如何使用 shell 脚本提取 xml 属性? [英] How do I extract xml properties using shell scripts?

问题描述

我已经使用 grep 从源文档中提取了标签，但是，现在我似乎无法弄清楚如何轻松地从字符串中提取属性.此外，我想避免使用标准安装中通常不存在的任何程序.

I have already extracted the tag from the source document using grep but, now I cant seem to figure out how to easily extract the properties from the string. Also I want to avoid having to use any programs that would not usually be present on a standard installation.

$tag='<img src="http://imgs.xkcd.com/comics/barrel_cropped_(1).jpg" title="Don't we all." alt="Barrel - Part 1" />'  

我需要得到以下变量

$src="http://imgs.xkcd.com/comics/barrel_cropped_(1).jpg"
$title="Don't we all."
$alt="Barrel - Part 1"

推荐答案

您可以使用 xmlstarlet.然后，您甚至不必自己提取元素:

You can use xmlstarlet. Then, you don't even have to extract the element yourself:

$ echo $tag|xmlstarlet sel -t --value-of '//img/@src'
http://imgs.xkcd.com/comics/barrel_cropped_(1).jpg

你甚至可以把它变成一个函数

You can even turn this into a function

$ get_attribute() {
  echo $1 | xmlstarlet sel -t -o """ -v $2 -o """
  }
$ src=get_attribute $tag '//img/@src'

如果不想多次重新解析文档，也可以这样做:

If you don't want to reparse the document several times, you can also do:

$ get_values() {
   eval file=\${$#}
   eval $#=    
   cmd="xmlstarlet sel "
   for arg in $@
   do
      if [ -n $arg ]
      then
        var=${arg%%\=*}
        expr=${arg#*=}
        cmd+=" -t -o \"$var="\" -v $expr -o \""\" -n"
      fi
   done
   eval $cmd $file
  }
$ eval $(get_values src='//img/@src' title='//img/@title' your_file.xml)
$ echo $src
http://imgs.xkcd.com/comics/barrel_cropped_(1).jpg
$ echo $title
Don't we all.

我确信有更好的方法来删除 shell 函数的最后一个参数，但我不知道.

I'm sure there's a better way to remove the last argument to a shell function, but I don't know it.

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