I am doing a range search with sed. I want to parse the log data from the date and time 2016-09-29 01:00 to 2016-09-29 01:30. That is why I have been using the following command,

$ sed -n '/2016-09-29 01:/,/2016-09-29 01:30:.*$/p'

But problem is if 1:30 is not available in log then it returns all the logs to the end.

So how can work with this so that if 1:30 doesn't exist it will go to the just next record not till end.

Things to consider: Logs contains stack trace so lines contain stack trace doesn't start with the date.

1 Answer 1


Not so strange. sed is a stream editor, it processes lines as they come. A range like /a/,/b/ means the lines are selected as soon as a is found, and no-longer selected after b is found. If b is never found, it never stops selecting lines.

Here, you should rather use awk instead. Assuming those timestamps are at the beginning of the line:

awk '$0 >= "2016-09-29 01:" && $0 < "2016-09-29 01:30"'

Note that it will only select lines that have a timestamp in the range, so would excludes lines that don't have a timestamp even if they are between lines that have timestamps in the range.

Another approach to work around that would be:

awk -v start='2016-09-29 01:' -v end='2016-09-29 01:30' '
  $0 >= start && $0 <= end, /^[0-9]{4}([ :-][0-9]{2}){5}/ && $0 >= end'

That is use a range like in sed, but enter the range on the first line that is between the 2 dates, and leave it only when we find a line with a timestamp that is greater than the end date.

  • Worked like charm. But then I saw that it doesn't include stack traces.
    – arif
    Commented Aug 29, 2018 at 11:22
  • @muhammad, see edit. Commented Aug 29, 2018 at 11:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .