为 JSON 集类型定义 `Reads`

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问题描述

如何为我的 People 案例类创建一个 play.api.libs.Reads?

How can I create a play.api.libs.Reads for my People case class?

scala> type Id = Long
defined type alias Id

scala> case class People(names: Set[Id])
defined class People

scala>   implicit val PeopleReads: Reads[People] = (
     |     (__  "names").read[Set[Id]])(People)
<console>:21: error: overloaded method value read with alternatives:
  (t: Set[Id])play.api.libs.json.Reads[Set[Id]] <and>
  (implicit r: play.api.libs.json.Reads[Set[Id]])play.api.libs.json.Reads[Set[Id]]
 cannot be applied to (People.type)
           (__  "names").read[Set[Id]])(People)

推荐答案

(...)(People) 语法专为构建参数列表(好吧，从技术上讲，它是一个带有 and 的 Builder，而不是一个列表)，并且想要将 People 构造函数提升到 Reads 以便您可以将其应用于这些参数.

The (...)(People) syntax is designed for when you've built up a list of arguments (well, technically it's a Builder, not a list) with and and want to lift the People constructor into the applicative functor for Reads so that you can apply it to those arguments.

例如，如果您的 People 类型如下所示:

For example, if your People type looked like this:

case class People(names: Set[Id], home: String)

你可以写:

implicit val PeopleReads: Reads[People] = (
  (__  "names").read[Set[Id]] and
  (__  "home").read[String]
)(People)

不过，在您的情况下，People 的构造函数只有一个参数，而您没有使用和，因此您没有 Builder[Reads[Set[Id] ~ String]，你刚刚得到了一个普通的Reads[Set[Id]].

In your case, though, the constructor for People has a single argument, and you haven't used and, so you don't have a Builder[Reads[Set[Id] ~ String], you've just got a plain old Reads[Set[Id]].

这很好，因为这意味着你不需要奇怪的应用函子语法——你只需要map:

This is nice, because it means you don't need the weird applicative functor syntax—all you need is map:

implicit val PeopleReads = (__  "names").read[Set[Id]].map(People)

大功告成.

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