如何打印文件集到一个文件，每行一个文件名？ [英] How can I print a fileset to a file, one file name per line?

问题描述

我有一个人口稠密的文件集，我需要打印的文件名匹配到一个文本文件中。

我试过这样：

 ＆LT;文件集ID =myfilesetDIR =../声音＆GT;
    ＆LT;包括姓名=* WAV。/＆GT;
    ＆LT;包括姓名=* OGG。/＆GT;
＆LT; /文件集＆GT;＆LT;属性名=声音REFID =myfileset/＆GT;
＆LT;回声文件=sounds.txt＆GT; $ {声音}＆LT; /回声＆GT;
 

它打印在一行中的所有文件，用分号隔开。我需要有每行一个文件。我怎么能做到这一点，而不诉诸调用操作系统命令或编写Java code？

更新

嗯，本来应该更具体的 - 列表中不能包含目录。我标志着ChssPly76作为公认的答案，无论如何，因为 pathconvert 的命令是我错过了什么。剥去目录和只列出文件名，我用了扁平化映射。

下面是我最终的脚本：

 ＆LT;文件集ID =sounds_filesetDIR =../声＆GT;
    ＆LT;包括姓名=* WAV。/＆GT;
    ＆LT;包括姓名=* OGG。/＆GT;
＆LT; /文件集＆GT;＆所述; pathconvert pathsep =与＆amp; #xA;财产=声音REFID =sounds_fileset＆GT;
    ＆LT;映射器类型=扁平化/＆GT;
＆LT; / pathconvert＆GT;＆LT;回声文件=sounds.txt＆GT; $ {声音}＆LT; /回声＆GT;
 

解决方案

使用 PathConvert 任务：

 ＆LT;文件集ID =myfilesetDIR =../声音＆GT;
    ＆LT;包括姓名=* WAV。/＆GT;
    ＆LT;包括姓名=* OGG。/＆GT;
＆LT; /文件集＆GT;＆LT; pathconvert pathsep =$ {} line.separator属性=声音REFID =myfileset＆GT;
    ＆LT;！ - 如果你想要的路径剥离添加这一点 - ＆GT;
    ＆所述;映射器＆GT;
        ＆LT; flattenmapper /＆GT;
    ＆LT; /映射器＆GT;
＆LT; / pathconvert＆GT;
＆LT;回声文件=sounds.txt＆GT; $ {声音}＆LT; /回声＆GT;
 

I have a populated fileset and I need to print the matching filenames into a text file.

I tried this:

<fileset id="myfileset" dir="../sounds">
    <include name="*.wav" />
    <include name="*.ogg" />
</fileset>

<property name="sounds" refid="myfileset" />
<echo file="sounds.txt">${sounds}</echo>

which prints all the files on a single line, separated by semicolons. I need to have one file per line. How can I do this without resorting to calling OS commands or writing Java code?

UPDATE:

Ah, should have been more specific - the list must not contain directories. I'm marking ChssPly76's as the accepted answer anyway, since the pathconvert command was exactly what I was missing. To strip the directories and list only the filenames, I used the "flatten" mapper.

Here is the script that I ended up with:

<fileset id="sounds_fileset" dir="../sound">
    <include name="*.wav" />
    <include name="*.ogg" />
</fileset>

<pathconvert pathsep="&#xA;" property="sounds" refid="sounds_fileset">
    <mapper type="flatten" />
</pathconvert>

<echo file="sounds.txt">${sounds}</echo>

解决方案

Use the PathConvert task:

<fileset id="myfileset" dir="../sounds">
    <include name="*.wav" />
    <include name="*.ogg" />
</fileset>

<pathconvert pathsep="${line.separator}" property="sounds" refid="myfileset">
    <!-- Add this if you want the path stripped -->
    <mapper>
        <flattenmapper />
    </mapper>
</pathconvert>
<echo file="sounds.txt">${sounds}</echo>

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