Regex to replace every character with a space

Question

Hello All,

Odd request this and my regex is failing me. I need to take a column and basically have the text string have as many " " as it does letters. (Appreciating this will be an empty field full of dynamic spaces depending on cell string length).

I did try REGEX_Replace([My data],"\.", " ") but I still get the text, What's the correct regex?

Thank you

Data:

My DataLength of StringTo Check8My Birthday11Happy Birthday14

IraWatt · Accepted Answer

Hey @Bobbins,

😄 that is an odd use case! You dont need the backslash as that escapes the special character. What you want is this:

REGEX_Replace([My Data], ".", " ")

If you want to learn more about Regex the community has some really quick interactive videos on getting to grips with it here https://community.alteryx.com/t5/Interactive-Lessons/tkb-p/interactive-lessons/label-name/Parsing%20Data

Any questions or issues please ask

Ira Watt
Technical Consultant
Watt@Bulien.com

DataNath · Accepted Answer

@Bobbins you just need to remove the \ where you have \. as your target. Escaping it with the backslash means Alteryx is looking for fullstops, whereas a . on it's own indicates any character:

MarqueeCrew · Accepted Answer

@Bobbins ,

isn't this easy to read?

Padleft("", length([field]," ")

no RegEx to harm the workflow

cheers,

mark