Shell Programming and Scripting

Hi Guys,

Good day! I hope you could help me on this, I have a file that conatins output upon executing

Code:

cat /var/log/messages

, then what I want is to get the logs that has been generated only starting from 24-hours earlier at the time of actual execution of the script. Is this possible?


Best Regards,
rymnd_12345

Code:

for x in `ls`; do [[ `stat -c%y $x | awk '{print $1}'` == `date -d "yesterday" +%F` ]] && echo $x; done

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The above code will only print those whose date is yesterday. Try the below:

Code:

for x in `ls`; do [ `stat -c%y $x | cut -c1-16 | sed 's/[- :]//g'` -ge `date -d"yesterday" +%Y%m%d%H%M` ] && echo $x; done

Probably useful.

What does "get the logs" mean? Please give and example. Please mention what Operating System and version you have and what Shell you use.

Hi methyl,

"get the logs" means that isn't it that after executing the command

Code:

cat /var/log/messages > output-file

I want to get only logs starting from yesterday (24hrs earlier) up to the current system time the script has been executed.

example if today is Dec 19 15:00:00 and then I executed the script, it will give an output of the logs starting from Dec 18 15:00:00 up to Dec 19 15:00:00.

Please advice,
rymnd_12345

@rymnd_12345: Did you try solution posted in post #2?

Hi balajesuri,

I have just read the 2nd script that you posted, now it works. Thanks so much! Thanks ULF


Br,
rymnd_12345