如何在bash中從XML獲取修訂號？

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我需要從 <entry> <string>https://anonsvn.jboss.org/repos/exo-jcr/jcr/trunk</string> <long>4687</long> </entry>如何在bash中從XML獲取修訂號？

來源

2011-07-29 Roman

這是無效的xml – linuts

得到修訂（4687）如果url不會改變，並且<修訂>總是來找它，那麼你可以使用以下命令：

$ grep -A 1 '<url>https://anonsvn.jboss.org/repos/exo-jcr/jcr/trunk' test.xhtml | 
     tail -1 | 
     sed -e 's/[^0-9]//g'

來源

2011-07-29 10:38:32 holygeek

Thaaanks很多！ – Roman

的xgrep或xpath實用程序可以幫助您：

xgrep -x '/build/actions/hudson.scm.SubversionTagAction/tags/entry/hudson.scm.SubversionSCM_-SvnInfo[url="https://anonsvn.jboss.org/repos/exo-jcr/jcr/trunk"]/revision/text()' sample.xml

或者：

xpath -q -e '/build/actions/hudson.scm.SubversionTagAction/tags/entry/hudson.scm.SubversionSCM_-SvnInfo[url="https://anonsvn.jboss.org/repos/exo-jcr/jcr/trunk"]/revision/text()' sample.xml

在處理XML內容時總是使用XML解析器。如果你沒有，你可能會遇到奇怪的問題，因爲這個經典答案說明：RegEx match open tags except XHTML self-contained tags

來源

2011-07-29 10:54:21

如何在bash中從XML獲取修訂號？

回答

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