蚂蚁创建文件集，从（绝对值）的属性dirsets [英] Ant create filesets, dirsets from (absolute) properties
问题描述
在蚂蚁，我已经定义了定义一些路径（绝对）需要在我的编译系统。
一些属性大多数的Ant任务需要的文件集，因此建立从属性文件集我必须做的：
＆LT;文件集DIR =包括=$ {} myprop/＆GT;
问题是， myprop 是绝对的，我应该忽略dir属性（这是不可能的），这样的有一种方法来定义（绝对）性能并有效地使用它们来创建文件集，dirsets等。或在蚂蚁的这是一个不好的做法（只是相对的路径应使用） <？ / p>
感谢。
有是关于绝对与相对与蚂蚁没有特别的规定。你应该避免硬编码绝对路径，如果可能的话 - 让你的构建更便携。如果必须使用绝对路径，构建的某些部分，试图建立他们 - 从 $ {BASEDIR} 项目的说 - 这样的变化更容易使
两件事情，可能会转换绝对路径相对路径用于文件集之类的是有用的：
-
属性任务支持的属性=相对，可用于从一个绝对路径转换为相对的。 -
pathconvert任务可以使用做同样的，当有一个属性几个绝对路径。
例如：
＆LT;！ - 绝对路径 - ＆GT;
＆LT;属性名=myprop值=$ {BASEDIR} / X/＆GT;
＆LT;属性名=myprop2值=$ {} BASEDIR / X：$ {BASEDIR} / Y/＆GT;＆LT;！ - 转换单个路径相对于BASEDIR - ＆GT;
＆LT;属性名=myprop_rel位置=$ {} myprop相对=YES/＆GT;＆LT;！ - 经路径/ pathconvert去多路径属性 - ＆GT;
＆LT;路径ID =mypath2PATH =$ {} myprop2/＆GT;
＆所述; pathconvert属性=myprop2_relREFID =mypath2pathsep =，＆GT;
＆LT;从地图=$ {BASEDIR} /到=/＆GT;
＆LT; / pathconvert＆GT;＆LT;文件集ID =my_filesetDIR =。包括=$ {} myprop2_rel/＆GT;
的在回应评论：的我不知道一个简单的方法来获得在蚂蚁文件集的最长的共同preFIX目录，
但这里有一个宏观包装脚本任务已经做了你可以称之为收缩包装。
您能够在输入文件集的引用调用它，和属性的名称来设置。
＆LT; macrodef NAME =共preFIX-dir的＆GT;
＆LT;属性名=REFID/＆GT;
＆LT;属性名=outputproperty/＆GT;
＆LT;＆连续GT;
＆LT; SCRIPT LANGUAGE =JavaScript的＆GT;＆LT;！[CDATA [ srcFiles = project.getReference（@ {} REFID）
.getDirectoryScanner（项目）
.getIncludedFiles（）; 如果（srcFiles.length大于0）{
preFIX =+ srcFiles [0];
对于（i = 0; I＆LT; srcFiles.length;我++）{
而（prefix.length＆放大器;＆安培; srcFiles [I] .substr（0，prefix.length）= preFIX！）{
preFIX = prefix.substr（0，prefix.length - 1）;
} 如果（！prefix.length）{
打破;
}
}
}
其他{
preFIX =;
}
preFIX = prefix.substring（0，prefix.lastIndexOf（'/'））;
project.setProperty（@ {} outputproperty，preFIX）; ]]＆GT;＆LT; / SCRIPT＆GT;
＆LT; /顺序＆GT;
＆LT; / macrodef＆GT;
（这是从<派生href=\"http://stackoverflow.com/questions/1336207/finding-common-$p$pfix-of-array-of-strings/1336357#1336357\">php实施。）
您可以用类似尝试一下：
＆LT;属性名=my.dir位置=...您的目录在这里....../＆GT;
＆LT;文件集ID =my.fsDIR =$ {} my.dir＆GT;
＆LT;！ - 在此定义的文件集 - ＆GT;
＆LT; /文件集＆GT;
＆LT;回声消息=输入文件集：$ {的toString：my.fs}'/＆GT;＆LT;共preFIX-DIR REFID =my.fsoutputproperty =preFIX/＆GT;
＆LT;回声消息=最长公共preFIX是：$ {preFIX}'/＆GT;
的 pathconvert 任务现在可以用来创建一个收缩包装文件集：
＆LT; pathconvert财产=缩水REFID =my.fspathsep =，＆GT;
＆LT;从地图=$ {my.dir} / $ {preFIX} /到=/＆GT;
＆LT; / pathconvert＆GT;
＆LT;文件集ID =my.fs.outDIR =$ {my.dir} / $ {preFIX}包括=$ {}缩水/＆GT;
＆LT;回声消息=收缩包装的文件集：$ {的toString：my.fs.out}'/＆GT;
In ant, I have defined some properties that define some paths (absolute) needed in my build system.
