做到从UNIX命令行基本的XML解析最简单的方法 [英] Simplest way to do basic xml parsing from unix command line

问题描述

我在寻找某种属性的XML文件。例如，包含以下模式的文件：

I'm searching for xml files that have certain properties. For example, files that contain the following pattern:

<param-value>
  <name>Hosts</name>
  <description>some description</description>
  <value></value>
</param-value>

有关这样的文件，我想分析的另一个变量的值，如：

For such files, I'd like to parse the value of another tag, such as:

<param-value>
  <name>Roles</name>
  <description>some description</description>
  <value>asdf</value>
</param-value>

和打印出来的文件名以ASDF一起。什么是命令行实现这一点最简单的方法？

And print out the file name along with "asdf". What's the simplest way to accomplish this from the command line?

一种方法我想只是使用grep的-l选项匹配的文件过滤掉，然后使用的xargs grep来提取角色的价值。然而，grep的不能很好与多行正则表达式工作。我看到了另一个问题，那表明它可能与-Pzo选项来完成，但没有任何运气得到它在我的情况下工作。有没有更简单的方法呢？

One approach I was thinking of was just using grep with the -l option to filter the matching files out, and then using xargs grep to extract the value of Roles. However, grep doesn't work well with multi-line regexes. I saw another question that showed it could be done with the -Pzo options, but didn't have any luck getting it to work in my case. Is there a simpler approach?

推荐答案

下面的Linux命令使用XPath在XML文件中访问指定的值

The following linux command uses XPath to access specified values within the XML file

for xml in `find . -name "*.xml"`
do  
echo $xml `xmllint --xpath "/param-value/value/text()" $xml`| awk 'NF>1'
done

示例输出匹配的XML文件：

Example output for matching XML files:

./test1.xml asdf
./test4.xml 1234

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