bash 正则表达式匹配行尾
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Regex to match the end of line
提问by steve
I am looking for BASH regex to pull the 'db' agruments from the below commands. The order of the arguments is not guaranteed however. For some reason I cannot get it to work completely.
我正在寻找 BASH 正则表达式来从以下命令中提取“db”参数。但是，不能保证参数的顺序。出于某种原因，我无法让它完全工作。
What I have so far
到目前为止我所拥有的
regex="--db (.*)($| --)"
[[ $@ =~ $regex ]]
DB_NAMES="${BASH_REMATCH[1]}"
# These are example lines
somecommand --db myDB --conf /var/home # should get "myDB"
somecommand --db myDB anotherDB manymoreDB --conf /home # should get "myDB anotherDB manymoreDB"
somecommand --db myDB # should get "myDB"
somecommand --db myDB anotherDB # should get "myDB anotherDB"
Any suggestion on the regex?
关于正则表达式的任何建议？
回答by axiac
The problem is that bashuses a flavor of regexthat does not include non-greedy repetition operators (*?, +?). Because *is greedy and there is no way to tell it to not be greedy, the first parenthesized subexpression ((.*)) matches everything up to the end of line.
问题是bash使用regex不包括非贪婪重复运算符 ( *?, +?)的风味。因为*is greedy 并且没有办法告诉它不贪婪，所以第一个带括号的子表达式 ( (.*)) 匹配到行尾的所有内容。
You can work around this if you know for that the values you want to capture do not contain a certain character and replace .with the character class that excludes that character.
如果您知道要捕获的值不包含特定字符并替换.为排除该字符的字符类，则可以解决此问题。
For example, if the values after --dbdo not contain dashes (-) you can use this regex:
例如，如果后面的值--db不包含破折号 ( -)，您可以使用regex：
regex='--db ([^-]*)($| --)'
It matches all the examples posted in the question.
它匹配问题中发布的所有示例。
回答by Martin Konecny
The following works:
以下工作：
regex="--db[[:space:]]([[:alnum:][:space:]]+)([[:space:]]--|$)"
[[ "$@" =~ $regex ]]
There were two issues:
有两个问题：
- Character classes such as [:space:] should be used to represent whitespace
(.*)is greedy and will go as far as your last--literal. Since bash doesn't support non-greedy matching, we have to match using[[:alnum:][:space:]]which will guarantee we stop at the next--.
- 应该使用 [:space:] 等字符类来表示空格
(.*)是贪婪的，并且会达到你最后的--文字。由于 bash 不支持非贪婪匹配，我们必须匹配 using[[:alnum:][:space:]]这将保证我们在下一个--.
回答by Downgoat
By default, RegEx tries to get the most matches possible, use a non-greedy (lazy)quantifier. You might also want to put --first so the engine will use that first
默认情况下，RegEx 尝试获得尽可能多的匹配，使用非贪婪（惰性）量词。您可能还想--放在第一位，以便引擎首先使用它
--db[[:space:]](.*?)([[:space:]]--|$)
Demo
演示
如果您不想要
----，则可以使用非捕获组--db[[:space:]](.*?)(?:[[:space:]]--|$)
^^ Notice the ?:
Demo
演示
回答by KatonahMike
I think you want to match on non-space characters to catch the first grouping:
我认为您想匹配非空格字符以捕获第一个分组：
regex="--db (\S+)( --|$)"
