引言:因为我记得在几年前回答了这个问题后，我经常提到这个问题，展示了一个动态访问Maven POM信息的动态版本(例如，也在测试期间)，今天我发现了一个类似的问题，涉及从另一个模块B访问模块a的Maven信息。

I thought about it for a moment and spontaneously had the idea to use a special annotation, applying it to a package declaration in package-info.java. I also created a multi-module example project on GitHub. I do not want to repeat the whole answer, so please see solution B in this answer. The Maven setup involves Templating Maven Plugin, but could also be solved in a more verbose way using a combination of resource filtering and adding generated sources directory to the build via Build Helper Maven. I wanted to avoid that, so I simply used Templating Maven.

在使用spring引导时，这个链接可能有用:https://docs.spring.io/spring-boot/docs/2.3.x/reference/html/howto.html#howto-properties-and-configuration

使用spring-boot-starter-parent，你只需要在你的应用配置文件中添加以下内容:

# get values from pom.xml
pom.version=@project.version@

之后的值是这样的:

@Value("${pom.version}")
private String pomVersion;

为了补充@kieste所发布的内容，我认为如果使用Spring-boot，这是在代码中提供Maven构建信息的最佳方式:http://docs.spring.io/spring-boot/docs/current/reference/htmlsingle/#production-ready-application-info上的文档非常有用。

您只需要激活执行器，并在应用程序中添加所需的属性。属性或application.yml

Automatic property expansion using Maven

You can automatically expand info properties from the Maven project using resource filtering. If you use the spring-boot-starter-parent you can then refer to your Maven ‘project properties’ via @..@ placeholders, e.g.

project.artifactId=myproject
project.name=Demo
project.version=X.X.X.X
project.description=Demo project for info endpoint
info.build.artifact=@project.artifactId@
info.build.name=@project.name@
info.build.description=@project.description@
info.build.version=@project.version@

假设你正在使用Java，你可以:

Create a .properties file in (most commonly) your src/main/resources directory (but in step 4 you could tell it to look elsewhere). Set the value of some property in your .properties file using the standard Maven property for project version: foo.bar=${project.version} In your Java code, load the value from the properties file as a resource from the classpath (google for copious examples of how to do this, but here's an example for starters). In Maven, enable resource filtering. This will cause Maven to copy that file into your output classes and translate the resource during that copy, interpreting the property. You can find some info here but you mostly just do this in your pom: <build> <resources> <resource> <directory>src/main/resources</directory> <filtering>true</filtering> </resource> </resources> </build>

你还可以获取其他标准属性，如project.name, project.description，甚至是你放在pom <properties>中的任意属性，等等。资源过滤与Maven概要文件相结合，可以在构建时为您提供可变的构建行为。当您在运行时使用-PmyProfile指定配置文件时，可以启用属性，然后可以在构建中显示这些属性。

如果使用Spring和Maven，这非常简单，不需要配置。 根据“使用Maven自动扩展属性”官方文档，您可以使用资源过滤自动扩展Maven项目中的属性。如果你使用spring-boot-starter-parent，你就可以用@.引用Maven的“项目属性”。@占位符，示例如下:

project.version=@project.version@
project.artifactId=@project.artifactId@

你可以在任何类中使用@Value注释来检索它:

@Value("${project.artifactId}@${project.version}")
private String RELEASE;

我希望这能有所帮助!

使用这个库可以获得简单的解决方案。向清单中添加您需要的任何内容，然后按字符串查询。

 System.out.println("JAR was created by " + Manifests.read("Created-By"));

http://manifests.jcabi.com/index.html