如何在scala repl中导入play

Question

如何在Scala repl中导入播放？

scala> import play.api.libs.json._
<console>:11: error: not found: value play
       import play.api.libs.json._

Answer 1

1)设置简单的构建工具(sbt){它很容易 - 从这里下载 - http://www.scala-sbt.org/download.html和说明 - http://www.scala-sbt.org/0.13/ docs/Installing-sbt-on-Windows.html }

2)创建一个build.sbt包含以下内容的空文件夹

//your-test-project/build.sbt

scalaVersion := "2.11.8"                                                                               

resolvers += "Typesafe repository" at "http://repo.typesafe.com/typesafe/releases/"                    

libraryDependencies += "com.typesafe.play" %% "play" % "2.5.12"

3)然后只需sbt console在文件夹的根目录下执行,该文件夹将下载播放并使其可用于您的控制台.

$ ls -l ~/.ivy2/cache/com.typesafe.play/play_2.11/jars/
total 15392
-rw-r--r--  1 as18  185223974  4107407 Jan 22 15:59 play_2.11-2.5.12.jar

然后你很高兴去.

scala> import play.api.libs.json._
import play.api.libs.json._

scala> val json: JsValue = Json.parse("""{ "compiler" : "scala", "ratings" : 5 }""")
json: play.api.libs.json.JsValue = {"compiler":"scala","ratings":5}

scala> val compiler = ( json \ "compiler" )
compiler: play.api.libs.json.JsLookupResult = JsDefined("scala")

此外,如果你已经拥有它,你可以直接提供jar,如下所示

scala -cp ~/.ivy2/cache/com.typesafe.play/play_2.11/jars/play_2.11-2.5.12.jar

scala> import play.api.libs._
import play.api.libs._