Hey Folks,
This one currently has me stumped. I have to check for the existence of a
carriage return (Enter key, hex 0D) in text data. First I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If Mid(txtComment.Text, i, 1) = CStr(13) Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This doesn't find the CR because CStr(13) equals a two character string
("13")
Then I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If CInt(Mid(txtComment.Text, i, 1)) = 13 Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This code traps to BogusComment when the first (non CR) character in the
string is checked.
Will using a KeyPress event to fix this?
Stephen Saunders
9  6801 
I found that:
Sub txtComment_KeyPress(KeyAscii As Integer)
If KeyAscii = 13 Then
'Issue message
MsgBox "Enter is not allowed in message text"
KeyAscii = 0
End If
End Sub
works.
Sorry about asking. I hope this helps someone else.
Stephen Saunders
"Stephen Saunders" <kd****@earthlink.net> wrote in message
news:6p***************@newsread3.news.atl.earthlin k.net... Hey Folks,
This one currently has me stumped. I have to check for the existence of a
carriage return (Enter key, hex 0D) in text data. First I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If Mid(txtComment.Text, i, 1) = CStr(13) Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This doesn't find the CR because CStr(13) equals a two character string
("13")
Then I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If CInt(Mid(txtComment.Text, i, 1)) = 13 Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This code traps to BogusComment when the first (non CR) character in the
string is checked.
Will using a KeyPress event to fix this?
Stephen Saunders
I found that:
Sub txtComment_KeyPress(KeyAscii As Integer)
If KeyAscii = 13 Then
'Issue message
MsgBox "Enter is not allowed in message text"
KeyAscii = 0
End If
End Sub
works.
Sorry about asking. I hope this helps someone else.
Stephen Saunders
"Stephen Saunders" <kd****@earthlink.net> wrote in message
news:6p***************@newsread3.news.atl.earthlin k.net... Hey Folks,
This one currently has me stumped. I have to check for the existence of a
carriage return (Enter key, hex 0D) in text data. First I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If Mid(txtComment.Text, i, 1) = CStr(13) Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This doesn't find the CR because CStr(13) equals a two character string
("13")
Then I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If CInt(Mid(txtComment.Text, i, 1)) = 13 Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This code traps to BogusComment when the first (non CR) character in the
string is checked.
Will using a KeyPress event to fix this?
Stephen Saunders
"Stephen Saunders" <kd****@earthlink.net> wrote in message
news:6p***************@newsread3.news.atl.earthlin k.net... Hey Folks,
This doesn't find the CR because CStr(13) equals a two character string
("13")
Try this Steve..
Put a text box and label on a form and a command button:
Private Sub Command1_Click()
Dim t$
Dim x
t$ = Text1.Text
x = InStr(t$, Chr$(13))
If x > 0 Then
Label1.Caption = "CR found at pos " & x
Else
Label1.Caption = "No CR found"
End If
End Sub
"Stephen Saunders" <kd****@earthlink.net> wrote in message
news:6p***************@newsread3.news.atl.earthlin k.net... Hey Folks,
This doesn't find the CR because CStr(13) equals a two character string
("13")
Try this Steve..
Put a text box and label on a form and a command button:
Private Sub Command1_Click()
Dim t$
Dim x
t$ = Text1.Text
x = InStr(t$, Chr$(13))
If x > 0 Then
Label1.Caption = "CR found at pos " & x
Else
Label1.Caption = "No CR found"
End If
End Sub
"Stephen Saunders" <kd****@earthlink.net> skrev i en meddelelse
news:6p***************@newsread3.news.atl.earthlin k.net... Hey Folks,
This one currently has me stumped. I have to check for the
existence of a carriage return (Enter key, hex 0D) in text data.
First I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If Mid(txtComment.Text, i, 1) = CStr(13) Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This doesn't find the CR because CStr(13) equals a two character
string
CStr(13) is NOT a CR, CStr(13) is "13" ! CHR(13) is a CR ! ;-)
And what's wrong with
IF instr(1, txtComment.Text, vbCR) > 0 the
MsgBox "Comment cannot contain the enter character"
txtComment.Text = Replace(txtComment.Text, vbCR, "")
End IF
--
/\ preben nielsen
\/\ pr**@post.tele.dk
"Stephen Saunders" <kd****@earthlink.net> skrev i en meddelelse
news:6p***************@newsread3.news.atl.earthlin k.net... Hey Folks,
This one currently has me stumped. I have to check for the
existence of a carriage return (Enter key, hex 0D) in text data.
First I tried:
'Check comment to not contain the CR (13) character
For i = 1 To Len(txtComment.Text)
On Error GoTo BogusComment2
If Mid(txtComment.Text, i, 1) = CStr(13) Then
MsgBox "Comment cannot contain the enter character"
'Strip the CR from the comment
txtComment.Text = Left(txtComment.Text, (i - 1)) + _
Right(txtComment.Text, (Len(txtComment.Text) - i))
Exit Sub
End If
On Error GoTo 0
Next i
This doesn't find the CR because CStr(13) equals a two character
string
CStr(13) is NOT a CR, CStr(13) is "13" ! CHR(13) is a CR ! ;-)
And what's wrong with
IF instr(1, txtComment.Text, vbCR) > 0 the
MsgBox "Comment cannot contain the enter character"
txtComment.Text = Replace(txtComment.Text, vbCR, "")
End IF
--
/\ preben nielsen
\/\ pr**@post.tele.dk
Of course, to be clearer, it would be better to just use vbCr and be
done with it.
Of course, to be clearer, it would be better to just use vbCr and be
done with it.
Here's a solution
dim i as integer
private sub Command1_Click()
open "orig.txt" for input as #1
'This would be the file you are trying to find the CR
while not eof(1)
a = input$(1, 1)
i = i + 1
if a$ = chr(13) then
MsgBox (msg = "CR Found at" & i & "Char")
'This will give you the number of char's reached until CR was found
end if
wend
end sub
I could get more detailed but this should be enough for you. If not let me know.
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