Extracting the Date in a File

[KamenRider]

HI Guys,

Can someone help me modify the formula below to get the date on the filename. Example

0000220454_20230710_FE_v01.xks

0000110383_20230721_FE_v01.xks

The formula is

DateTimeParse(REGEX_Replace([FileName], '[^\d{8}]', ''),'%Y%m%d')

I also welcome new ideas.

Thank,

Kamen

[Prometheus]

@KamenRider I used a RegEx tool's Parse function with this expression: ^\d{10}\_(\d{8}) then used a Formula tool and applied your expression to the field the RegEx tool created: DateTimeParse(REGEX_Replace([RegExOut1], '[^\d{8}]', ''),'%Y%m%d').

 

 

[binu_acs · Accepted solution]

@KamenRider One way of doing this

[Prometheus]

@KamenRider @binu_acs 

Oh yes. This is great because it creates a nested expression to parse and create the date in one.