Java 使用 Jackson 对 Scala 案例类进行（反）序列化
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Using Hymanson to (De)-serialize a Scala Case Class
提问by Kevin Meredith
I tested out the serialization of a Scala case class using Hymanson.
我使用 Hymanson 测试了 Scala 案例类的序列化。
DeserializeTest.java
反序列化测试.java
public static void main(String[] args) throws Exception { // being lazy to catch-all
final ObjectMapper mapper = new ObjectMapper();
final ByteArrayOutputStream stream = new ByteArrayOutputStream();
mapper.writeValue(stream, p.Foo.personInstance());
System.out.println("result:" + stream.toString());
}
}
Foo.scala
富斯卡
object Foo {
case class Person(name: String, age: Int, hobbies: Option[String])
val personInstance = Person("foo", 555, Some("things"))
val PERSON_JSON = """ { "name": "Foo", "age": 555 } """
}
When I ran the above mainof the Java class, an exception was thrown:
当我运行上面main的Java类时，抛出了一个异常：
[error] Exception in thread "main" org.codehaus.Hymanson.map.JsonMappingException:
No serializer found for class p.Foo$Person and no properties discovered
to create BeanSerializer (to avoid exception,
disable SerializationConfig.Feature.FAIL_ON_EMPTY_BEANS) )
How can I (de)-serialize Scala case classes?
我如何（反）序列化 Scala 案例类？
采纳答案by Lionel Port
Hymanson is expecting your class to be a JavaBean, which means its expects the class to have a getX() and/or setX() for every property.
Hymanson 希望您的类是 JavaBean，这意味着它希望该类的每个属性都有一个 getX() 和/或 setX()。
Option 1
选项1
You can create JavaBean classes in Scala using the annotation BeanProperty.
您可以使用注释BeanProperty在 Scala 中创建 JavaBean 类。
Example
例子
case class Person(
@BeanProperty val name: String,
@BeanProperty val age: Int,
@BeanProperty val hobbies: Option[String]
)
In this case a val will mean only a getter is defined. If you want setters for deserialization you defined the properties as var.
在这种情况下，val 意味着只定义了一个 getter。如果你想要反序列化的 setter，你可以将属性定义为 var。
Option 2
选项 2
While option 1 will work, if you really want to use Hymanson there are wrappers that allow it to deal with Scala classes like FasterXML's scala modulewhich might be a better approach. I haven't used it as I've just been using the Json library built in to play.
虽然选项 1 可以工作，但如果您真的想使用 Hymanson，则可以使用包装器来处理 Scala 类，例如FasterXML 的 scala 模块，这可能是更好的方法。我没有使用它，因为我只是使用内置的 Json 库来播放。
回答by Priyank Desai
Found a solution that works with Hymanson and scala case classes.
找到了一个适用于 Hymanson 和 scala case 类的解决方案。
I used a scala module for Hymanson - Hymanson-module-scala.
我为 Hymanson 使用了一个 scala 模块 - Hymanson-module-scala。
libraryDependencies ++= Seq(
"com.fasterxml.Hymanson.core" % "Hymanson-databind" % "2.5.3",
"com.fasterxml.Hymanson.module" %% "Hymanson-module-scala" % "2.2.2"
)
I had to annotate fields in my case class with @JsonProperty.
我必须用@JsonProperty 注释我的案例类中的字段。
This is what my case class looks like:
这是我的案例类的样子：
case class Person(@JsonProperty("FName") FName: String, @JsonProperty("LName") LName: String)
And this is how I deserialize:
这就是我反序列化的方式：
val objectMapper = new ObjectMapper() with ScalaObjectMapper
objectMapper.registerModule(DefaultScalaModule)
val str = """{"FName":"Mad", "LName": "Max"}"""
val name:Person = objectMapper.readValue[Person](str)
Serialization is easier:
序列化更容易：
val out = new ByteArrayOutputStream()
objectMapper.writeValue(out, name)
val json = out.toString
Would like to clarify that I am using
想澄清一下我正在使用
com.fasterxml.Hymanson.databind.ObjectMapper
In the question, it seems he is using
在问题中，他似乎正在使用
org.codehaus.Hymanson.map.ObjectMapper
which won't work with ScalaObjectMapper.
这不适用于 ScalaObjectMapper。
回答by faisal00813
Based on Priyank Desai's answer I have created a generic function to convert json stringto case class
根据 Priyank Desai 的回答，我创建了一个通用函数来转换json string为case class
import com.fasterxml.Hymanson.annotation.JsonProperty
import com.fasterxml.Hymanson.databind.ObjectMapper
import com.fasterxml.Hymanson.module.scala.DefaultScalaModule
import com.fasterxml.Hymanson.module.scala.experimental.ScalaObjectMapper
def jsonToType[T](json:String)(implicit m: Manifest[T]) :T = {
val objectMapper = new ObjectMapper() with ScalaObjectMapper
objectMapper.registerModule(DefaultScalaModule)
objectMapper.readValue[T](json)
}
Usage:
用法：
case class Person(@JsonProperty("name") Name:String, @JsonProperty("age") Age:Int)
val personName = jsonToType[Person](jsonString).name
回答by Keshav Lodhi
I have created a generic function to convert
JSON String to Case Class/ObjectandCase Class/Object to JSON String.Please find a working and detailed answer which I have provided using generics here.
我创建了一个通用函数来转换
JSON String to Case Class/Object和Case Class/Object to JSON String。请在此处找到我使用泛型提供的有效且详细的答案。
