bash 两个分隔符之间的 grep 子字符串
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grep substring between two delimiters
提问by Ulrik
I have a lot of bashscripts that use perlexpressions within grepin order to extract a substring between two delimiters. Example:
我有很多bash脚本使用其中的perl表达式grep来提取两个分隔符之间的子字符串。例子：
echo BeginMiddleEnd | grep -oP '(?<=Begin).*(?=End)'
The problem is, when I ported these scripts to a platform running busybox, 'integrated' grepdoes not recognize -P switch. Is there a clean way to do this using grepand regular expressions?
问题是，当我将这些脚本移植到运行的平台时busybox，“集成”grep无法识别 -P 开关。有没有一种干净的方法可以使用grepand来做到这一点regular expressions？
Edit:
There is no perl, sedor awkon that platform. It's a lightweight linux.
编辑：该平台上没有perl,sed或awk。这是一个轻量级的linux。
回答by anubhava
You can use awkwith custom field separator like this to get same output:
您可以awk像这样使用自定义字段分隔符来获得相同的输出：
echo 'BeginMiddleEnd' | awk -F 'Begin|End' '{print }'
Middle
回答by Vytenis Bivainis
Assuming there's no more than one occurrence per line, you can use
假设每行不超过一次，您可以使用
sed -nr 's/.*Begin(.*)End.*//p'
With grep and non-greedy quantifier you could also print more than one per line.
使用 grep 和非贪婪量词，您还可以每行打印多个。
回答by Mark Setchell
Use bashbuilt-in parameter substitution:
使用bash内置参数替换：
# grab some string from grep output
f=BeginMiddleEnd
middleend=${f/Begin/} # do some substitution to lose "Begin"
echo $middleend
MiddleEnd
beginmiddle=${f%%End} # strip from right end to lose "End"
echo $beginmiddle
BeginMiddle
Loads more examples here.
在此处加载更多示例。
