On 3/25/2008 9:03 PM, Seb wrote:
> On Wed, 26 Mar 2008 02:45:40 +0100,
> Janis Papanagnou <Janis_Papanagnou@[EMAIL PROTECTED]
> wrote:
>
> [...]
>
>
>>You can write two gensub()s, where you replace the third
>>[[:digit:]]{2} subpattern by 0[[:digit:]] and [1-9][[:digit:]], resp.,
>>and add the respective century information before the replacement \\3.
>
>
> Good idea, this seemed to work:
>
>
> awk --re-interval '{
> sub(/ end,/, " ending,")
> if ($0 ~ /[[:digit:]]{2}\/[[:digit:]]{2}\/[[:digit:]]{2}/) {
>
$0=gensub(/([[:digit:]]{2})\/([[:digit:]]{2})\/([1-9])([[:digit:]])/,
> "19\\3\\4-\\1-\\2", "g")
> $0=gensub(/([[:digit:]]{2})\/([[:digit:]]{2})\/0([[:digit:]])/,
> "200\\3-\\1-\\2", "g")
> }
> print
> }'
>
> The regexps are a bit nasty though. Thanks!
I don't think you need the test, just do the gensub() and the pattern will
either match or it won't. Maybe you could simplify it a bit too, e.g.:
awk --re-interval '{
sub(/ end,/, " ending,")
$0=gensub(/([[:digit:]]{2})\/([[:digit:]]{2})\/([[:digit:]]{2})/,
"19\\3\\4-\\1-\\2", "g")
$0=gensub(/19(0[[:digit:]]-)/,"20\\1", "g")
print
}' file
Ed.
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