[英]Scala: ObjectMapper not able to map JSON to member class
I have json file with content: 我有内容的json文件:
{
"ruleName": "rule1",
"steps": [{
"stepIdentifer": "SI1"
}, {
"stepIdentifer": "SI2"
}]
}
I am trying to map it to scala class (Rule) using following code: 我正在尝试使用以下代码将其映射到scala类(Rule):
import java.io.FileInputStream
import com.fasterxml.jackson.module.scala.DefaultScalaModule
import com.fasterxml.jackson.databind.{DeserializationFeature, ObjectMapper}
import com.fasterxml.jackson.module.scala.experimental.ScalaObjectMapper
import com.google.gson.GsonBuilder
def main(args: Array[String]): Unit = {
val file:String = "<file_path>";
val stream = new FileInputStream(file)
val mapper = new ObjectMapper with ScalaObjectMapper
mapper.registerModule(DefaultScalaModule)
mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
val rule: Rule = mapper.readValue[Rule](stream)
val gson = new GsonBuilder().setPrettyPrinting().create()
println(gson.toJson(rule)) // PRINT_STATEMENT
}
Output from print statement is: 打印语句的输出为:
{
"ruleName": "rule1",
"steps": {}
}
Json file contains "steps", but in output, it is not mapped with member class RuleStep. Json文件包含“步骤”,但是在输出中,它没有与成员类RuleStep映射。
Scala Class definition of Rule class is as follow: 规则类的Scala类定义如下:
class Rule {
var ruleName: String = null;
var steps:List[RuleStep] = null;
}
Scala Class definition of RuleStep class is as follow: RuleStep类的Scala类定义如下:
class RuleStep {
var stepIdentifer: String = null
}
I am not able to understand what I missed? 我无法理解我错过了什么? What should I do to match member class (RuleStep) with nested Json (attribute key: "steps")?
我该怎么做才能将成员类(RuleStep)与嵌套的Json(属性键:“ steps”)匹配?
Versions: 版本:
Scala = 2.11
libraryDependencies += "com.google.code.gson" % "gson" % "2.6.2"
libraryDependencies += "com.fasterxml.jackson.core" % "jackson-databind" % "2.6.2"
libraryDependencies += "com.fasterxml.jackson.core" % "jackson-core" % "2.6.2"
Probably Gson doesn't work with Scala classes properly. Gson可能无法正确处理Scala类。 There was a similar problem .
出现了类似的问题 。 But
mapper.writeValueAsString(rule)
works well and returns: 但是
mapper.writeValueAsString(rule)
可以正常工作并返回:
{"ruleName":"rule1","steps":[{"stepIdentifer":"SI1"},{"stepIdentifer":"SI2"}]}
Also you can use other JSON libraries which are more convenient to use like spray-json or even circe which is based on functional paradigm 您还可以使用其他更方便使用的JSON库,例如spray-json甚至基于函数范式的circe
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