如何获取ant构建xml文件&x27;她自己的名字?
假设我有一个名为如何获取ant构建xml文件&x27;她自己的名字?,ant,filenames,Ant,Filenames,假设我有一个名为build\u dev\u linux.xml的文件 我的问题是 如何找到ant脚本XML文件自己的名称,build\u dev\u linux.XML 所以我可以把它放在XML文件中的变量或属性上。?Ant定义了一个有用的内置属性列表: basedir the absolute path of the project's basedir (as set with the basedir attribute of &
build\u dev\u linux.xml
的文件
我的问题是
如何找到ant脚本XML文件自己的名称,build\u dev\u linux.XML
所以我可以把它放在XML文件中的变量或属性上。?Ant定义了一个有用的内置属性列表:
basedir the absolute path of the project's basedir (as set
with the basedir attribute of <project>).
ant.file the absolute path of the buildfile.
ant.version the version of Ant
ant.project.name the name of the project that is currently executing;
it is set in the name attribute of <project>.
ant.project.default-target
the name of the currently executing project's
default target; it is set via the default
attribute of <project>.
ant.project.invoked-targets
a comma separated list of the targets that have
been specified on the command line (the IDE,
an <ant> task ...) when invoking the current
project.
ant.java.version the JVM version Ant detected; currently it can hold
the values "1.2", "1.3",
"1.4", "1.5" and "1.6".
ant.core.lib the absolute path of the ant.jar file.
为什么不能直接设置包含名称的属性以减少处理点的次数。我不知道从哪里开始,现在我想我有了一个。:
也可能有用,请参阅
<basename file="${ant.file}" property="buildfile"/>