Maven 插件:在打包过程中操作资源
在我的 Maven 项目中,我在资源中有一个 xml 文件。根据某些输入参数,我希望在打包到 jar 或 war 之前对文件进行调整。当然,原始文件是不能动的。
创建多个 xml 文件并选择合适的文件不是一种选择,例如,对于 spring 配置文件,因为 xml 文件中可能存在多种内容组合。
所以,我想到创建一个 Maven 插件,在打包之前操作文件。也许,当maven将文件复制到目标文件夹时但在maven将文件打包到jar/war之前,我需要操作该文件。
@Mojo(name = "manipulate-xml", defaultPhase = LifecyclePhase.PREPARE_PACKAGE)
public class MyMojo extends AbstractMojo {
@Parameter(defaultValue = "${project}", required = true, readonly = true)
MavenProject project;
@Parameter(property = "option")
String option;
public void execute() throws MojoExecutionException {
if (option.equals("optionA")) {
// get file from target and manipulate
} else if (option.equals("optionB")) {
// get file from target and manipulate
}
}
}
然后,我可以将 Maven 插件嵌入到我的项目中并使用构建该项目
mvn clean package -Doption=optionA
但是,现在我陷入了困境。我不知道如何从目标获取文件,即使这是正确的方法。
另外,是否可以在打包过程中阻止某些依赖项被打包到jar/war中?
我很感激任何帮助。
in my maven project, I've got a xml file in resources. Depending on some input parameter I want the file to be adapted before packaged into a jar or war. Of course, the original file shall not be touched.
It is not an option to create multiple xml-files and select a suitable one, for example, with spring profiles as there can be numerous combinations of contents in the xml file.
So, I thought of creating a maven plugin, that manipulates the file before packaging. Probably, I need to manipulate the file, when maven has copied the file to the target folder but before maven packages the file into the jar/war.
@Mojo(name = "manipulate-xml", defaultPhase = LifecyclePhase.PREPARE_PACKAGE)
public class MyMojo extends AbstractMojo {
@Parameter(defaultValue = "${project}", required = true, readonly = true)
MavenProject project;
@Parameter(property = "option")
String option;
public void execute() throws MojoExecutionException {
if (option.equals("optionA")) {
// get file from target and manipulate
} else if (option.equals("optionB")) {
// get file from target and manipulate
}
}
}
Then, I could embedded the maven plugin into my project and build the project with
mvn clean package -Doption=optionA
However, now I am stuck. I do not know, how to get the file from target and even if this is the right approach.
Besides, is it possible during the packaging to prevent some dependencies from being packaged into the jar/war?
I appreciate any help.
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根据操作的含义,您可以使用 Maven 资源插件的可能性 (https://maven.apache.org/plugins/maven-resources-plugin/index.html)。
如果您需要修改 xml 中的一些简单值,请使用 xml 中的属性并让资源插件在构建期间替换它们。构建的值可以位于 pom.xml 中,也可以通过
-Dproperty=value
提供给 maven。如果您想选择不同的文件,请定义多个 Maven 配置文件,在每个配置文件中您可以配置资源插件以仅复制所需的文件,然后在构建中选择正确的配置文件。
如果内置的可能性还不够,您甚至可以为资源插件编写自己的过滤器,这可能比编写自定义的成熟 Maven 插件更容易。
Depending on what manipulating means, you can use the possibilities of the maven resources plugin (https://maven.apache.org/plugins/maven-resources-plugin/index.html).
If you need to modify some simple values inside the xml, use properties in the xml and let the resources plugin replace them during build. The values for the build can be either in the pom.xml or given to maven via
-Dproperty=value
.If you want to select a different files, define multiple maven profiles, in each you can configure the resources plugin to copy only the wanted files and then select the correct profile in the build.
If the built-in possibilities are not enough, you might even program your own filter for the resources plugin, that might be easier than writing a custom full fledged maven plugin.