I could have sworn I have stripped CRLF in the past but not sure why the following isn't working:
myString = "ABC" & vbCrLf & "DEF"
str1 = Replace(myString, vbLf, "")
str2 = Replace(str1, vbCrLf, "")
str3 = Replace(str2, vbNewLine, "")
MsgBox str3
The code above doesn't work the result is:
ABC
DEF
myString = "ABC" & vbCrLf & "DEF"
str1 = Replace(myString, Chr(13), "")
str2 = Replace(str1, Chr(10), "")
MsgBox str2
The code above does work the result is:
ABCDEF
Solution: Thanks @ Mat for the answer (The problem on the first code was the order I was trying to remove the items) VbCrLf & VbNewLine is the same and trying to remove the combo vbCr+VbLf after removing VbLf won't work
The premise is flawed:
myString = "ABC" & vbCrLf & "DEF"
The string is made of "ABC", vbCrLf
, and "DEF".
vbCrLf
is vbCr
and vbLf
, which on any Windows box is vbNewLine
.
When you do:
str1 = Replace(myString, vbLf, "")
You replace vbLf
and leave the vbCr
character in place.
str2 = Replace(str1, vbCrLf, "")
Then you replace vbCrLf
but vbLf
is already gone so vbCrLf
isn't in the string.
str3 = Replace(str2, vbNewLine, "")
Then you replace vbNewLine
which is basically doing the exact same thing as the previous instruction, and the result is a string that's been stripped of vbLf
but still contains vbCr
.
This code works as expected:
Sub Test()
Dim foo As String
foo = "foo" & vbCrLf & "bar"
Debug.Print foo
foo = Replace(foo, vbNewLine, vbNullString)
Debug.Print foo
End Sub
As does this:
Sub Test()
Dim foo As String
foo = "foo" & vbNewLine & "bar"
Debug.Print foo
foo = Replace(foo, vbNewLine, vbNullString)
Debug.Print foo
End Sub
Or this:
Sub Test()
Dim foo As String
foo = "foo" & vbNewLine & "bar"
Debug.Print foo
foo = Replace(foo, vbCrLf, vbNullString)
Debug.Print foo
End Sub
Or even this:
Sub Test()
Dim foo As String
foo = "foo" & vbNewLine & "bar"
Debug.Print foo
foo = Replace(foo, vbCr, vbNullString)
foo = Replace(foo, vbLf, vbNullString)
Debug.Print foo
End Sub
Your second snippet works as intended, because you do remove both vbCr
(Chr(13)
) and vbLf
(Chr(10)
) characters. Simple as that.