访问对象Scala中的属性 [英] Accessing attributes in objects Scala

问题描述

我需要从json中获取的对象类型的示例可以在以下示例（src）中找到：

An example of the sort of objects I need to grab from my json can be found in the following example(src):

{
  "test": {
    "attra": "2017-10-12T11:17:52.971Z",
    "attrb": "2017-10-12T11:20:58.374Z"
  },
  "dummyCheck": false,
  "type": "object",
  "ruleOne": {
    "default": 2557
  },
  "ruleTwo": {
    "default": 2557
  }
}

从上面的例子中我想访问ruleOne下的默认值。
我已经尝试过以下几种不同的事情，但我似乎在苦苦挣扎。我可以抓住像dummyCheck这样的值。键入我需要去的地方的最佳方法是什么？
如何获取以下值的示例：

From the example above I want to access the default value under "ruleOne". I've tried messing about with several different things below but I seem to be struggling. I can grab values like "dummyCheck" ok. What's the best way to key into where I need to go? Example of how I am trying to get the value below:

import org.json4s._
import org.json4s.native.JsonMethods._
import org.json4s.DefaultFormats

implicit val formats = DefaultFormats
val test = parse(src)
println((test \ "ruleOne.default").extract[Integer])

编辑：
进一步扩展上述内容：

To further extend what is above:

  def extractData(data: java.io.File) = {
    val json = parse(data)
    val result = (json \ "ruleOne" \ "default").extract[Int]
    result
  }

如果我将上述内容扩展为通过传入调用的函数：

If I was to extend the above into a function that is called by passing in:

extractData(src)

这只会给我RuleOne.default ..是否有一种方法可以扩展它，以便我可以动态地将多个字符串参数传递给parse（如splat）

That would only ever give me RuleOne.default.. is there a way I could extend it so that I could dynamically pass it multiple string arguments to parse (like a splat)

  def extractData(data: java.io.File, path: String*) = {
    val json = parse(data)
    val result = (json \ path: _*).extract[Int]
    result
  }

因此消耗它就像

extractData(src, "ruleOne", "default")

推荐答案

这适用于json4s-jackson％3.6.0-M2，但它应该与 native 后端完全相同。

This here works with "json4s-jackson" % "3.6.0-M2", but it should work in exactly the same way with native backend.

val src = """
  |{
  |  "test": {
  |    "attra": "2017-10-12T11:17:52.971Z",
  |    "attrb": "2017-10-12T11:20:58.374Z"
  |  },
  |  "dummyCheck": false,
  |  "type": "object",
  |  "ruleOne": {
  |    "default": 2557
  |  },
  |  "ruleTwo": {
  |    "default": 2557
  |  }
  |}""".stripMargin

import org.json4s._
import org.json4s.jackson.JsonMethods._
import org.json4s.DefaultFormats

implicit val formats = DefaultFormats
val test = parse(src)
println((test \ "ruleOne" \ "default").extract[Int])

输出：

2557

使其与本地，只需替换

import org.json4s.jackson.JsonMethods._

by

import org.json4s.native.JsonMethods._

并确保您拥有正确的依赖关系。

and make sure that you have the right dependencies.

编辑

这是 vararg 将字符串参数转换为路径的方法：

Here is a vararg method that transforms string parameters into a path:

def extract(json: JValue, path: String*): Int = {
  path.foldLeft(json)(_ \ _).extract[Int]
}

有了这个，你现在可以这样做：

With this, you can now do:

println(extract(test, "ruleOne", "default"))
println(extract(test, "ruleTwo", "default"))

请注意，它接受 JValue ，而不是文件，因为带有 File 的版本会不必要地测试，而 JValue -version可以用解析后的字符串常量进行测试。

Note that it accepts a JValue, not a File, because the version with File would be unnecessarily painful to test, whereas JValue-version can be tested with parsed string constants.

这篇关于访问对象Scala中的属性的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！