I got a problem with my vba UserForm, here is the code:
Private Sub CommandButton1_Click()
Dim r As Variant
Dim C As Integer

C = TextBox1.Text

If IsNumeric(C) = False Then
MsgBox "Please enter an integer."

Else
For r = 1 To 700
If Cells(r, C).Value = "0" Then
Rows(r).Hidden = True
End If
Next
End If
End Sub

I'd like to make it when people key in a non-numeric character in the
textbox, it will promt out the message,however, when i Dim C As
Integer, once we key in a non numeric char it will cause a run time
error,since it is not an integer as i declared and if i Dim c As
Variant, the Cells(r,c) will not work as Cells() required an integer
inside.
Can anyone tell me how to solve the problem?

You could declare C as a Variant and that should work. You could also
declare it as a String and use CLng(C) for the Cell argument (I dislike
Variants, so the latter would be my choice).

As an aside... more than like your use of IsNumeric will work okay for you,
but you may find this previous posting of mine of some interest...

I usually try and steer people away from using IsNumeric to "proof"
supposedly numeric text. Consider this (also see note below):

ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)")

Most people would not expect THAT to return True. IsNumeric has some "flaws"
in what it considers a proper number and what most programmers are looking
for.

I had a short tip published by Pinnacle Publishing in their Visual Basic
Developer magazine that covered some of these flaws. Originally, the tip was
free to view but is now viewable only by subscribers.. Basically, it said
that IsNumeric returned True for things like -- currency symbols being
located in front or in back of the number as shown in my example (also
applies to plus, minus and blanks too); numbers surrounded by parentheses as
shown in my example (some people use these to mark negative numbers);
numbers containing any number of commas before a decimal point as shown in
my example; numbers in scientific notation (a number followed by an upper or
lower case "D" or "E", followed by a number equal to or less than 305 -- the
maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for
Hexadecimal, &O or just & in front of the number for Octal).

NOTE:
======
In the above example and in the referenced tip, I refer to $ signs and
commas and dots -- these were meant to refer to your currency, thousands
separator and decimal point symbols as defined in your local settings --
substitute your local regional symbols for these if appropriate.

As for your question about checking numbers, here are two functions that I
have posted in the past for similar questions..... one is for digits only
and the other is for "regular" numbers:

Function IsDigitsOnly(Value As String) As Boolean
IsDigitsOnly = Len(Value) 0 And _
Not Value Like "*[!0-9]*"
End Function

Function IsNumber(ByVal Value As String) As Boolean
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9.]*" And _
Not Value Like "*.*.*" And _
Len(Value) 0 And Value < "." And _
Value < vbNullString
End Function

Here are revisions to the above functions that deal with the local settings
for decimal points (and thousand's separators) that are different than used
in the US (this code works in the US too, of course).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

I'm not as concerned by the rejection of entries that include one or more
thousand's separators, but we can handle this if we don't insist on the
thousand's separator being located in the correct positions (in other words,
we'll allow the user to include them for their own purposes... we'll just
tolerate their presence).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
Dim TS As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Get local setting for thousand's separator
' and eliminate them. Remove the next two lines
' if you don't want your users being able to
' type in the thousands separator at all.
TS = Mid$(Format$(1000, "#,###"), 2, 1)
Value = Replace$(Value, TS, "")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

Rick


wrote in message
...
I got a problem with my vba UserForm, here is the code:
Private Sub CommandButton1_Click()
Dim r As Variant
Dim C As Integer

C = TextBox1.Text

If IsNumeric(C) = False Then
MsgBox "Please enter an integer."

Else
For r = 1 To 700
If Cells(r, C).Value = "0" Then
Rows(r).Hidden = True
End If
Next
End If
End Sub

I'd like to make it when people key in a non-numeric character in the
textbox, it will promt out the message,however, when i Dim C As
Integer, once we key in a non numeric char it will cause a run time
error,since it is not an integer as i declared and if i Dim c As
Variant, the Cells(r,c) will not work as Cells() required an integer
inside.
Can anyone tell me how to solve the problem?

On 1ÔÂ30ÈÕ, ÏÂÎç6ʱ02·Ö, "Rick Rothstein \(MVP - VB\)"
wrote:
You could declare C as a Variant and that should work. You could also
declare it as a String and use CLng(C) for the Cell argument (I dislike
Variants, so the latter would be my choice).

As an aside... more than like your use of IsNumeric will work okay for you,
but you may find this previous posting of mine of some interest...

I usually try and steer people away from using IsNumeric to "proof"
supposedly numeric text. Consider this (also see note below):

ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)")

Most people would not expect THAT to return True. IsNumeric has some "flaws"
in what it considers a proper number and what most programmers arelooking
for.

