在 bash 脚本中拆分字符串
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Splitting a string in a bash script
提问by Ian Burris
Possible Duplicate:
Split string based on delimiter in bash?
可能的重复：
基于 bash 中的分隔符拆分字符串？
I have a bunch of files named test<numbers or letters>.<number>.outso like test1.1024.outor test2.2.out. Is there some way I can use a regular expression like ^test.*?..(.*).out$to parse out the middle number on each file and then be able to access the group?
我有一大堆命名的文件中test<numbers or letters>.<number>.out如此喜欢test1.1024.out或test2.2.out。有什么方法可以使用正则表达式^test.*?..(.*).out$来解析每个文件的中间数字，然后才能访问该组？
回答by rob mayoff
for f in test*.out; do
number=${f#test*.}
number=${number%.*}
echo $f has middle number $number
done
回答by Swiss
sed -r 's/^test[[:alnum:]]*\.([[:digit:]]+)\.out$//'
Alternatively (A shorter version of the BASH for loop):
或者（BASH for 循环的较短版本）：
awk -F '.' '{print }'
回答by Niall Byrne
Depending on the version of Bash this could work too:
根据 Bash 的版本，这也可以工作：
test=test1.1024.out
if [[ $test =~ ^test[A-Za-z0-9]+\.([0-9]+)\.out$ ]]; then
echo ${BASH_REMATCH[1]}
fi
