错误:未知参数'idea-shell'

时间:2018-05-21 13:49:48

标签: scala intellij-idea playframework sbt

我正在使用 Intellij Idea社区版。我跑的时候

new https://github.com/sbt/scala-seed.g8

new PlayFramework/play-scala-seed.g8
来自 sbt shell

它给了我以下错误:

Error: Unknown argument 'idea-shell' g8 0.7.2 Usage: giter8 [options] <template>   <template>            git or file URL, or github user/repo   -b, --branch <value>  Resolve a template within a given branch   -f, --force           Force overwrite of any existing files in output directory   --version             Display version number  
--paramname=paramval  Set given parameter value and bypass interaction EXAMPLES Apply a template from github
    g8 foundweekends/giter8 Apply using the git URL for the same template
    g8 git://github.com/foundweekends/giter8.git Apply template from a remote branch
    g8 foundweekends/giter8 -b some-branch Apply template from a local repo
    g8 file://path/to/the/repo Apply given name parameter and use defaults for all others.
    g8 foundweekends/giter8 --name=template-test

2 个答案:

答案 0 :(得分:0)

这是因为在IntelliJ Scala插件中如何实现sbt shell。此外,new命令实际上并不意味着从实时shell会话中运行。

从终端运行new

sbt new https://github.com/sbt/scala-seed.g8

答案 1 :(得分:0)

您应该从终端使用它,如: sbt new scala/scala-seed.g8

如果您仍有问题:

  • 也许你必须从一个不是SBT项目的目录(没有build.sbt

  • 运行它
  • 当然您需要安装sbt