I had a short tip published by Pinnacle Publishing in their Visual Basic
Developer magazine that covered some of these flaws. Originally, the tip was
free to view but is now viewable only by subscribers.. Basically, it said
that IsNumeric returned True for things like -- currency symbols being
located in front or in back of the number as shown in my example (also
applies to plus, minus and blanks too); numbers surrounded by parentheses as
shown in my example (some people use these to mark negative numbers);
numbers containing any number of commas before a decimal point as shown in
my example; numbers in scientific notation (a number followed by an upper or
lower case "D" or "E", followed by a number equal to or less than 305 -- the
maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for
Hexadecimal, &O or just & in front of the number for Octal).

NOTE:
======
In the above example and in the referenced tip, I refer to $ signs and
commas and dots -- these were meant to refer to your currency, thousands
separator and decimal point symbols as defined in your local settings --
substitute your local regional symbols for these if appropriate.

As for your question about checking numbers, here are two functions that I
have posted in the past for similar questions..... one is for digits only
and the other is for "regular" numbers:

Function IsDigitsOnly(Value As String) As Boolean
IsDigitsOnly = Len(Value) 0 And _
Not Value Like "*[!0-9]*"
End Function

Function IsNumber(ByVal Value As String) As Boolean
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9.]*" And _
Not Value Like "*.*.*" And _
Len(Value) 0 And Value < "." And _
Value < vbNullString
End Function

Here are revisions to the above functions that deal with the local settings
for decimal points (and thousand's separators) that are different than used
in the US (this code works in the US too, of course).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

I'm not as concerned by the rejection of entries that include one or more
thousand's separators, but we can handle this if we don't insist on the
thousand's separator being located in the correct positions (in other words,
we'll allow the user to include them for their own purposes... we'll just
tolerate their presence).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
Dim TS As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Get local setting for thousand's separator
' and eliminate them. Remove the next two lines
' if you don't want your users being able to
' type in the thousands separator at all.
TS = Mid$(Format$(1000, "#,###"), 2, 1)
Value = Replace$(Value, TS, "")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

Rick

wrote in message

...



I got a problem with my vba UserForm, here is the code:
Private Sub CommandButton1_Click()
Dim r As Variant
Dim C As Integer

C = TextBox1.Text

If IsNumeric(C) = False Then
MsgBox "Please enter an integer."

Else
For r = 1 To 700
If Cells(r, C).Value = "0" Then
Rows(r).Hidden = True
End If
Next
End If
End Sub

I'd like to make it when people key in a non-numeric character in the
textbox, it will promt out the message,however, when i Dim C As
Integer, once we key in a non numeric char it will cause a run time
error,since it is not an integer as i declared and if i Dim c As
Variant, the Cells(r,c) will not work as Cells() required an integer
inside.
Can anyone tell me how to solve the problem?- Òþ²Ø±»ÒýÓÃÎÄ×Ö -

- ÏÔʾÒýÓõÄÎÄ×Ö -

Have you tested my code above? the problem i met is i can't declare C
as variant, it still cannot work in the function : Cells( )

If Cells(r, clng(C)).Value = "0" Then

wrote:

On 1ÔÂ30ÈÕ, ÏÂÎç6ʱ02·Ö, "Rick Rothstein \(MVP - VB\)"
wrote:
You could declare C as a Variant and that should work. You could also
declare it as a String and use CLng(C) for the Cell argument (I dislike
Variants, so the latter would be my choice).

As an aside... more than like your use of IsNumeric will work okay for you,
but you may find this previous posting of mine of some interest...

I usually try and steer people away from using IsNumeric to "proof"
supposedly numeric text. Consider this (also see note below):

ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)")

Most people would not expect THAT to return True. IsNumeric has some "flaws"
in what it considers a proper number and what most programmers arelooking
for.

I had a short tip published by Pinnacle Publishing in their Visual Basic
Developer magazine that covered some of these flaws. Originally, the tip was
free to view but is now viewable only by subscribers.. Basically, it said
that IsNumeric returned True for things like -- currency symbols being
located in front or in back of the number as shown in my example (also
applies to plus, minus and blanks too); numbers surrounded by parentheses as
shown in my example (some people use these to mark negative numbers);
numbers containing any number of commas before a decimal point as shown in
my example; numbers in scientific notation (a number followed by an upper or
lower case "D" or "E", followed by a number equal to or less than 305 -- the
maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for
Hexadecimal, &O or just & in front of the number for Octal).

