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I have a list of dates which runs from:

01/02/01
01/03/01
01/04/01
01/05/01
01/06/01
01/07/01
01/08/01
01/09/01
01/10/01
01/11/01
01/12/01
...
...
12/05/91
12/06/91
12/07/91
12/08/91
12/09/91
12/10/91
12/11/91
12/12/91
12/13/91

That is it runs from 1901 through 1991.
I need to replace the last two digits of the year with 4 digits of the year, so that 12/13/91 shows as 12/13/1991 and also end up with the forward slashes "/" removed.

What would be the best approach to do this? There are 3 other columns to the left of this column, for what it's worth.

Please and thank you in advance.

Use "Posix" regular expression syntax:

Configure | Preferences | Editor

[X] Use POSIX regular expression syntax

Search | Replace... (<F8>):

Find what: \<([0-9]{2})/([0-9]{2})/([0-9]{2})\>
Replace with: \1\219\3

[X] Regular expression

Replace All

That is a perverse way to store a date!

It is for a data pool file wherein a load testing application feeds the date to an application which puts the "/" back in - so I need the raw data just as someone would type it from the keyboard, e.g., 09271951, as the application automatically adds the forward slashes.

someone

... if they are American.

And does it do what you want?