Date: Wed, 20 Feb 2013 04:35:18 +0200 From: "bw.mail.lists" <bw.mail.lists@gmail.com> To: freebsd-questions@freebsd.org Subject: Re: convert date and time to epoch in awk Message-ID: <51243666.5000408@gmail.com> In-Reply-To: <00df01ce0efe$c98f8030$5cae8090$@freebsd.org> References: <CAHM0YgvadWLRUNvyQzTj0b=YkfZAyzRRCTEyjByqXM9yXsynhg@mail.gmail.com> <00df01ce0efe$c98f8030$5cae8090$@freebsd.org>
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>> So, is there a way to compare two dates in FreeBSD's awk or convert
a date
>> to epoch? Or some other fast way to select the last 10 minutes from
a log
>> file? An example would be appreciated, if possible.
>
>
> Converting a date to an epoch is easy with date(1) (note: awk can
make a system
> call and read back the stdout into a variable).
>
> For example, if I want to convert the date:
>
> Fri 01 Feb 2013
>
> into an epoch using:
>
> date -j -f "%a %d %b %Y" "Fri 01 Feb 2013" +%s
>
> The output of which is the following epoch:
>
> 1359763497
>
> Doing this all from awk:
>
> echo "Fri 01 Feb 2013" | awk '
> {
> mydate = $0
> "date -j -f \"%a %d %b %Y\" \"" mydate "\" +%s" | getline myepoch
> print mydate " = " myepoch
> }'
>
> Hope this helps.
>
It does, exactly what I was looking for, many thanks.
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