Hello,

I have log file with entries:

2005-10-1 Server error at......
exception at
2005-10-2 Starting something
Something started
Client requested ...........
2005-10-3 Server encountered error at.....
Something happended


I would like to print only lines containing errors, but only those,
which are from 2005-10-3 or later.
How can i achieve this? grep -i -e "error" myfile.txt prints all lines
with error, but i would like to print
only lines with errors from date 2005-10-3 to today.

Thanks in advance

T0me3

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On 2006-01-21, T0me3 wrote:
> Hello,
>
> I have log file with entries:
>
> 2005-10-1 Server error at......
> exception at
> 2005-10-2 Starting something
> Something started
> Client requested ...........
> 2005-10-3 Server encountered error at.....
> Something happended
>
>
> I would like to print only lines containing errors, but only those,
> which are from 2005-10-3 or later.
> How can i achieve this? grep -i -e "error" myfile.txt prints all lines
> with error, but i would like to print
> only lines with errors from date 2005-10-3 to today.

Assuming that the lines are in chronological order:

awk '/^[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9]/  &
#123;
split($1,d,"-")
date = sprintf("%04d%02d%02d", d[1], d[2], d[3])
}
/error/ && date > 20051003 { print }
'

--
Chris F.A. Johnson, author   |    <http://cfaj.freeshell.org>
Shell Scripting Recipes:     |  My code in this post, if any,
A Problem-Solution Approach  |          is released under the
2005, Apress                 |     GNU General Public Licence

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T0me3 wrote:
> I have log file with entries:
>
> 2005-10-1 Server error at......
> exception at
> 2005-10-2 Starting something
> Something started
> Client requested ...........
> 2005-10-3 Server encountered error at.....
> Something happended
>
> I would like to print only lines containing errors, but only those,
> which are from 2005-10-3 or later.

If your log entries are chronological, how about something like:
sed -ne '/^2005-10-3/,${/error/p;}'

One can also make the part within {} more complex to handle, e.g.
multiple regular expressions, e.g.:
sed -ne '/^2005-10-3/,${/error/{p;n;};/fail/p;}'

[ Post a follow-up to this message ]

2006-01-21, 03:41(-08), T0me3:
> Hello,
>
> I have log file with entries:
>
> 2005-10-1 Server error at......
> exception at
> 2005-10-2 Starting something
> Something started
> Client requested ...........
> 2005-10-3 Server encountered error at.....
> Something happended
>
>
> I would like to print only lines containing errors, but only those,
> which are from 2005-10-3 or later.
> How can i achieve this? grep -i -e "error" myfile.txt prints all lines
> with error, but i would like to print
> only lines with errors from date 2005-10-3 to today.
[...]

I would suggest that you modify your application that logs those
lines so that it formats the time stamps as 2005-10-03 instead.
This way, a string comparison would be enough:

awk '
$1 "" > "2005-10-03" && /error/' < file.log


--
Stéphane

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