Trim and Copy numbers st passed the date

Question

I have a column of dates with added digits to the end of each date. 
I need to be able to trim the trailing digits off the end of the string/year and place them into a new column label overage.

Date4/12/2022527/8/2022245/10/2022483/17/2022658/12/2022984/27/20221710/10/20229912/1/2022781/13/202296

Goal

DateOverage4/12/2022527/8/2022245/10/202248

GMG0241 · Answer

Following on from @cjaneczko, If you want an all in one, you can use REGEX_REPLACE([Date],"\d+\/\d+\/\d{4}(.*)","$1") where the regex expression in the brackets is outputted. In the example I provided, it gets any character (digit, string, whatever) but you could change it to something like  "\d+\/\d+\/\d{4}(\d+)" if you specifically only wanted to grab digits

cjaneczko · Answer

You can use a formula tool. To extract the date you can use Regex \d+\/\d+\/\d{4}. For the trailing digits you can use a right(FieldName,2)