NOTE:
======
In the above example and in the referenced tip, I refer to $ signs and
commas and dots -- these were meant to refer to your currency, thousands
separator and decimal point symbols as defined in your local settings --
substitute your local regional symbols for these if appropriate.

As for your question about checking numbers, here are two functions that I
have posted in the past for similar questions..... one is for digits only
and the other is for "regular" numbers:

Function IsDigitsOnly(Value As String) As Boolean
IsDigitsOnly = Len(Value) 0 And _
Not Value Like "*[!0-9]*"
End Function

Function IsNumber(ByVal Value As String) As Boolean
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9.]*" And _
Not Value Like "*.*.*" And _
Len(Value) 0 And Value < "." And _
Value < vbNullString
End Function

Here are revisions to the above functions that deal with the local settings
for decimal points (and thousand's separators) that are different than used
in the US (this code works in the US too, of course).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

I'm not as concerned by the rejection of entries that include one or more
thousand's separators, but we can handle this if we don't insist on the
thousand's separator being located in the correct positions (in other words,
we'll allow the user to include them for their own purposes... we'll just
tolerate their presence).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
Dim TS As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Get local setting for thousand's separator
' and eliminate them. Remove the next two lines
' if you don't want your users being able to
' type in the thousands separator at all.
TS = Mid$(Format$(1000, "#,###"), 2, 1)
Value = Replace$(Value, TS, "")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

Rick

wrote in message

...



I got a problem with my vba UserForm, here is the code:
Private Sub CommandButton1_Click()
Dim r As Variant
Dim C As Integer

C = TextBox1.Text

If IsNumeric(C) = False Then
MsgBox "Please enter an integer."

Else
For r = 1 To 700
If Cells(r, C).Value = "0" Then
Rows(r).Hidden = True
End If
Next
End If
End Sub

I'd like to make it when people key in a non-numeric character in the
textbox, it will promt out the message,however, when i Dim C As
Integer, once we key in a non numeric char it will cause a run time
error,since it is not an integer as i declared and if i Dim c As
Variant, the Cells(r,c) will not work as Cells() required an integer
inside.
Can anyone tell me how to solve the problem?- Òþ²Ø±»ÒýÓÃÎÄ×Ö -

- ÏÔʾÒýÓõÄÎÄ×Ö -

Have you tested my code above? the problem i met is i can't declare C
as variant, it still cannot work in the function : Cells( )

--

Dave Peterson

ps. Did you really want to test for the number 0 or the text '0?

if isempty(cells(r, clng(c).value) then
'skip it???
else
if cells(r, clng(c)).value = 0 then
rows(r).delete
end if



Dave Peterson wrote:

If Cells(r, clng(C)).Value = "0" Then

wrote:

On 1ÔÂ30ÈÕ, ÏÂÎç6ʱ02·Ö, "Rick Rothstein \(MVP - VB\)"
wrote:
You could declare C as a Variant and that should work. You could also
declare it as a String and use CLng(C) for the Cell argument (I dislike
Variants, so the latter would be my choice).

As an aside... more than like your use of IsNumeric will work okay for you,
but you may find this previous posting of mine of some interest...

I usually try and steer people away from using IsNumeric to "proof"
supposedly numeric text. Consider this (also see note below):

ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)")

Most people would not expect THAT to return True. IsNumeric has some "flaws"
in what it considers a proper number and what most programmers arelooking
for.

I had a short tip published by Pinnacle Publishing in their Visual Basic
Developer magazine that covered some of these flaws. Originally, the tip was
free to view but is now viewable only by subscribers.. Basically, it said
that IsNumeric returned True for things like -- currency symbols being
located in front or in back of the number as shown in my example (also
applies to plus, minus and blanks too); numbers surrounded by parentheses as
shown in my example (some people use these to mark negative numbers);
numbers containing any number of commas before a decimal point as shown in
my example; numbers in scientific notation (a number followed by an upper or
lower case "D" or "E", followed by a number equal to or less than 305 -- the
maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for
Hexadecimal, &O or just & in front of the number for Octal).

NOTE:
======
In the above example and in the referenced tip, I refer to $ signs and
commas and dots -- these were meant to refer to your currency, thousands
separator and decimal point symbols as defined in your local settings --
substitute your local regional symbols for these if appropriate.

