以编程方式获取Play中的路线!框架2.5.x

fra*_*eff 7 routing scala playframework swagger

play-swagger模块中,我们使用play.routes.compiler.RoutesFileParser来从Play获取所有已注册的路由,并进一步将它们解析为API文档:

val parsedRoutes = RoutesFileParser.parse(new File(app.classloader.getResource(routesFile).toURI))
val routes = parsedRoutes.right.get.collect {
  case (route: PlayRoute) => {
    logger.debug(s"Adding route '$route'")
    Seq(route.copy(path = route.path.copy(parts = StaticPart(prefix + "/") +: route.path.parts)))
  }
  case (include: PlayInclude) => {
    logger.debug(s"Processing route include $include")
    parseRoutesHelper(playRoutesClassNameToFileName(include.router), include.prefix)
  }
}.flatten
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这会导致以下错误: java.lang.NoClassDefFoundError: play/routes/compiler/RoutesFileParser

似乎RoutesFileParser在Play中变得私密!2.5.

获取所有Play路线的正确方法是什么,如上例所示?

col*_*ebb 1

这绝对是一个依赖问题。检查依赖树。

如果尚不存在,请将以下内容添加到构建文件中:

"com.typesafe.play" %% "routes-compiler" % "2.5.0"
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然后这将编译并运行良好:

val routesURI = app.classloader.getResource("routes").toURI
val parsedRoutes = play.routes.compiler.RoutesFileParser.parse(new File(routesURI))
println(parsedRoutes)
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(其中 app 是 play.api.Application 的运行实例)