UKC

Hi,

I need some help with a regular expression in vi.

Have thousands of dates in the format yyyy-mm-dd. I need to put double quotes around each of them to make "yyyy-mm-dd". Can't work out what the regexr should be to do it. Finding the pattern is ok using /([0-9]{4}-[0-9]{2}-[0-9]{2})/g but can't then get the quotes around the matched pattern

Anybody got any ideas?

Thanks in advance.

In reply to Richard White:
Place the different parts of the regex in parentheses to group them, then use $1 for the first match, $2 for the second, etc.

So your replacement is /([0-9]{4})-([0-9]{2})-([0-9]{2})/"$1"-"$2"-"$3"/g

Slightly easier to read is:

/(d+)-(d+)-(d+)/"$1"-"$2"-"$3"/g

Though that doesn't limit the number of digits in each group so you could have 12334567-9-21 as a valid input.

In reply to Bob:
Er... I don't think that's what he wanted? Won't that produce something like "yyyy"-"mm"-"dd"?

/([0-9]{4}-[0-9]{2}-[0-9]{2})/"$1"/g

Maybe. Haven't done this in ages.

In reply to Richard White:

:%s/d{4}-d{2}-d{2}/""/g

http://vimregex.com/#substitute

In reply to Jack B:

Oops, yes you're right. I misread the request. You need the grouping operator around the whole regex then do "$1" to quote it.

In reply to All:

Thanks for your help. Got it working using the provided examples.

Good to see the art of the regular expression is not dead yet!

Rich.