SBT：排除资源子目录 [英] SBT: Exclude resource subdirectory

问题描述

我一直在使用Gradle来处理我的大部分Scala项目，但我想评估SBT作为替代品的适用性。我在Gradle中完成的一件事是从最终构建中排除某个资源目录（例如，使用CoffeeScript编写将作为最终资源包含的JavaScript文件）。

在Gradle中，我会这样做：

  sourceSets {
 main {
资源{
 exclude'com / example / export / dev'//排除开发资源
} 
} 
} 
  pre> 
 
 
这将从最终版本中排除资源包 com.example.export.dev 包。 / p> 
 
 
我如何在SBT中做同样的事情？我已经在编译 -  =（编译中的resourceDirectory）.value /com / example / export / dev中尝试了
  unmanagedResourceDirectories in Compile  - 
  
但是这并没有做任何事情（我理解为什么，但这并没有什么帮助） 。并且SBT网站上的文档仅讨论排除文件模式（在类路径，来源，和资源）。
 
 
 
作为一个更具描述性的图片，假设我们有以下资源目录结构：
  com 
 \ ---示例
 \ --- export 
 \ --- dev 
 \ --- 
  
在最后的输出中，我想要：
 
 
  com 
 \ ---例子
 \ ---出口
 \ ---东西
  
 
 
解决方案
 From  https://github.com/sbt/sbt-jshint/issues/14 ： 
 
 < pre $  excludeFilter in unmanagedResources：= {
 val public =（（compile中的resourceDirectory）.value /com/example/export/dev）。 getCanonicalPath 
新的SimpleFileFilter（_。getCanonicalPath startsWith public）
} 
  

I've been using Gradle for most of my Scala projects, but I want to evaluate the suitability of SBT as a replacement. One of the things I've done in Gradle is to exclude a certain resource directory from the final build (for example, using CoffeeScript to write JavaScript files that will be included as final resources).

In Gradle, I'd do this by:

sourceSets {
    main {
        resources {
            exclude 'com/example/export/dev' // exclude development resources
        }
    }
}

And this would exclude the resource package com.example.export.dev package from the final build.

How would I do the same in SBT? I've tried

unmanagedResourceDirectories in Compile -= (resourceDirectory in Compile).value / "com/example/export/dev"

but that doesn't do a thing (I understand why, but that doesn't really help). And the documentation on the SBT web site only talks about excluding file patterns (at Classpaths, sources, and resources).

As a more descriptive image, say we have the following resource directory structure:

com
\---example
     \---export
         \---dev
         \---something

In the final output, I want:

com
\---example
    \---export
        \---something

解决方案

From https://github.com/sbt/sbt-jshint/issues/14:

excludeFilter in unmanagedResources := {
  val public = ((resourceDirectory in Compile).value / "com" / "example" / "export" / "dev").getCanonicalPath
  new SimpleFileFilter(_.getCanonicalPath startsWith public)
}

这篇关于SBT：排除资源子目录的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！