Maven解压缩以忽略原始文件夹

我的一个项目具有这样的结构:

One of my projects has a structure like this:

.\ca\<modulename>\dist\<modulename>.zip

我想制作一个仅包含zip文件的软件包，因此，如果我有子模块mod1，mod2，modn，我将得到:

I want to make a package that just has the zip files in a single package, so if I have submodules mod1, mod2, modn, I'll end up with:

ZIP CONTENT:
  mod1.zip
  mod2.zip
  ...
  modn.zip

到目前为止，我所做的是创建一个maven项目，并使用maven dependency和assembly插件达到了这一目的:

What I've done so far is create a maven project and, using maven dependency and assembly plugins I got to this:

<plugins>

            <plugin>
                <groupId>org.apache.maven.plugins</groupId>
                <artifactId>maven-dependency-plugin</artifactId>
                <executions>
                    <execution>
                        <id>unpack</id>
                        <phase>generate-sources</phase>
                        <goals>
                            <goal>unpack</goal>
                        </goals>

                        <configuration>
                            <artifactItems>
                                <artifactItem>
                                    <groupId>es.xunta.emprego.cntxes</groupId>
                                    <artifactId>myproject-cas</artifactId>
                                    <version>${cas.version}</version>
                                    <overWrite>true</overWrite>
                                    <outputDirectory>target/dependency</outputDirectory>
                                    <type>zip</type>
                                    <includes>ca/**/dist/*.zip</includes>
                                </artifactItem>
                            </artifactItems>
                        </configuration>
                    </execution>
                </executions>
            </plugin>


            <plugin>
                <artifactId>maven-assembly-plugin</artifactId>
                <configuration>
                    <descriptor>src/main/assembly/zip.xml</descriptor>
                    <appendAssemblyId>false</appendAssemblyId>
                </configuration>
                <executions>
                    <execution>
                        <id>make-assembly</id>
                        <!-- this is used for inheritance merges -->
                        <phase>package</phase>
                        <!-- append to the packaging phase. -->
                        <goals>
                            <goal>single</goal>
                        </goals>
                    </execution>
                </executions>
            </plugin>
        </plugins>

此为assembly/zip.xml:

<?xml version="1.0" encoding="UTF-8" ?>
<assembly>
    <id>bin</id>
    <formats>
        <!-- formato de salida del empaquetado -->
        <format>zip</format>
    </formats>
    <includeBaseDirectory>false</includeBaseDirectory>
    <fileSets>
        <fileSet>
            <outputDirectory>/</outputDirectory>
            <directory>target/dependency</directory>
            <includes>
                <include>/**/dist/*.zip</include>
            </includes>
            <useDefaultExcludes>true</useDefaultExcludes>
        </fileSet>
    </fileSets>
</assembly>  

但这尊重我原始zip的文件夹结构，因此结果是:

But this is respecting my original zip's folder structure, so the result is:

ZIP CONTENT
   ca\mod1\dist\mod1.zip
   ca\mod2\dist\mod2.zip
   ...
   ca\modn\dist\modn.zip

这似乎是个小问题，但是模块的数量很大，而且必须浏览每个文件夹非常麻烦.我一直在为assembly和dependency苦苦挣扎，但还没有找到实现我想要的方法的方法.请注意，我正在使用通配符(ca/**/dist/*.zip)，因为在编译之前我不知道将存在的文件名.

It might seem a small issue but the number of modules is big and gathering the different zip files is very annoying having to browse through every folder. I've been struggling with assembly and dependency but haven't found a way to achieve what I want. Note that I'm using wildcards (ca/**/dist/*.zip) because I don't know prior to compilation the name of the files that will be there.

有什么需要帮助的吗?

Any help please?