On Mar 21, 12:26*pm, joel wrote:
The return in the cell is code as an ascii character Linefeed which is
a 10. In the code below I set a variable to character 10. *You can also
use vblf. *You can use any string method with the linefeed like REPLACE
to remove the character or replace the character with other characters

LF = Chr(10)
or
LF = vbLF

with
Set c = .Columns("A").Find(what:=LF, _
LookIn:=xlValues, lookat:=xlPart)
if not c is nothing
FirstAddr = c.address
do
c = replace(c,LF, " *")
set c = .Columns("A").findnext(after:=c)
while not c is nothing and c.address < FirstAddr

end if
end with

The code below is complicated. *I just wanted to show a couple of
different methods. *I don't know how many cells you are trying to replce
the Linefeed with two spaces. *You can achieve the same thing as
follows

LF = vbLF

with sheets("sheet1")
.columns("A").replace(LF," *")

end with

--
joel
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Hi Joel,

great help. Thank you very much for your professional advice. It works
as desired.

Regards, Andreas