我一直在使用简单的副本从目录中获取一个特殊文件:
<copy todir="target/failures">
<fileset dir="target/reports" includes="**/*FAILED.txt"/>
</copy>
现在,我想将包含此文件的整个文件夹复制到失败文件夹。 目录结构如下所示:
target
reports
folder1
folder2
folder3
failures
因此,如果在folder1中发现故障,我想将整个内容复制到故障,然后继续浏览其余文件夹。看起来它应该很简单,但我似乎无法找到一个内置的任务来实现这一点,任何想法?
答案 0 :(得分:0)
使用condition来决定是否执行复制目标。
<project name="demo" default="copy">
<fileset id="failures" dir="target/reports" includes="**/*FAILED.txt"/>
<condition property="failures.found">
<resourcecount refid="failures" when="greater" count="0" />
</condition>
<target name="copy" if="failures.found">
<copy todir="target/failures" overwrite="true">
<fileset id="failures" dir="target/reports"/>
</copy>
</target>
</project>
更强大,更灵活的解决方案使用groovy ANT task:
<project name="demo" default="copy">
<path id="build.path">
<pathelement location="/path/to/task/jars/groovy-all-2.1.1.jar"/>
</path>
<target name="copy">
<taskdef name="groovy" classname="org.codehaus.groovy.ant.Groovy" classpathref="build.path"/>
<fileset id="failures" dir="target/reports" includes="**/*FAILED.txt"/>
<groovy>
project.references.failures.each {
def failFile = new File(it.toString())
def failFolder = new File(failFile.parent)
ant.copy(todir:"target/failures/${failFolder.name}", overwrite:true) {
fileset(dir:failFolder)
}
}
</groovy>
</target>
</project>