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  • It works, thank you. Is there a way to keep line contains $date_pattern? Cause it removes 01/Jan/2025:00:00 and keep values. I want to keep this date too Commented Sep 26 at 10:34
  • @Aleksey: sed "0,\~$date_pattern~{\~$date_pattern~p;d}" seems to work. Commented Sep 26 at 12:42
  • ... or bring the last command outside the quotes and shell escape the ! like sed "0,\~$date_pattern~"{//\!d} file ... OP seems to be using GNU sed which uses the tilde to set action for every N line starting from a certain line number like seq 6 | sed '0~3s/./x/' i.e. first~step so expecting 0~N and not 0,~ ... hence the error reported in the OP invalid usage of line address 0 ... // is the last match. Commented Sep 26 at 12:59
  • @Raffa: No, they use ~ instead of / for an address, not the first~step form. That's why there's a comma after the zero. Commented Sep 26 at 14:09
  • 1
    That $in = 1 if $d + 30 * 24 * 60 * 60 > localtime doesn't set $in back to 0 when a timestamp is found for earlier dates. So you might as well skip the time parsing after the first timestamp from 30 days ago has been seen. Or use $in = $d + 30 * 24 * 60 * 60 > localtime Commented Oct 3 at 8:51