regex: how to replace all occurrences of a string within another string, if the original string matches some filter

Question

i need to replace all occurrences of a string within another string, if the original string matches some filter
i can only use a single regex using an s command, because i need to send the assembled command to a 3rd party API

i have tried to use positive lookahead as to not consume the string in which i want to replace characters, but somehow i can not get the replacing to work as expected.

here is what i have tried so far and what was the outcome: (note that the filter - here [0-9]+ is just an example and will be passed in from the call site and i can not directly influence it.

expected result: 9999997890

perl -e '$x = "4564567890"; $x =~ s/(?=^[0-9]+$)456/999/g; print $x'

actual result: 9994567890

this replaces only the first occurrence of 456. why is this happening?
even less understandable for me is that if i change the filter lookahead to (?=.*), both occurrences of 456 are being replaced. why does changing the filter have any effect on the replacing portion of the regex?

i seem to be missing some very basic point about how mixing filtering and replacing stuff in one s command works.

@WiktorStribiżew this seems to work...can you write it as an answer so i can accept it? also, would you care to explain why your regex works and mine does not? :D — Armin Walland
– Armin Walland, Commented Dec 11, 2019 at 10:27
Probably simpler to understand if you could use eg ^.*?\D.*(*SKIP)(*F)|456 to skip strings that don't contain only digits. — bobble bubble
– bobble bubble, Commented Dec 11, 2019 at 10:38

Wiktor Stribiżew · Accepted Answer · 2019-12-11 10:36:22Z

2

Your regex only replaces the 456 that is at the start of the string that only consists of digits.

You may use

s/(?:\G(?!^)|^(?=\d+$))\d*?\K456/999/g

See the regex demo

Pattern details

(?:\G(?!^)|^(?=\d+$)) - a custom boundary that matches either the end of the previous successful match (\G(?!^)) or (|) the start of string (^) that only contains digits ((?=\d+$))
\d*? - 0+ digits, but as few as possible
\K - omit the currently matched chars
456 - a 456 substring.

The idea is:

Use the \G based pattern to pre-validate the string: (?:\G(?!^)|^(?=<YOUR_VALID_LINE_FORMAT>$))
Then adjust the consuming pattern after the above one.

answered Dec 11, 2019 at 10:36

Wiktor Stribiżew

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2 Comments

justhalf Over a year ago

You can combine the non-capturing group I think: s/(?=\G(?!^)|^\d+$)\d*?\K456/999/g

Wiktor Stribiżew Over a year ago

@justhalf Yes, that is also working here, just I - personally - do not like using zero-width assertions inside zero-width assertions.

bobble bubble · Accepted Answer · 2019-12-11 11:02:01Z

1

Alternatively you can probably use (*SKIP)(*F) to skip strings not composed only of digits .

s/^\d*\D.*(*SKIP)(*F)|456/999/g

See this demo at regex101 or your demo at tio.run

The left part ^\d*\D.* tries to match any \D non digit. If found, skips .* rest of the string and fails | OR matches the specified substring 456.

answered Dec 11, 2019 at 11:02

bobble bubble

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