Most of ant tasks need filesets, so to build a fileset from a property I would have to do:
<fileset dir="" includes="${myprop}"/>
The problem is that myprop is absolute and I should omit the dir attribute (which is not possible), so is there a way to define (absolute) properties and efficiently use them to create filesets, dirsets, etc.. or in ant is this a bad practice (only relative paths should be used)?
Thanks.
There's no particular rules about absolute versus relative with Ant. You should avoid hard-coding absolute paths if possible - to make your build more portable. If you have to use absolute paths for some part of the build, try to construct them - say from the ${basedir} of the project - so that changes are easier to make.
Two things that may be useful for converting absolute paths to relative paths for use in filesets and the like:
- the
propertytask supports an attributerelative=that can be used to convert from an absolute path to relative. - the
pathconverttask can be used to do the same, when there are several absolute paths in one property.
Examples:
<!-- Absolute paths -->
<property name="myprop" value="${basedir}/x" />
<property name="myprop2" value="${basedir}/x:${basedir}/y" />
<!-- Convert single path to relative to basedir -->
<property name="myprop_rel" location="${myprop}" relative="yes"/>
<!-- Go via path/pathconvert for the property with multiple paths -->
<path id="mypath2" path="${myprop2}" />
<pathconvert property="myprop2_rel" refid="mypath2" pathsep=",">
<map from="${basedir}/" to="" />
</pathconvert>
<fileset id="my_fileset" dir="." includes="${myprop2_rel}" />
In response to a comment: I'm not aware of a simple way to get the longest common prefix directory of a fileset in Ant, but here's a macro wrapping a script task that does what you might call 'shrink wrapping'. You call it with a reference to the 'input' fileset, and the name of the property to set.
<macrodef name="common-prefix-dir">
<attribute name="refid" />
<attribute name="outputproperty" />
<sequential>
<script language="javascript"><![CDATA[
srcFiles = project.getReference( "@{refid}" )
.getDirectoryScanner( project )
.getIncludedFiles( );
if ( srcFiles.length > 0 ) {
prefix = "" + srcFiles[0];
for ( i = 0; i < srcFiles.length; i++ ) {
while ( prefix.length && srcFiles[i].substr( 0, prefix.length ) != prefix ) {
prefix = prefix.substr( 0, prefix.length - 1);
}
if ( !prefix.length ) {
break;
}
}
}
else {
prefix = "";
}
prefix = prefix.substring( 0, prefix.lastIndexOf( '/' ) );
project.setProperty( "@{outputproperty}", prefix );
]]></script>
</sequential>
</macrodef>
(This was derived from a php implementation.)
You can try it out with something like:
<property name="my.dir" location="...your dir here..." />
<fileset id="my.fs" dir="${my.dir}">
<!-- define fileset here -->
</fileset>
<echo message="Input fileset is: '${toString:my.fs}'" />
<common-prefix-dir refid="my.fs" outputproperty="prefix" />
<echo message="Longest common prefix is: '${prefix}'" />
The pathconvert task can now be used to create a shrink-wrapped fileset:
<pathconvert property="shrink" refid="my.fs" pathsep=",">
<map from="${my.dir}/${prefix}/" to="" />
</pathconvert>
<fileset id="my.fs.out" dir="${my.dir}/${prefix}" includes="${shrink}" />
<echo message="Shrink-wrapped fileset is: '${toString:my.fs.out}'" />
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