我想将包含在“”中的所有文件夹包含在将在软件包中创建的文件夹中,该文件夹将被称为“公共”。
更明确地说,我希望树结构采用以下形式:
mypackage.jar
| common
| library
| public
| layout
| application
| rss
| config
| package.properties
| .htaccess
目前,该交付件具有以下形式:
mypackage.jar
| library
| public
| layout
| application
| rss
| config
| package.properties
| .htaccess
我不知道如何执行此操作,这似乎很简单。抱歉,这是我第一次完成Maven脚本。
非常感谢您的时间:)
这是我的代码:
<build>
<directory>target</directory>
<finalName>${project.artifactId}-${project.version}</finalName>
<outputDirectory>${project.build.directory}/pdc</outputDirectory>
<resources>
<resource>
<directory>${project.basedir}</directory>
<includes>
<include>library/</include>
<include>public/</include>
<include>layout/</include>
<include>application/</include>
<include>rss/</include>
<include>config/include.php</include>
<include>config/constant.php</include>
<include>.htaccess</include>
<include>package.properties</include>
</includes>
</resource>
</resources>
<plugins>
<plugin>
<artifactId>maven-compiler-plugin</artifactId>
<version>2.3.2</version>
</plugin>
<plugin>
<groupId>org.apache.maven.plugins</groupId>
<artifactId>maven-jar-plugin</artifactId>
<version>2.4</version>
<configuration>
<outputDirectory>target</outputDirectory>
</configuration>
</plugin>
</plugins>
</build>
答案 0 :(得分:1)
如下添加targetPath
。
<project>
<build>
...
<resources>
<resource>
<targetPath>common</targetPath>
...
</resource>
</resources>
<testResources>
...
</testResources>
...
</build>
</project>