Need to count words and put /

[mmustkee]

Hi Team,

I have data in which we need to place "/" after each 3 letter.  We have "/" in many of the field but sometimes it is 6 letters together without "/". Please refer table below

Date	Result
ssn/ssd/fdcdcf/cdf/klnsdn	ssn/ssd/fdc/dcf/cdf/kln/sdn
ssn/ssd/fdcdcf/cdf/kln/dfc	ssn/ssd/fdc/dcf/cdf/kln/dfc
ssd/dfrsdr/dsr	ssd/dfr/sdr/dsr
dfrssn/sdf/dfrtgy/kly	dfr/ssn/sdf/dfr/tgy/kly

[FrederikE]

Hey @mmustkee,

You can achive this using RegEx in a Formula tool:

REGEX_Replace(Replace([Date], "/", ""), "(...)", "$1/")

See the attached WF. 

[mmustkee]

@FrederikE 

Thanks for your solution.

We can not replace "/" with "" as  we have huge data and sometimes we have incorrect data. so only thing I want is to place "/" where ever we have 6 letters together without space and without "/"

[FrederikE · Accepted solution]

Hey @mmustkee,

Alright. Then

REGEX_Replace([Date], "(\w{3})(\w{3})", "$1/$2")

This should do :)