在正则表达式匹配中包含以下行

Question

I'm trying to set up a regular expression in PHP so that it matches both single and multiple lines. What I'm trying to do with the data below is match all of those entries that are within the 1p range.

[24-Jun-2021 09:15:24 America/New_York] One line data here
[24-Jun-2021 09:15:27 America/New_York] One line data here
[24-Jun-2021 09:15:31 America/New_York] One line data here
[24-Jun-2021 13:49:21 America/New_York] One line data here
[24-Jun-2021 13:49:27 America/New_York] One line data here
[24-Jun-2021 13:49:27 America/New_York] One line data here
[24-Jun-2021 13:49:28 America/New_York] Start of multi-line data here
multi-line data here
multi-line data here
multi-line data here
multi-line data here
multi-line data here
multi-line data here
end of multi-line data here
[24-Jun-2021 13:51:16 America/New_York] One line data here
[24-Jun-2021 14:51:25 America/New_York] One line data here

The regex I'm using (and it could probably be written better as is but my regex-fu is weak) is

/(\\[24-Jun-2021\\s13:.+)\\n\\[\\d\\d/mi

(again, I'm specifically looking for that date/time and is why I have the date and time hard coded in the regex). And the line of php I'm using is

preg_match_all('/(\[24-Jun-2021\s13:.+)\n\[\d\d/mi', $string, $array);

That pattern has no problem matching all of those entries that are only "One line data here". But what I'm needing is so that it matches everything for the 1p hour, including the entirety of the multi-line entry. I've tried a variety of different patterns but they either resulted in either no matches at all or still just those single line matches. I've tried adding the s modifier to the end of the pattern (ie, /mis ) but all that does is match everything from the first 1p entry all the way down to the end of the string and that's definitely not what I'm wanting.

I've been beating my head against the wall for several hours. I've been searching to try to find something that might help but I keep coming up empty. I'm hoping that someone knows how to do this.

thnx,
Christoph

Answer 1

Using /s will make the dot match a newline. In the pattern that you posted /(\\[24-Jun-2021\\s13:.+?)\\n\\[\\d\\d/mis it will capture as least as possible lines in group 1 and then match \\n\\[\\d\\d .

Because it matches that at the start of the string, it will not match the multi-line data here part, see this example .

You might overcome this issue using a positive lookahead (?=\\n\\[\\d\\d) , asserting that part on the next line instead of matching it.

\[24-Jun-2021\s13:.+?(?=\n\[\d\d)

See a regex demo

The line in php (without the /m flag as there are no anchors in the pattern):

preg_match_all('/\[24-Jun-2021\s13:.+?(?=\n\[\d\d)/is', $string, $array);

---

As you want to start the match with the specific date followed by all lines that do not start with a date pattern (and using a non greedy quantifier .*? increases backtracking), I would suggest a pattern to match all following lines that to not start with \\n\\[\\d\\d .

As the specific date is at the start if the string, you can add an anchor ^ and the /m flag.

^\[24-Jun-2021\s13:.+(?:\R(?!\[\d\d).*)*

The pattern matches:

• ^ Start of string
• \\[24-Jun-2021\\s13:.+ Match the specific date and the rest of the line
• (?: Non capture group
  • \\R(?!\\[\\d\\d) Match a newline and assert not [ and 2 digits directly to the right
  • .* If that is the case, match the whole line
• )* Close non capture group, and optionally repeat to match all lines

Regex demo | Php demo

The line in php (without the /s flag):

preg_match_all('/^\[24-Jun-2021\s13:.+(?:\R(?!\[\d\d).*)*/im', $string, $array);