如何在命令行中定义复杂的Maven属性 [英] How to define complex Maven properties in the comand line

问题描述

我正在使用Maven cobertura插件进行检索我的单元测试代码涵盖了这一点，我正在通过命令行使用它:

I'm using the Maven cobertura plugin to retrieve my unit test code covering and I'm using it through the command line:

mvn cobertura:cobertura

我想做的是从命令行配置排除项.如您从官方文档中看到的，我们可以配置一个 仪器用户属性.

What I would like to do is configure the exclusions from command line. As you can see from the official documentation, we can configure an instrumentation user property.

此Instrumentation Configuration对象具有以下结构:

This Instrumentation Configuration object has the below structure:

<instrumentation>
  <excludes>
    <exclude>com/example/dullcode/**/*.class</exclude>
  </excludes>
</instrumentation>

有什么方法可以仅使用命令行形式的类似上述配置复杂对象

Is there any way to configure a complex object like the above using only the command line in the form of

-Dcobertura.instrumentation.excludes.<something>=com/example/dullcode/**/*.class

?

推荐答案

否，您不能在命令行上定义一个复杂的参数.但是，您可以实施一个技巧来实现此目的:定义一个在命令行中覆盖的Maven属性.

No, you can't define a complex parameter on the command line. But you can implement a trick to make this work: define a Maven property that you override on the command-line.

您可以使用以下方式配置插件:

You can configure the plugin with:

<instrumentation>
  <excludes>
    <exclude>${cobertura.instrumentation.exclude}</exclude>
  </excludes>
</instrumentation>

然后在命令行上具有

-Dcobertura.instrumentation.exclude=com/example/dullcode/**/*.class

将正确排除那些类.而且，如果您未指定system属性，则不会排除任何内容.

will correctly exclude those classes. And if you don't specify the system property, nothing will be excluded.

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