如何在bash命令行中的perl搜索和替换中引用匹配的模式？

Question

I want to do search and replace(foo to foobar) with bash using perl command:

sudo perl -0777 -i -pe s/'foo'/'foobar'/gs a.txt

but I don't always know what is 'foo', so I want a variable kind of thing which stores the matched pattern.

Also can I get a substring of the matched pattern? Like 'foo' is replaced with 'oobar'(foo becomes oo)?

Answer 1

使用表达式评估修饰符e 。

perl -0777 -i -pe's/(foo)/substr($1, 1) . "bar"/egs' a.txt

Answer 2

$& refers to the entire matched pattern and instead of taking substring of that I used negative lookbehind (?<!u) this means any character other than u :

sudo perl -i -0777 -pe s/(?<!u)'oo'/'u$&'/gs

This will match not only any foo but any occurrence of oo but never uoo and replace it with uoobar .

Answer 3

In the example you gave, you don't need to use the matched text.

perl -pe's/foo\K/bar/g'

In the scenario you described, you don't need to use the matched text.

perl -pe's/f\Koo/oobar/g'

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That said, $& contains the matched text.

perl -pe's/foo/$&bar/g'

And $1 contains the text matched by the earliest capture, $2 contains the text matched by the second earliest capture, etc.

perl -pe's/(f)oo/$1oobar/g'

And /e can be used to treat the replacement expression as code to execute for each match.

perl -pe's/foo/ substr($&,0,1)."oobar" /eg'

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• There's no point in using /s since the pattern doesn't contain . .
• There's no point in using -0777 since your pattern can't span lines.
• The quotes you used were useless, and it's less noisy to quote the entire program instead of individual sections of it.