Print all entries in file after (and including) a specific time

Question

This is my date format in my file: [1/24/21 9:19:20:183 MYT].

I want to print all entries marked with timestamps from this moment onward to the end of the file. How can I do this using sed?

Answer 1

Do you want to only print the date or the entire line containing the date?

To print the entire line, I would use grep '\[1/24/21 9:19:20:183 MYT\]' file, but if you insist on using sed, it can be accomplished with sed -n '/\[1\/24\/21 9:19:20:183 MYT\]/p' file.

EDIT: Ugly backslashes can be avoided with grep -F, as suggested by user freddy in the first comment below.

The -n option ensures that lines are only displayed if they are explicitly printed by the sed program. The p command explicitly prints the line.

To only print the date, grep is also the better solution. You just add the -o or --only-matching option. It should also be possible with sed, but requires more skill than I have.

EDIT: OP actually has a different requirement and modified the question. You can give a line range to sed. "from the line that contains [1/24/21 9:19:20:183 MYT] to the end" is written as /\[1\/24\/21 9:19:20:183 MYT\]/,$. The dollar sign indicates the last line. So:

sed -n '/\[1\/24\/21 9:19:20:183 MYT\]/,$p' file

Answer 2

If it doesn't have to be sed, the following awk program would work:

awk -v start="[1/24/21 9:19:20:183 MYT]" 'index($0,start)==1{f=1} f' file

This will parse the file until it finds a line starting with the string as specified in the start variable. Once found, it sets a flag f to 1. Lines are only printed if this flag is set.

This assumes a strict time ordering of entries, i.e. you cannot have "earlier" timestamps occuring after the line with the start timestamp. Although improbable, this is possible in case of buffered logging.