java - How to break the line before a date pattern using Shell script? -


i have text file that:

13/03/2015 09:21:02 descrição de log (sessão terminada);name.lastname -- 127.0.0.1:66666 -> 0.0.0.0 -- 00:01 -- 13/03/2015 09:21:02 descrição de log (autenticação realizada com sucesso);name.lastname.....

i know how can break line every time have date. pattern dd/mm/yyyy. text like:

13/03/2015 09:21:02 descrição de log (sessão terminada);name.lastname -- 127.0.0.1:66666 -> 0.0.0.0 -- 00:01 --              13/03/2015 09:21:02 descrição de log (autenticação realizada com sucesso);name.lastname.....

can me that? can java code, sed, awk, whatever...

thanks.

with gnu sed:

sed -r 's, ([0-9]{2}/[0-9]{2}/[0-9]{4} ),\n\1,g' filename

the [0-9]{2}/[0-9]{2}/[0-9]{4} part of regex matches date pattern; whole regex matches surrounded spaces avoid false positives , because example data suggests input meets format. , used separator instead of / in s command because search regex contains slashes.


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