Shell Programming and Scripting

Yes this can be done..
Good option, but it giving each time some output... and not sure at what valuue of 'p' there will be an existing date in log file.

So after each execution of command its giving some output.

CAN YOU GIVE SOME SAMPLE DATA and you required output so that i can come to some conclusion

Yes Raj, this can be done but problem is that we are not sure that at what value of 'p' there will be an existing date in logfile.
So for each value of 'p' command will get executed and hence resulting in some unexpected response.

---------- Post updated at 02:27 AM ---------- Previous update was at 02:21 AM ----------

sample data
:

2012 Aug 01 : sdfjsdfkljsdlfgjlsdfjgldjflgjsdfgasdfg
2012 Aug 05 10:12:dfkjsdlfksdfksdfks;dkf;sdkf;ksdfjsdf
2012 Aug 05 10:13: kfgsdklfjlsedjkflsdjkfljsdfgsda
2012 Aug 30 10:11: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf

So if i am asking data of last 30 days it should give whole data from 2012 Aug 01 to 2012 Aug 30.
if last 25 days then:
2012 Aug 05 10:12:dfkjsdlfksdfksdfks;dkf;sdkf;ksdfjsdf
2012 Aug 05 10:13: kfgsdklfjlsedjkflsdjkfljsdfgsda
2012 Aug 30 10:11: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf

But if i running it for last 29 days it should give
2012 Aug 05 10:12:dfkjsdlfksdfksdfks;dkf;sdkf;ksdfjsdf
2012 Aug 05 10:13: kfgsdklfjlsedjkflsdjkfljsdfgsda
2012 Aug 30 10:11: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf

but not of 2012 Aug 01

Code:

$cat file
2012 Aug 01 : sdfjsdfkljsdlfgjlsdfjgldjflgjsdfgasdfg
2012 Aug 05 10:12:dfkjsdlfksdfksdfks;dkf;sdkf;ksdfjsdf
2012 Aug 05 10:13: kfgsdklfjlsedjkflsdjkfljsdfgsda
2012 Aug 30 10:11: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf
2012 Aug 31 10:12: sdhfsdkfsdjflsjdfjsdf

#I don't know how <= is not working for the end date so add one more date to your end date...

$ awk -F ":" '{ if ($1 >= "2012 Aug 01" && $1 <= "2012 Aug 31" ) print }' file
2012 Aug 01 : sdfjsdfkljsdlfgjlsdfjgldjflgjsdfgasdfg
2012 Aug 05 10:12:dfkjsdlfksdfksdfks;dkf;sdkf;ksdfjsdf
2012 Aug 05 10:13: kfgsdklfjlsedjkflsdjkfljsdfgsda
2012 Aug 30 10:11: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf

$ awk -F ":" '{ if ($1 >= "2012 Aug 01" && $1 <= "2012 Sep 1" ) print }' file
2012 Aug 01 : sdfjsdfkljsdlfgjlsdfjgldjflgjsdfgasdfg
2012 Aug 05 10:12:dfkjsdlfksdfksdfks;dkf;sdkf;ksdfjsdf
2012 Aug 05 10:13: kfgsdklfjlsedjkflsdjkfljsdfgsda
2012 Aug 30 10:11: sdhfsdkfsdjflsjdfjsdf
2012 Aug 30 10:12: sdhfsdkfsdjflsjdfjsdf
2012 Aug 31 10:12: sdhfsdkfsdjflsjdfjsdf

and the code to extract last nth day's date is

Code:

START=`TZ="GMT+$((24*N))" date +"%Y %b %d"`

---------- Post updated at 02:50 AM ---------- Previous update was at 02:44 AM ----------

Yes Pamu, i can do this too, good option but in case if date format changed.
will this work?
Specially in awk -F ":" the ":" symbol. beacuse in another file this can be "-" or ";"

-F ":" this used as field separator. If date format changed by that means field separator changed then we have to implement that to the code. But this is not major change you can do it easily.

for START variable..

You can use below code to assign date variable to one day ahead of today's date...

Code:

START=$(date --date="tomorrow" +"%Y %b %d")

whether the log will be of one month only or will be of another month also