For replace you can use vbCrLf:
Replace(string, vbCrLf, "")
You can also use chr(13)+chr(10).
I seem to remember in some odd cases that chr(10) comes before chr(13).
For replace you can use vbCrLf:
Replace(string, vbCrLf, "")
You can also use chr(13)+chr(10).
I seem to remember in some odd cases that chr(10) comes before chr(13).