Jackson Scala JSON serialize/deserialize 元组

Question

下面我只是尝试使用 jackson-module-scala 进行 Tuple2 的往返。使用 Scala 2.10.4。为什么不起作用？

可以看出，序列化程序选择将 Tuple2 编码为 JSON 数组。为什么解串器不能将 JSON 数组解码回 Tuple2？

wget http://central.maven.org/maven2/com/fasterxml/jackson/module/jackson-module-scala_2.10/2.4.4/jackson-module-scala_2.10-2.4.4.jar
wget http://central.maven.org/maven2/com/google/guava/guava/15.0/guava-15.0.jar
wget http://central.maven.org/maven2/com/fasterxml/jackson/core/jackson-core/2.4.4/jackson-core-2.4.4.jar
wget http://central.maven.org/maven2/com/fasterxml/jackson/core/jackson-databind/2.4.4/jackson-databind-2.4.4.jar
wget http://central.maven.org/maven2/com/fasterxml/jackson/core/jackson-annotations/2.4.4/jackson-annotations-2.4.4.jar

scala -cp jackson-module-scala_2.10-2.4.4.jar:guava-15.0.jar:jackson-core-2.4.4.jar:jackson-databind-2.4.4.jar:jackson-annotations-2.4.4.jar

val mapper = new com.fasterxml.jackson.databind.ObjectMapper()
mapper.registerModule(com.fasterxml.jackson.module.scala.DefaultScalaModule)
val writer = new java.io.StringWriter()
mapper.writeValue(writer, (1,2))
mapper.readValue(writer.toString,classOf[Tuple2[Integer,Integer]])

com.fasterxml.jackson.databind.JsonMappingException: Can not deserialize instance of scala.Tuple2 out of VALUE_NUMBER_INT token
    at [Source: [1,2]; line: 1, column: 2]
    at com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:148)
    at com.fasterxml.jackson.databind.DeserializationContext.mappingException(DeserializationContext.java:762)
    at com.fasterxml.jackson.databind.DeserializationContext.mappingException(DeserializationContext.java:758)
    at com.fasterxml.jackson.module.scala.deser.TupleDeserializer.deserialize(TupleDeserializerModule.scala:61)
    at com.fasterxml.jackson.module.scala.deser.TupleDeserializer.deserialize(TupleDeserializerModule.scala:15)
    at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:3066)
    at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2161)

Answer 1

查看 DeserializerTest 特征似乎需要将 TypeReference 传递给 readValue 方法。这应该有效：

mapper.readValue[(Int,Int)](writer.toString, new TypeReference[(Int,Int)]{})

Jackson Scala JSON serialize/deserialize 元组

Jackson Scala JSON serialize/deserialize Tuple

serialization

json

scala

tuples

jackson