On Tue, 15 Nov 2016, Rich Shepard wrote:

> On Tue, 15 Nov 2016, Paul Heinlein wrote:
>
>> I'm late to this party, but in this case, you can use GNU date, e.g.,
>
> Welcome to the party, Paul. There's still some microbrew on tap; 
> help yourself.
>
>> for D in "2014/04/10" "2015/05/11" "2016/06/12"; do
>>  date -d $D +%F
>> done
>
> How do I generalize this so I can pass a 4,000-row filename and date 
> will act on only the date field within that row?

Use sed. :-)

The alternative is to write a little script using a CSV parser library 
(e.g., Perl's Text::CSV module), but this is probably not what you 
wanted to spend your time doing.

-- 
Paul Heinlein <> [email protected] <> http://www.madboa.com/
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