正则表达式以特定模式开始新匹配

Question

Hello im kinda new to regex and have a small, maybe simple question.

I have the given text:

17.11.2020 15:32 typical Pat. seems sleeping
Additional test

17.11.2020 15:32 typical Pat. seems sleeping
Additional test

17.11.2020 15:32 typical Pat. seems sleeping
Additional test

My current regex (\d{2}.\d{2}.\d{4}\s\d{2}:\d{2})\s?(.*) matches only till sleeping but reates 3 matches correctly. But i need the Additional test text also in the second group. i tried something like (\d{2}.\d{2}.\d{4}\s\d{2}:\d{2})\s?([,.:\w\s]*) but now i have only one huge match because the second group takes everything until the end.

How can i match everything until a new line with a date starts and create a new match from there on?

Answer 1

If you are sure there is only one additional line to be matched you can use

(?m)^(\d{2}\.\d{2}\.\d{4}\s\d{2}:\d{2})\s*(.*(?:\n.*)?)

See the regex demo . Details:

• (?m) - a multiline modifier
• ^ - start of a line
• (\d{2}\.\d{2}\.\d{4}\s\d{2}:\d{2}) - Group 1: a datetime string
• \s* - zero or more whitespaces
• (.*(?:\n.*)?) - Group 2: any zero or more chars other than a newline char as many as possible and then an optional line, a newline followed with any zero or more chars other than a newline char as many as possible.

If there can be any amount of lines, you may consider

(?m)^(\d{2}\.\d{2}\.\d{4}[\p{Zs}\t]\d{2}:\d{2})[\p{Zs}\t]*(?s)(.*?)(?=\n\d{2}\.\d{2}\.\d{4}|\z)

See this regex demo . Here,

• (?m)^(\d{2}\.\d{2}\.\d{4}[\p{Zs}\t]\d{2}:\d{2}) - matches the same as above, just \s is replaced with [\p{Zs}\t] that only matches horizontal whitespace
• [\p{Zs}\t]* - 0+ horizontal whitespace chars
• (?s) - now, . will match any chars including a newline
• (.*?) - Group 2: any zero or more chars, as few as possible
• (?=\n\d{2}\.\d{2}\.\d{4}|\z) - up to the leftmost occurrence of a newline, followed with a date string, or up to the end of string.

Answer 2

You are using \s repeatedly using the * quantifier with the character class [,.:\w\s]* and \s also matches newlines and will match too much.

You can just match the rest of the line using (.*\r?\n.*) which would not match a newline, then match a newline and the next line in the same group.

^(\d{2}.\d{2}.\d{4}\s\d{2}:\d{2})\s?(.*\r?\n.*)

Regex demo

If multiple lines can follow, match all following lines that do not start with a date like pattern.

^(\d{2}\.\d{2}\.\d{4})\s*(.*(?:\r?\n(?!\d{2}\.\d{2}\.\d{4}).*)*)

Explanation

• ^ Start of the string
• ( Capture group1
• \d{2}\.\d{2}\.\d{4} Match a date like pattern
• ) Close group 1
• \s* Match 0+ whitespace chars (Or match whitespace chars without newlines [^\S\r\n]* )
• ( Capture group 2
  • .* Match the whole line
  • (?:\r?\n(?.\d{2}\.\d{2}\.\d{4}).*)* Optionally repeat matching the whole line if it does not start with a date like pattern
• ) Close group 2

Regex demo