Question

我有以下使用reactivemongo驱动程序的函数，实际上写得很好。

def writeDocument() = {
    val document = BSONDocument(
      "firstName" -> "Stephane",
      "lastName" -> "Godbillon",
      "age" -> 29)

    val future = collection.insert(document)

    future.onComplete {
      case Failure(e) => throw e
      case Success(result) => {
        println("successfully inserted document with result = " + result)
      }
    }
  }

但该函数的局限性在于JSON被硬编码到BSONDocument中。如何更改它以便我可以将任何JSON字符串传递给函数？

简而言之：如何将JSON字符串转换为BSONDocument？

更新2：

package controllers

//import play.api.libs.json._
//import reactivemongo.bson._
//import play.api.libs.json.Json

import scala.util.{Success, Failure}
import reactivemongo.api._
//import scala.concurrent.ExecutionContext.Implicits.global


import play.modules.reactivemongo.json.collection._
import reactivemongo.play.json._

object Mongo {

  //val collection = connect()

  def collection: JSONCollection = {
    val driver = new MongoDriver
    val connection = driver.connection(List("localhost"))
    val db = connection("superman")
    db.collection[JSONCollection]("IncomingRequests")
  }


  // TODO: Make this work with any JSON String
  def writeDocument() = {

    val jsonString = """{
                       | "guid": "alkshdlkasjd-ioqweuoiquew-123132",
                       | "title": "Hello-2016",
                       | "year": 2016,
                       | "action": "POST",
                       | "start": "2016-12-20",
                       | "stop": "2016-12-30"}"""


    val document = Json.parse(jsonString)
    val future = collection.insert(document)
    future.onComplete {
      case Failure(e) => throw e
      case Success(result) => {
        println("successfully inserted document with result = " + result)
      }
    }
  }

}

现在的问题是import reactivemongo.play.json._被视为未使用的导入（在我的IntelliJ上），我仍然会收到以下错误

[info] Compiling 9 Scala sources and 1 Java source to /Users/superman/target/scala-2.11/classes...
[error] /Users/superman/app/controllers/Mongo.scala:89: No Json serializer as JsObject found for type play.api.libs.json.JsValue. Try to implement an implicit OWrites or OFormat for this type.
[error] Error occurred in an application involving default arguments.
[error]     val future = collection.insert(document)
[error]                                   ^
[error] one error found
[error] (compile:compileIncremental) Compilation failed
[error] application - 

! @6oo00g47n - Internal server error, for (POST) [/validateJson] ->

play.sbt.PlayExceptions$CompilationException: Compilation error[No Json serializer as JsObject found for type play.api.libs.json.JsValue. Try to implement an implicit OWrites or OFormat for this type.
Error occurred in an application involving default arguments.]
        at play.sbt.PlayExceptions$CompilationException$.apply(PlayExceptions.scala:27) ~[na:na]
        at play.sbt.PlayExceptions$CompilationException$.apply(PlayExceptions.scala:27) ~[na:na]
        at scala.Option.map(Option.scala:145) ~[scala-library-2.11.6.jar:na]
        at play.sbt.run.PlayReload$$anonfun$taskFailureHandler$1.apply(PlayReload.scala:49) ~[na:na]
        at play.sbt.run.PlayReload$$anonfun$taskFailureHandler$1.apply(PlayReload.scala:44) ~[na:na]
        at scala.Option.map(Option.scala:145) ~[scala-library-2.11.6.jar:na]
        at play.sbt.run.PlayReload$.taskFailureHandler(PlayReload.scala:44) ~[na:na]
        at play.sbt.run.PlayReload$.compileFailure(PlayReload.scala:40) ~[na:na]
        at play.sbt.run.PlayReload$$anonfun$compile$1.apply(PlayReload.scala:17) ~[na:na]
        at play.sbt.run.PlayReload$$anonfun$compile$1.apply(PlayReload.scala:17) ~[na:na]

Answer 1

首先，您可以使用reactivemongo将模型类序列化为BSON。检查文档以了解具体方法。

如果您想通过播放json从BSONDocument制作String，可以使用

val playJson: JsValue = Json.parse(jsonString)
val bson: BSONDocument = play.modules.reactivemongo.json.BSONFormats.reads(playJson).get

修改

我在这里的文档中找到了更多：

http://reactivemongo.org/releases/0.11/documentation/tutorial/play2.html

你可以导入这两个

import reactivemongo.play.json._ import play.modules.reactivemongo.json.collection._

而不是使用默认的Collection实现（与之交互 BSON结构+ BSONReader / BSONWriter），我们使用专门的与JsObject + Reads / Writes一起使用的实现。

所以你创建这样的专门集合（必须是def，而不是val）：

def collection: JSONCollection = db.collection[JSONCollection]("persons")

从现在开始，您可以使用play json而不是BSON，因此只需将Json.parse(jsonString)作为要插入的文档传入即可。您可以在链接中看到更多示例。

编辑2 我得到你的代码编译：

包裹控制器

import play.api.libs.concurrent.Execution.Implicits._ import play.api.libs.json._ import play.modules.reactivemongo.json.collection.{JSONCollection, _} import reactivemongo.api.MongoDriver import reactivemongo.play.json._ import play.api.libs.json.Reads._ import scala.util.{Failure, Success} object Mongo { def collection: JSONCollection = { val driver = new MongoDriver val connection = driver.connection(List("localhost")) val db = connection("superman") db.collection[JSONCollection]("IncomingRequests") } def writeDocument() = { val jsonString = """{ | "guid": "alkshdlkasjd-ioqweuoiquew-123132", | "title": "Hello-2016", | "year": 2016, | "action": "POST", | "start": "2016-12-20", | "stop": "2016-12-30"}""" val document = Json.parse(jsonString).as[JsObject] val future = collection.insert(document) future.onComplete { case Failure(e) => throw e case Success(result) => println("successfully inserted document with result = " + result) } } }

重要的进口是

import play.api.libs.json.Reads._

您需要JsObject，而不仅仅是JsValue

val document = Json.parse(jsonString).as[JsObject]

Answer 2

ReactiveMongo release note说明：

使用Play JSON支持时，如果发生上一个错误，则为必须确保使用importmongo.play.json._导入导入默认的BSON / JSON转换。

如何将JSON字符串转换为BSONDocument

2 个答案: