Reputation: 74
I have 100K rows file and every row contain date 12/13/2019. I want to replace that date with 12/20/2019. But when I am entering the command like :%s/12/13/2019/12/20/2019/g. It gives an error that couldn't find pattern. Format for date is (MM/DD/YYYY)
Upvotes: 0
Views: 55
Reputation: 196886
A substitution is made of several parts:
:<range>s/<search>/<replace>/<flags>
Between those parts, you have /
as default separator. Since /
separates the <search>
part from the <replace>
part, any /
in your search pattern is going to be interpreted as a separator, leading to undesirable results.
One solution is to escape your slashes with an anti-slash:
:%s/12\/13\/2019/12\/20\/2019/g
Another one (my favourite) is to use an alternative separator:
:%s@12/13/2019@12/20/2019@g
Reference:
:help :s
:help pattern-delimiter
Upvotes: 1