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From: Rakesh Sharma on 8 Jan 2010 16:32
On Jan 8, 5:04 am, master44 <trp...(a)gmail.com> wrote:
> I looking to write a bash shell script to parse a log file for the
> first occurance of a date and then once that date is found, the
> contents of the rest of the log file will be copied to a new file.
>
> Here is a simplified example of a log file (test.log)
>
> 2010-01-05 ....
> 2010-01-06 ....
> 2010-01-07 ....
> 2010-01-07 ....
> 2010-01-07 ....
>
> So for example I want to search for the first occurance of 2010-01-07
> and then copy the rest of the file to anther file (test2.log)
>
> So the new file (test2.log) would look like this:
>
> 2010-01-07 ....
> 2010-01-07 ....
> 2010-01-07 ....
>
> I am thinking this can be done with sed and I have tried the
> following, but something is off...
>
> grep "2010-01-07|sed 's/^0//g'`" test.log|tail +2|head -1 > test2.log
>
> If anyone has done this before or has any ideas, I'd appreciate it!
>
> Thanks


you can use the tool "sed" for this.

sed -e '
/2010-01-07/!d
:loop
n
bloop
' yourfile

or,

sed -e '/2010-01-07/,$!d' yourfile

--Rakesh
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