How to match a string without considering new lines in Python Regex?

Question

I have a string here

Some text here

Maybe something here too

Work hard in silence

let your success


make noise

rest of the paragraph here with some

new lines too

My goal is to find and remove Work hard in silence let your success and also the text before it.

How do I tell regex to find in target string without taking new lines into account?

Final result should be the remaining string keeping the new lines in it.

make noise

rest of the paragraph here with some

new lines too

If you know what the string is going to be, and where the newlines are, why use regex? This is not a good usecase of regex. — FHTMitchell, Jun 15 '18 at 12:47
[`r"\s+".join([re.escape(x) for x in searchphrase.split()])`](https://ideone.com/4zDUFi) — Wiktor Stribiżew, Jun 15 '18 at 12:47
@FHTMitchell I posted an example, I am scraping a source actually, I never know where the new lines will be — Umair Ayub, Jun 15 '18 at 12:47

wiesion · Answer 1 · 2018-06-15T13:13:13.383

0

This is very basic RegExp, you need Work\s+hard\s+in\s+silence\s+let\s+your\s+success\s+make\s+noise

Your current RegExp is searching for a literal match, that means each space character must be a space character - it is not implicitly any kind of whitespace character. You need to use \s, which matches any whitespace character (equal to [\r\n\t\f\v ]). The quantifier + expects at least one whitespace.

EDIT

I am not sure if it was not in the initial requirements - but if you want to keep multiline mode active and match everything until the end of your searched sentence:

[\S\s]*Work\s+hard\s+in\s+silence\s+let\s+your\s+success\s*

edited Jun 15 '18 at 13:13

answered Jun 15 '18 at 12:50

wiesion

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what about the text before that phrase? – depperm Jun 15 '18 at 12:51
@depperm put `.*` at the beginning of the expr. – FHTMitchell Jun 15 '18 at 12:51
I know how, I was bringing to attention the OP asked and wiesion didn't address it – depperm Jun 15 '18 at 12:52
@FHTMitchell you mean `.*(Work\s+hard\s+in\s+silence\s+let\s+your\s+success\s+make\s+noise)`??? – Umair Ayub Jun 15 '18 at 12:54
@Umair oops yes -- edited – FHTMitchell Jun 15 '18 at 12:55
@FHTMitchell that doesnt match the string before the string I want to match – Umair Ayub Jun 15 '18 at 12:56
@FHTMitchell it doesnt match the lines before `Work hard in silence` – Umair Ayub Jun 15 '18 at 12:58
@FHTMitchell yeah see https://regex101.com/r/dzImVc/3 – Umair Ayub Jun 15 '18 at 12:59
Ah I see, sorry. Still half asleep. You need to *not* run in multi-line mode, you want single-line mode (`re.DOTALL`) and do `^.*` https://regex101.com/r/dzImVc/4 – FHTMitchell Jun 15 '18 at 13:02
@FHTMitchell hmm, works there but not in my code ... see https://repl.it/@UmairAyub/regextest – Umair Ayub Jun 15 '18 at 13:10
1

i see, updated my answer – wiesion Jun 15 '18 at 13:13
Does match though https://repl.it/repls/CrispSurprisedHertz – FHTMitchell Jun 15 '18 at 13:17
on a sidenote, if you wish to remove all whitespaces until the first non-whitespace character, change your code to `r"[\S\s]*" + searchphrase + r"[^\S]+"` – wiesion Jun 15 '18 at 13:24
@wiesion I have problem with regex ... See https://regex101.com/r/dzImVc/5 what if I have multiple occurrences of my to-find string? – Umair Ayub Jun 15 '18 at 15:47
That depends on what you want to achieve - You want to stop on the first occurrence? – wiesion Jun 15 '18 at 15:53
If so, just use `^.*?Songdowon\s+International\s+Children\'s\s+Camp\s*` (no need for \S\s group anymore since you turned off multiline mode) - `.` means match any character, `*` quantifier means 0 or n times, and adding `?` to the quantifier means do it non-greedy (default is greedy). – wiesion Jun 15 '18 at 16:01

How to match a string without considering new lines in Python Regex?

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