlinRecord)}")
我得到:
Java record: {"id":9223372036854775807}
Kotlin record: {"id":"9223372036854775807"}
对于定义为 java class 的记录,长值未转换为字符串。
我的自定义序列化器:
public class LongToStringJSONSerializer extends JsonSerializer<Long> {
private boolean forPrimitive;
private LongToStringJSONSerializer(boolean forPrimitive) {
this.forPrimitive = forPrimitive;
}
@Override
public void serialize(Long longVal, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws
IOException {
jsonGenerator.writeObject(longVal == null ? "" : String.valueOf(longVal));
}
@Override
public Class<Long> handledType() {
return forPrimitive ? long.class : Long.class;
}
public static LongToStringJSONSerializer ofPrimitive() {
return new LongToStringJSONSerializer(true);
}
public static LongToStringJSONSerializer ofObject() {
return new LongToStringJSONSerializer(false);
}
}
任何人都可以解释这种行为以及需要采取什么措施来解决它?谢谢!
问题出在错误的序列化类型上。
对象映射器应该这样定义:
val mapper = ObjectMapper()
.registerModule(KotlinModule())
.registerModule(JavaTimeModule())
.registerModule(
SimpleModule()
.addSerializer(Long::class.javaPrimitiveType, LongToStringJSONSerializer.ofObject())
.addSerializer(Long::class.javaObjectType, LongToStringJSONSerializer.ofPrimitive())
)
我错过了Long::class.javaObjectType
在我的项目中有 java 和 kotlin 模块。在某些 kotlin 模块中,我需要自定义序列化。所有长值都应序列化为字符串。我用杰克逊。
示例:
科特林
data class KotlinRecord(val id: Long)
java
public class JavaRecord {
private Long id;
public Long getId() {
return id;
}
public JavaRecord setId(Long id) {
this.id = id;
return this;
}
}
当 ObjectMapper 在 java 模块中配置时,我像这样序列化值:
ObjectMapper mapper = new ObjectMapper()
.registerModule(new KotlinModule())
.registerModule(new JavaTimeModule())
.registerModule(
new SimpleModule()
.addSerializer(Long.class, LongToStringJSONSerializer.ofObject())
.addSerializer(long.class, LongToStringJSONSerializer.ofPrimitive())
);
JavaRecord javaRecord = new JavaRecord().setId(Long.MAX_VALUE);
KotlinRecord kotlinRecord = new KotlinRecord(Long.MAX_VALUE);
System.out.println("Java record: " + mapper.writeValueAsString(javaRecord));
System.out.println("Kotlin record: " + mapper.writeValueAsString(kotlinRecord));
我得到以下结果:
Java record: {"id":"9223372036854775807"}
Kotlin record: {"id":"9223372036854775807"}
没关系。这就是我需要的。
但是如果我像这样在 kotlin 模块中做同样的事情:
val mapper = ObjectMapper()
.registerModule(KotlinModule())
.registerModule(JavaTimeModule())
.registerModule(
SimpleModule()
.addSerializer(Long::class.java, LongToStringJSONSerializer.ofObject())
.addSerializer(Long::class.javaPrimitiveType, LongToStringJSONSerializer.ofPrimitive())
)
val javaRecord = JavaRecord().setId(Long.MAX_VALUE)
val kotlinRecord = KotlinRecord(Long.MAX_VALUE)
println("Java record: ${mapper.writeValueAsString(javaRecord)}")
println("Kotlin record: ${mapper.writeValueAsString(kotlinRecord)}")
我得到:
Java record: {"id":9223372036854775807}
Kotlin record: {"id":"9223372036854775807"}
对于定义为 java class 的记录,长值未转换为字符串。
我的自定义序列化器:
public class LongToStringJSONSerializer extends JsonSerializer<Long> {
private boolean forPrimitive;
private LongToStringJSONSerializer(boolean forPrimitive) {
this.forPrimitive = forPrimitive;
}
@Override
public void serialize(Long longVal, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws
IOException {
jsonGenerator.writeObject(longVal == null ? "" : String.valueOf(longVal));
}
@Override
public Class<Long> handledType() {
return forPrimitive ? long.class : Long.class;
}
public static LongToStringJSONSerializer ofPrimitive() {
return new LongToStringJSONSerializer(true);
}
public static LongToStringJSONSerializer ofObject() {
return new LongToStringJSONSerializer(false);
}
}
任何人都可以解释这种行为以及需要采取什么措施来解决它?谢谢!
问题出在错误的序列化类型上。
对象映射器应该这样定义:
val mapper = ObjectMapper()
.registerModule(KotlinModule())
.registerModule(JavaTimeModule())
.registerModule(
SimpleModule()
.addSerializer(Long::class.javaPrimitiveType, LongToStringJSONSerializer.ofObject())
.addSerializer(Long::class.javaObjectType, LongToStringJSONSerializer.ofPrimitive())
)
我错过了Long::class.javaObjectType