As for your question about checking numbers, here are two functions that I
have posted in the past for similar questions..... one is for digits only
and the other is for "regular" numbers:

Function IsDigitsOnly(Value As String) As Boolean
IsDigitsOnly = Len(Value) 0 And _
Not Value Like "*[!0-9]*"
End Function

Function IsNumber(ByVal Value As String) As Boolean
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9.]*" And _
Not Value Like "*.*.*" And _
Len(Value) 0 And Value < "." And _
Value < vbNullString
End Function

Here are revisions to the above functions that deal with the local settings
for decimal points (and thousand's separators) that are different than used
in the US (this code works in the US too, of course).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

I'm not as concerned by the rejection of entries that include one or more
thousand's separators, but we can handle this if we don't insist on the
thousand's separator being located in the correct positions (in other words,
we'll allow the user to include them for their own purposes... we'll just
tolerate their presence).

Function IsNumber(ByVal Value As String) As Boolean
Dim DP As String
Dim TS As String
' Get local setting for decimal point
DP = Format$(0, ".")
' Get local setting for thousand's separator
' and eliminate them. Remove the next two lines
' if you don't want your users being able to
' type in the thousands separator at all.
TS = Mid$(Format$(1000, "#,###"), 2, 1)
Value = Replace$(Value, TS, "")
' Leave the next statement out if you don't
' want to provide for plus/minus signs
If Value Like "[+-]*" Then Value = Mid$(Value, 2)
IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Not Value Like "*" & DP & "*" & DP & "*" And _
Len(Value) 0 And Value < DP And _
Value < vbNullString
End Function

Rick

wrote in message

...



I got a problem with my vba UserForm, here is the code:
Private Sub CommandButton1_Click()
Dim r As Variant
Dim C As Integer

C = TextBox1.Text

If IsNumeric(C) = False Then
MsgBox "Please enter an integer."

Else
For r = 1 To 700
If Cells(r, C).Value = "0" Then
Rows(r).Hidden = True
End If
Next
End If
End Sub

I'd like to make it when people key in a non-numeric character in the
textbox, it will promt out the message,however, when i Dim C As
Integer, once we key in a non numeric char it will cause a run time
error,since it is not an integer as i declared and if i Dim c As
Variant, the Cells(r,c) will not work as Cells() required an integer
inside.
Can anyone tell me how to solve the problem?- Òþ²Ø±»ÒýÓÃÎÄ×Ö -

- ÏÔʾÒýÓõÄÎÄ×Ö -

Have you tested my code above? the problem i met is i can't declare C
as variant, it still cannot work in the function : Cells( )

--

Dave Peterson

--

Dave Peterson

On 1月31日, 上�9时33分, Dave Peterson wrote:
ps. Â*Did you really want to test for the number 0 or the text '0?

if isempty(cells(r, clng(c).value) then
Â* 'skip it???
else
Â* Â*if cells(r, clng(c)).value = 0 then
Â* Â* Â* rows(r).delete
end if





Dave Peterson wrote:

If Cells(r, clng(C)).Value = "0" Then

wrote:

On 1ÔÂ30ÈÕ, �ÂÎç6ʱ02·Ö, "Rick Rothstein \(MVP - VB\)"
wrote:
You could declare C as a Variant and that should work. You could also
declare it as a String and use CLng(C) for the Cell argument (I dislike
Variants, so the latter would be my choice).

As an aside... more than like your use of IsNumeric will work okay for you,
but you may find this previous posting of mine of some interest...

I usually try and steer people away from using IsNumeric to "proof"
supposedly numeric text. Consider this (also see note below):

Â* Â* ReturnValue = IsNumeric("($1,23,,3.4,,,5,,E67$)")

Most people would not expect THAT to return True. IsNumeric has some "flaws"
in what it considers a proper number and what most programmers arelooking
for.

I had a short tip published by Pinnacle Publishing in their Visual Basic
Developer magazine that covered some of these flaws. Originally, the tip was
free to view but is now viewable only by subscribers.. Basically, it said
that IsNumeric returned True for things like -- currency symbols being
located in front or in back of the number as shown in my example (also
applies to plus, minus and blanks too); numbers surrounded by parentheses as
shown in my example (some people use these to mark negative numbers);
numbers containing any number of commas before a decimal point as shown in
my example; numbers in scientific notation (a number followed by an upper or
lower case "D" or "E", followed by a number equal to or less than 305 -- the
maximum power of 10 in VB); and Octal/Hexadecimal numbers (&H for
Hexadecimal, &O or just & in front of the number for Octal).

NOTE:
======
In the above example and in the referenced tip, I refer to $ signs and
commas and dots -- these were meant to refer to your currency, thousands
separator and decimal point symbols as defined in your local settings --
substitute your local regional symbols for these if appropriate.

