如何使用标准 Scala 类在 Scala 中解析 JSON? [英] How to parse JSON in Scala using standard Scala classes?
问题描述
我在 Scala 2.8 中使用内置的 JSON 类来解析 JSON 代码.由于最小化依赖性,我不想使用 Liftweb 之一.
I am using the build in JSON class in Scala 2.8 to parse JSON code. I don't want to use the Liftweb one or any other due to minimizing dependencies.
我这样做的方式似乎过于迫切,有没有更好的方法?
The way I am doing it seems too imperative, is there a better way to do it?
import scala.util.parsing.json._
...
val json:Option[Any] = JSON.parseFull(jsonString)
val map:Map[String,Any] = json.get.asInstanceOf[Map[String, Any]]
val languages:List[Any] = map.get("languages").get.asInstanceOf[List[Any]]
languages.foreach( langMap => {
val language:Map[String,Any] = langMap.asInstanceOf[Map[String,Any]]
val name:String = language.get("name").get.asInstanceOf[String]
val isActive:Boolean = language.get("is_active").get.asInstanceOf[Boolean]
val completeness:Double = language.get("completeness").get.asInstanceOf[Double]
}
推荐答案
这是一个基于提取器的解决方案,它将执行类转换:
This is a solution based on extractors which will do the class cast:
class CC[T] { def unapply(a:Any):Option[T] = Some(a.asInstanceOf[T]) }
object M extends CC[Map[String, Any]]
object L extends CC[List[Any]]
object S extends CC[String]
object D extends CC[Double]
object B extends CC[Boolean]
val jsonString =
"""
{
"languages": [{
"name": "English",
"is_active": true,
"completeness": 2.5
}, {
"name": "Latin",
"is_active": false,
"completeness": 0.9
}]
}
""".stripMargin
val result = for {
Some(M(map)) <- List(JSON.parseFull(jsonString))
L(languages) = map("languages")
M(language) <- languages
S(name) = language("name")
B(active) = language("is_active")
D(completeness) = language("completeness")
} yield {
(name, active, completeness)
}
assert( result == List(("English",true,2.5), ("Latin",false,0.9)))
在 for 循环开始时,我人为地将结果包装在一个列表中,以便在最后生成一个列表.然后在 for 循环的其余部分中,我使用生成器(使用 <-
)和值定义(使用 =
)将使用 unapply 方法的事实.
At the start of the for loop I artificially wrap the result in a list so that it yields a list at the end. Then in the rest of the for loop I use the fact that generators (using <-
) and value definitions (using =
) will make use of the unapply methods.
(旧答案已被删除 - 如果您好奇,请查看编辑历史记录)
(Older answer edited away - check edit history if you're curious)
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