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sed Equivalent for awk/grep

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# 1  
Old 10-11-2013
sed Equivalent for awk/grep
Any equivalent command using awk or grep?
Code:
sed -n "/^$(date --date='10 minutes ago' '+%b %_d %H:%M')/,\$p" /abc.log

Last edited by Franklin52; 10-11-2013 at 03:04 AM.. Reason: Please use code tags
# 2  
Old 10-11-2013
Try this:

Code:
awk -v tm="$(date --date="10 minutes ago" "+%b %_d %H:%M")" '$0 ~ "^"tm"*" {p++} p' /abc.log

---------- Post updated at 02:45 PM ---------- Previous update was at 02:43 PM ----------

or
Code:
awk "/^$(date --date="-10 min" "+%b %_d %H:%M")/{p++} p" /abc.log

# 3  
Old 10-11-2013
cool. both seems to work
# 4  
Old 10-11-2013
The "*" is too many in the first sample.
# 5  
Old 10-11-2013
Changing dates in log files ? .... why ?
# 6  
Old 10-11-2013
Quote:
Originally Posted by ctsgnb
Changing dates in log files ? .... why ?
This thread is about printing from a logfile, and so are the solutions.
# 7  
Old 10-11-2013
Quote:
Originally Posted by MadeInGermany
This thread is about printing from a logfile, and so are the solutions.
I was just wondering about the needs, because it seems weird to me to alter the dates of a log file before printing it .
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