[英]how to access parent project's classes in sub project in playFramework 2.3.x
我正在使用play 2.3.8並使用此gudie在我的項目中創建子項目,我創建了子項目“ mySubProject”,然后在月食中將項目導入了父項目myParentProject
和mySubProject
現在我首先有兩個問題->這是否正確首先,我進口myParentProject
在日食的話,我進口mySubProject
二線>在我mySubProject
我可以訪問的類myParentProject
並導入了包,但mySubProject
當我要訪問我的類/包myParentProject
它不會讓我這樣做,它顯示錯誤Object not found
是根項目myParentProject
文件
name := """myParentProject"""
version := "1.0-SNAPSHOT"
lazy val root = (project in file(".")).enablePlugins(PlayScala)
.aggregate(mySubProject)
.dependsOn(mySubProject)
scalaVersion := "2.11.1"
fork in run := true
javaOptions in run ++= Seq("-J-Xms1G", "-J-Xmx2G")
val appDependencies = Seq(
// Add your project dependencies here,
"org.scalatestplus" %% "play" % "1.2.0" % "test"
)
lazy val mySubProject = project
libraryDependencies ++= Seq("org.scalatest" %% "scalatest" % "2.2.1" % "test"withSources() withJavadoc(),
"com.esotericsoftware.kryo" % "kryo" % "2.10",
"org.mongodb" %% "casbah" % "2.8.0",
"org.slf4j" % "slf4j-api" % "1.6.4",
"org.elasticsearch" % "elasticsearch" % "1.6.0",
"org.codehaus.groovy" % "groovy-all" % "2.4.0",
"org.apache.lucene" % "lucene-expressions" % "4.10.4",
"org.scalatest" %% "scalatest" % "2.2.1" % "test"withSources() withJavadoc(),
"org.easymock" % "easymock" % "3.1" withSources() withJavadoc(),
"org.mockito" % "mockito-all" % "1.9.5",
"com.typesafe.akka" %% "akka-actor" % "2.3.6",
"ch.qos.logback" % "logback-core" % "1.0.9",
"com.github.nscala-time" %% "nscala-time" % "2.0.0",
"net.liftweb" %% "lift-json" % "2.6+",
"net.liftweb" %% "lift-json" % "2.6+",
"com.hazelcast" % "hazelcast" % "3.5",
"com.hazelcast" % "hazelcast-client" % "3.5",
"com.twitter" % "chill-bijection_2.11" % "0.7.0"
//,"com.codahale" % "jerkson_2.9.1" % "0.5.0"
)
這是子項目mySubProject
構建文件
name := """mySubProject"""
version := "1.0-SNAPSHOT"
scalaVersion := "2.11.1"
libraryDependencies ++= Seq("org.scalatest" %% "scalatest" % "2.2.1" % "test"withSources() withJavadoc(),
"org.slf4j" % "slf4j-api" % "1.6.4",
"org.elasticsearch" % "elasticsearch" % "1.6.0",
"org.codehaus.groovy" % "groovy-all" % "2.4.0",
"org.apache.lucene" % "lucene-expressions" % "4.10.4",
"org.easymock" % "easymock" % "3.1" withSources() withJavadoc(),
"org.mockito" % "mockito-all" % "1.9.5",
"com.typesafe.akka" %% "akka-actor" % "2.3.6",
"ch.qos.logback" % "logback-core" % "1.0.9",
"com.github.nscala-time" %% "nscala-time" % "2.0.0",
"net.liftweb" %% "lift-json" % "2.6+",
"net.liftweb" %% "lift-json" % "2.6+")
這是我的父項目myParentProject
的代碼,我創建了一個名為app / myPackagae / abc.scala的類,這是代碼
package myPackagae
import mySubProject._
class abc {
def helloAbc()={
println(" i am root project and i am class abc ")
}
val test=new Testing
test.helloTesting()
}
在mySubProject
我在mySubProject
創建了一個類,這里是代碼
package mySubProject
//import parentProjectPackage._
import myPackagae._ //here is an error not found: object myPackagae
class Testing {
def helloTesting() ={
println("i am a subproject or child project and i am class Testing")
}
//and here i want to access class abc and its method helloAbc() but eclipse is not importing
}
請指導我如何在子項目中導入根項目的包/類
這是一個古老的問題,但仍被認為是一個答案。
TLDR :
通常避免將根項目導入子項目,除非您確定根項目不依賴於子項目(例如,您可以將根路由設置為子模塊,這是一個依賴項)。
詳細答案 :
以下是我關注的項目結構:
- root
- app
- controllers
- models
- modules
- moduleone
- build.sbt
- app
- controllers.moduleone
- models.moduleone
- ...
- moduletwo
- build.sbt
- app
- controllers.moduletwo
- ...
- build.sbt
以下是根項目build.sbt文件。 注意子項目的dependsOn設置,它基本上將根項目作為依賴項添加到子項目,因此您可以將根項目類導入子項目。
lazy val modelone: Project = project.in(file("modules/moduleone"))
.enablePlugins(PlayScala)
.settings(
name := "module-one",
libraryDependencies ++= common
)
.dependsOn(root)
////*******************************
//// Root module
////*******************************
val root: Project = project.in(file("."))
.enablePlugins(PlayScala)
.settings(
name := "root",
libraryDependencies ++= common
)
在子項目中,您可以執行以下操作:
package models.moduleone
import models.SOMETHING_FROM_ROOT_PROJECT
...
將根項目和子項目定義為不同的包是一個好主意,這樣可以使導入更加清晰。
還要注意,您應該避免子項目和根項目之間的依賴循環。 我見過一些項目,其中根項目依賴於子項目的某些功能。 在這種情況下,您將為根項目添加dependsOn
,但是如果您的子項目也做同樣的事情,它將無法編譯。
一個更常見的情況是,您通常最終會在根路由中這樣做:
-> /api/moduleone moduleone.Routes
這會將相關流量路由到moduleone。 現在,需要做的是首先編譯moduleone,然后在根項目中使用該路由,這意味着根build.sbt必須為:
val root: Project = project.in(file("."))
.enablePlugins(PlayScala)
.dependsOn(moduleone)
.aggregate(moduleone)
.settings(
libraryDependencies ++= common
)
如果然后將根項目導入到moduleone中,它將變成一個循環,它將永遠不會編譯。
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.