As for your question about checking numbers, here are two functions that I
have posted in the past for similar questions..... one is for digits only
and the other is for "regular" numbers:

Â* Â* Â*Function IsDigitsOnly(Value As String) As Boolean
Â* Â* Â* Â* Â*IsDigitsOnly = Len(Value) 0 And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Not Value Like "*[!0-9]*"
Â* Â* Â*End Function

Â* Â* Â*Function IsNumber(ByVal Value As String) As Boolean
Â* Â* Â* Â* Â*' Â* Leave the next statement out if you don't
Â* Â* Â* Â* Â*' Â* want to provide for plus/minus signs
Â* Â* Â* Â* Â*If Value Like "[+-]*" Then Value = Mid$(Value, 2)
Â* Â* Â* Â* Â*IsNumber = Not Value Like "*[!0-9.]*" And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â*Not Value Like "*.*.*" And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â*Len(Value) 0 And Value < "." And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â* Â*Value < vbNullString
Â* Â* Â*End Function

Here are revisions to the above functions that deal with the local settings
for decimal points (and thousand's separators) that are different than used
in the US (this code works in the US too, of course).

Â* Â* Â*Function IsNumber(ByVal Value As String) As Boolean
Â* Â* Â* Â*Dim DP As String
Â* Â* Â* Â*' Â* Get local setting for decimal point
Â* Â* Â* Â*DP = Format$(0, ".")
Â* Â* Â* Â*' Â* Leave the next statement out if you don't
Â* Â* Â* Â*' Â* want to provide for plus/minus signs
Â* Â* Â* Â*If Value Like "[+-]*" Then Value = Mid$(Value, 2)
Â* Â* Â* Â*IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Not Value Like "*" & DP & "*" & DP & "*" And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Len(Value) 0 And Value < DP And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Value < vbNullString
Â* Â* Â*End Function

I'm not as concerned by the rejection of entries that include one or more
thousand's separators, but we can handle this if we don't insist on the
thousand's separator being located in the correct positions (in other words,
we'll allow the user to include them for their own purposes... we'll just
tolerate their presence).

Â* Â* Â*Function IsNumber(ByVal Value As String) As Boolean
Â* Â* Â* Â*Dim DP As String
Â* Â* Â* Â*Dim TS As String
Â* Â* Â* Â*' Â* Get local setting for decimal point
Â* Â* Â* Â*DP = Format$(0, ".")
Â* Â* Â* Â*' Â* Get local setting for thousand's separator
Â* Â* Â* Â*' Â* and eliminate them. Remove the next two lines
Â* Â* Â* Â*' Â* if you don't want your users being able to
Â* Â* Â* Â*' Â* type in the thousands separator at all.
Â* Â* Â* Â*TS = Mid$(Format$(1000, "#,###"), 2, 1)
Â* Â* Â* Â*Value = Replace$(Value, TS, "")
Â* Â* Â* Â*' Â* Leave the next statement out if you don't
Â* Â* Â* Â*' Â* want to provide for plus/minus signs
Â* Â* Â* Â*If Value Like "[+-]*" Then Value = Mid$(Value, 2)
Â* Â* Â* Â*IsNumber = Not Value Like "*[!0-9" & DP & "]*" And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Not Value Like "*" & DP & "*" & DP & "*" And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Len(Value) 0 And Value < DP And _
Â* Â* Â* Â* Â* Â* Â* Â* Â* Value < vbNullString
Â* Â* Â*End Function

Rick

wrote in message

...

I got a problem with my vba UserForm, here is the code:
Private Sub CommandButton1_Click()
Dim r As Variant
Dim C As Integer

C = TextBox1.Text

If IsNumeric(C) = False Then
MsgBox "Please enter an integer."

Else
For r = 1 To 700
If Cells(r, C).Value = "0" Then
Rows(r).Hidden = True
End If
Next
End If
End Sub

I'd like to make it when people key in a non-numeric character in the
textbox, it will promt out the message,however, when i Dim C As
Integer, once we key in a non numeric char it will cause a run time
error,since it is not an integer as i declared and if i Dim c As
Variant, the Cells(r,c) will not work as Cells() required an integer
inside.
Can anyone tell me how to solve the problem?- Òþ²Ø±»ÒýÓÃÎÄ×Ö -

- �ÔʾÒýÓõÄÎÄ×Ö -

Have you tested my code above? the problem i met is i can't declare C
as variant, it still cannot work in the function : Cells( )

--

Dave Peterson

--

Dave Peterson- éš�è—�被引用文å*— -

- 显示引用的文å*— -

thank you. Sorry about i don't know the function "Clng()".
I would like to test if the cell value number "0" then hide. However,
since it is userform, if user key in any non-integer accidentally, it
will occur error. So to prevent this, i have to test the input value
if not integer it will prompt out a message box.
Thanks for your help.