使用json4s在Scala应用中生成json [英] Producing json in a Scala app using json4s

问题描述

我正在尝试使用json4s在Scala应用程序中生成JSON.坦率地说，这是一些示例值，我将这些值汇总到我的Scalatra应用中对其进行测试:

I am attempting to produce JSON in a Scala app using json4s. Fairly straight forward, Here's some sample value I put together to test it in my Scalatra app:

import org.json4s._
import org.json4s.JsonDSL._


object JsonStub {
    val getPeople = 
        ("people" ->
            ("person_id" -> 5) ~
            ("test_count" -> 5))

}

在我的控制器中，我简单地拥有:

In my controller, I simply have:

import org.json4s._
import org.json4s.JsonDSL._
import org.json4s.{DefaultFormats, Formats}

class FooController(mongoDb: MongoClient)(implicit val swagger: Swagger) extends ApiStack with NativeJsonSupport with SwaggerSupport {

get ("/people", operation(getPeople)) {
        JsonStub.getPeople
    }   

}

但是我在浏览器中看到的输出如下:

The output I'm seeing in the browser however, is the following:

{
  "_1": "people",
  "_2": {
    "person_id": 5,
    "test_count": 5
  }
}

_1和_2键来自何处?我原本期望此输出:

Any clue where the _1 and _2 keys are coming from? I was expecting this output instead:

{
  "people":{
    "person_id": 5,
    "test_count": 5
  }
}

推荐答案

在输出中看到的是一个反射性序列化的元组，其中包含字段_1和_2.这是因为编译器为JsonStub.getPeople推断的返回类型为Tuple2[String, JObject].

What you're seeing in the output is a reflectively serialized tuple, which has fields _1 and _2. This is because the return type that the compiler has inferred for JsonStub.getPeople is Tuple2[String, JObject].

json4s DSL使用隐式转换将元组之类的值转换为JValue.但是，如果您不告诉编译器您想要JValue，它将不会应用转换.

The json4s DSL uses implicit conversions to turn values like the tuple into a JValue. But, if you don't tell the compiler you wanted a JValue, it won't apply the conversion.

理想情况下，这将导致编译错误，因为您尝试从不正确的类型中生成JSON.不幸的是，由于您的Web框架假定您要回退到基于反射的序列化，因此这意味着有另一种方法可以将元组转换为JSON，这不是您想要的.

Ideally, this would result in a compile error, because you tried to produce JSON from something that isn't the right type. Unfortunately, because your web framework assumes you want to fall back to reflection-based serialization, it means there is another way to turn the tuple into JSON, which isn't what you wanted.

如果您明确告诉编译器您需要JValue而不是Tuple2，则DSL的隐式转换将在正确的位置应用.

If you explicitly tell the compiler that you want a JValue and not a Tuple2, the DSL's implicit conversion will be applied in the correct place.

val getPeople: JValue = 
    ("people" ->
        ("person_id" -> 5) ~
        ("test_count" -> 5))

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