bash 如何从正则表达式中获取所有匹配项？
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How to get all matches from regex?
提问by michal kralik
I would like to get all occurrences of [0-9A-Z]+?for processing later.
I do have
我想得到所有出现的[0-9A-Z]+?以供稍后处理。我有
if [[ `cat file` =~ '[0-9A-Z]+?' ]]; then
echo $BASH_REMATCH;
fi
Which gives me first match, but how could I process all the matches in the file?
这给了我第一场比赛，但我如何处理文件中的所有比赛？
Thank you
谢谢
采纳答案by Ian Kelling
If you want to get just the matched text of a regular expression, I would use
如果您只想获取正则表达式的匹配文本，我会使用
grep -o 'regex' file
In the spirit of your code I would modify it to be
本着您的代码精神，我会将其修改为
while read line; do
[[ $line =~ regex ]] || continue
# do somethinw with $line or $BASH_REMATCH, perhaps put it in an array.
done < file
If you want to match multiple regexes on the same line, here is a way.
如果你想在同一行匹配多个正则表达式，这里有一个方法。
while read line; do
# do somethinw with $line
done < <(grep -o 'regex' file)
I assume your regex is supposed to be a simplified example of what your really doing. The ? is not helpfull and your quotes are matching a literal string.
我认为您的正则表达式应该是您真正在做什么的简化示例。这 ？没有帮助，并且您的引号与文字字符串匹配。
