Question

我有一个超级pom：

<resources>
  <resource>
    <filtering>true</filtering>
    <directory>(path to my project)\src\main\resources</directory>
    <includes>
      <include>**/application*.yml</include>
      <include>**/application*.properties</include>
    </includes>
  </resource>
  <resource>
    <directory>(path to my project)\src\main\resources</directory>
    <excludes>
      <exclude>**/application*.yml</exclude>
      <exclude>**/application*.properties</exclude>
    </excludes>
  </resource>
</resources>

案例是：

当我将下面的代码放在我的项目的pom中时，会自动覆盖super pom配置：

<build>
    <resources>
        <resource>
            <filtering>true</filtering>
            <directory>my-project/src/main/resources</directory>
            <excludes>
                <exclude>*.yml</exclude>
            </excludes>
        </resource>
    </resources>
</build>

我如何使用super pom配置和＆＃34; concat＆＃34;我的项目是：

<resources>
  <resource>
    <filtering>true</filtering>
    <directory>(path to my project)\src\main\resources</directory>
    <includes>
      <include>**/application*.yml</include>
      <include>**/application*.properties</include>
    </includes>
  </resource>
  <resource>
    <directory>(path to my project)\src\main\resources</directory>
    <excludes>
      <exclude>**/application*.yml</exclude>
      <exclude>*.yml</exclude> <!-- my project pom config -->
      <exclude>**/application*.properties</exclude>
    </excludes>
  </resource>
</resources>

Answer 1

当我学习Maven时，我也注意到子模块中定义的任何资源优先于父POM中的资源。让他们联合起来的唯一方法就是让他们联合起来。如上所述是在子节点中重复父配置，然后像您一样添加新选项。这是Maven 3.2，不确定以后的版本是否有更好的处理方法。 Plugins have an explicit way to say 'merge'，请参阅combine.children和相关属性。

如何在maven中继承构建选项？

1 个答案: