如何从maven build jar中排除一组包？ [英] How to exclude a set of packages from maven build jar?

问题描述

这就是我所需要的一切。其他细节：我有一个src / bootstrap / java文件夹和常规的src / main / java文件夹。出于显而易见的原因，每个人都需要去一个单独的罐子。我能够使用这个生成一个引导程序jar：

That's all I need. Additional details: I have a src/bootstrap/java folder and the regular src/main/java folder. Each one needs to go to a separate jar for obvious reasons. I was able to generate a bootstrap jar using this:

  <plugin>
    <artifactId>maven-jar-plugin</artifactId>
    <version>2.3.1</version>
    <executions>
      <execution>
        <id>only-bootstrap</id>
        <goals><goal>jar</goal></goals>
        <phase>package</phase>
        <configuration>
          <classifier>bootstrap</classifier>
          <includes>
            <include>sun/**/*</include>
          </includes>
        </configuration>
      </execution>
    </executions>
  </plugin>

但常规jar仍然包含bootstrap类。我正在使用这个答案。

But the regular jar still includes the bootstrap classes. I am compiling the bootstrap classes with this answer.

生成myproject.jar没有引导类的任何灯光？

Any light to generate a myproject.jar WITHOUT the bootstrap classes?

推荐答案

您必须使用default-jar作为ID：

You gotta use "default-jar" for the ID:

  <plugin>
    <artifactId>maven-jar-plugin</artifactId>
    <version>2.3.1</version>
    <executions>
      <execution>
        <id>only-bootstrap</id>
        <goals><goal>jar</goal></goals>
        <phase>package</phase>
        <configuration>
          <classifier>bootstrap</classifier>
          <includes>
            <include>sun/**/*</include>
          </includes>
        </configuration>
      </execution>
      <execution>
        <id>default-jar</id>
        <phase>package</phase>
        <goals>
          <goal>jar</goal>
        </goals>
        <configuration>
          <excludes>
            <exclude>sun/**/*</exclude>
          </excludes>
        </configuration>
      </execution>
    </executions>
  </plugin>

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