Hello Basically i need to extract the version info from the XML I thought of puttin the entire result to a XML file and then parse 开发者_StackOverflowit to get the revision information.
i tried to output the svn info to a xml file by specifying
$ svn info C:\Projects\Foo.xml --xml C:\Projects\FileInfo.xml
Im getting errors if i try to save
$ svn info --xml C:\Projects\Foo.xml
<?xml version="1.0"?>
<info>
<entry
kind="dir"
path="."
revision="1">
<url>https://rbins.com/trunk/Projects/Foo.xml</url>
<repository>
<root>https://rbins.com/trunk/Projects/</root>
<uuid>5e7d134a-54fb-0310-bd04-b611643e5c25</uuid>
</repository>
<wc-info>
<schedule>normal</schedule>
<depth>infinity</depth>
</wc-info>
<commit
revision="240">
<author>sally</author>
<date>2003-01-15T23:35:12.847647Z</date>
</commit>
</entry>
</info>
i do not receive error if i run
$ svn info C:\Projects\Foo.xml --xml
Help me where am i going wrong also suggest me if theres any alternative available to extract the revision number of the working copy
Thanks in advance
Use svnversion
instead:
>C:\mydir>svnversion
1652:1653
You need to redirect svn info
's output to the file, the --xml
option takes no parameter:
svn info C:\Projects\Foo.xml --xml > C:\Projects\FileInfo.xml
精彩评论