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如果 test=true，我想从检测代码中排除 cobertura。

现在，我在顶部有所有必需的陈述：

require 'buildr/java/cobertura'
require 'buildr/scala'

然后每次构建运行我都会得到这个：

使用 cobertura 数据文件 C:/usr/git_workspaces/reports/cobertura.ser 检测类

这意味着我的生产代码中有 cobertura 仪器。

这是我构建的下一部分

compile.with  CORE, SLF4J, LOG4J, WS_CLIENTS, JODA_TIME,
            [omitted for brevity]

compile.options.other = %w(-encoding UTF-8)

cobertura.exclude '[omitted for brevity]'

resources.filter.using *RESOURCES_FILTER

test.using :junit
# need this because of forking.  It does not appear to use the environmental variables defined above.
test.using :java_args => ["-XX:MaxPermSize=128M"]
test.with JUNIT, SCALATEST, MOCKITO, POWERMOCK, HAMCREST, SPRING.test

# Pakcage is below here but the code has already been instrumented...

compile.with 是实际发生编译的地方吗？所以我可以做一个 if 测试然后添加 cobertura ？

4

1 回答 1

2

如果您不需要“buildr/java/cobertura”，那么将没有检测。

我们像这样“解决”了（至少命令行参数必须包含“cobertura”，否则不会检测类）

ARGV.each do |a| 
  if (a.include?("cobertura")) 
    require "buildr/java/cobertura"
    break
  end
end

你可以这样做

require_cb = ture
ARGV.each do |a|   
    if (a.include?("test=yes"))
    require_cb = false      
    break
  end
end
require "buildr/java/cobertura" if require_cb

hth

于 2012-10-24T14:54:52.010 回答