Forum Discussion

Copper Contributor

Sep 06, 2023

Solved

fomula to use

mtarler
Sep 06, 2023
the second column looks like
=MID(A1, SEARCH("@",A1)-1, LEN(A1)-(SEARCH("@",A1)-1) )
or with Excel 365
=LET(in,A1:A100, at, SEARCH("@",in)-1, MID(in, at, LEN(in)-at)
and third column
=LEFT(A1,SEARCH("@",A1)-1)

MVP

Sep 06, 2023

And your question is?

Copper Contributor

Sep 06, 2023

can you help me

mtarler
Silver Contributor
Sep 06, 2023
the second column looks like
=MID(A1, SEARCH("@",A1)-1, LEN(A1)-(SEARCH("@",A1)-1) )
or with Excel 365
=LET(in,A1:A100, at, SEARCH("@",in)-1, MID(in, at, LEN(in)-at)
and third column
=LEFT(A1,SEARCH("@",A1)-1)
- mjizy01
  Copper Contributor
  Sep 06, 2023
  "=MID(A1,SEARCH("@",A1)-1,LEN(A1)-SEARCH("@",A1)-1)"
  =MID(A1,SEARCH("@",A1)-1,LEN(A1)-SEARCH("@",A1)+1)
  this is the right one. thanks for helping out
  - mtarler
    Silver Contributor
    Sep 06, 2023
    oops, a minus of a minus is a +. good catch. I'll edit the above to prevent confusion.