Suppose I have a string like this:

"32 01 22 88 03"

I need to remove all the leading zeros in the string. The result should be:

"32 1 22 88 3"

What's the easiest way to go about this? Thanks!

Mr. T

Hi,

Am Sat, 13 Oct 2018 04:08:32 -0700 (PDT) schrieb Tatsujin:

Suppose I have a string like this:

"32 01 22 88 03"

I need to remove all the leading zeros in the string. The result should be:

"32 1 22 88 3"

Find & Select = Replace and replace space and 0 (" 0") with space (" ")


Regards
Claus B.
--
Windows10
Office 2016

El sábado, 13 de octubre de 2018, 6:08:39 (UTC-5), Tatsujin escribió:
Suppose I have a string like this:

"32 01 22 88 03"

I need to remove all the leading zeros in the string. The result should be:

"32 1 22 88 3"

What's the easiest way to go about this? Thanks!

Mr. T

?Replace("32 01 22 88 03", "0", "")

El sábado, 13 de octubre de 2018, 6:08:39 (UTC-5), Tatsujin escribió:
Suppose I have a string like this:

"32 01 22 88 03"

I need to remove all the leading zeros in the string. The result should be:

"32 1 22 88 3"

What's the easiest way to go about this? Thanks!

Mr. T

?Replace("32 01 22 88 03", "0", "")

Oops! That's going to replac *ALL* zeros; - the task is to replace *LEADING*
ZEROS ONLY!!

--
Garry

Free usenet access at http://www.eternal-september.org
Classic VB Users Regroup!
comp.lang.basic.visual.misc
microsoft.public.vb.general.discussion

?Replace("32 01 22 88 03", "0", "")

Oops! That's going to replac *ALL* zeros; - the task is to replace *LEADING*
ZEROS ONLY!!


I devised the following solution. Maybe using regular expressions
is overkill, but it worked. Here it is:

Public Sub MyReplace()
' Include"Microsoft VBScript Regular Expressions 5.5" in Tools-References
Dim regEx As New VBScript_RegExp_55.RegExp

Dim s1 As String
Dim sFinal As String
Dim sExample As String

sExample = "01 07 08 22 88 06 04"

' Replace leading zeros (in middle of line)
regEx.Pattern = " 0"
regEx.Global = True
regEx.IgnoreCase = False
s1 = regEx.Replace(sExample, " ")

' Remove leading zeros (at beginning of line)
regEx.Pattern = "^0"
regEx.Global = True
regEx.IgnoreCase = False
sFinal = regEx.Replace(s1, "")

MsgBox sFinal

End Sub


- Robert Crandall

?Replace("32 01 22 88 03", "0", "")

Oops! That's going to replac *ALL* zeros; - the task is to replace *LEADING*
ZEROS ONLY!!


I devised the following solution. Maybe using regular expressions
is overkill, but it worked. Here it is:

Public Sub MyReplace()
' Include"Microsoft VBScript Regular Expressions 5.5" in Tools-References
Dim regEx As New VBScript_RegExp_55.RegExp

Dim s1 As String
Dim sFinal As String
Dim sExample As String

sExample = "01 07 08 22 88 06 04"

' Replace leading zeros (in middle of line)
regEx.Pattern = " 0"
regEx.Global = True
regEx.IgnoreCase = False
s1 = regEx.Replace(sExample, " ")

' Remove leading zeros (at beginning of line)
regEx.Pattern = "^0"
regEx.Global = True
regEx.IgnoreCase = False
sFinal = regEx.Replace(s1, "")

MsgBox sFinal

End Sub


- Robert Crandall

Yep, too much typing for me! I already have functions for various filtering
needs; here's one for removing leading zeros...


Function NoPad_Zeros$(sText$)
' Returns a string with no leading zeros
Dim vTmp, n&
Application.Volatile

vTmp = Split(sText, " ")
For n = LBound(vTmp) To UBound(vTmp)
vTmp(n) = CLng(vTmp(n))
Next 'n
NoPad_Zeros = Join(vTmp, " ")
End Function

...that you can call from code OR use as a cell formula.

In the IW:
?nopad_zeros("01 07 08 22 88 06 04")
Returns 1 7 8 22 88 6 4

In a cell:
A1 contains 01 07 08 22 88 06 04
B1 contains =nopad_zeros(A1)
displays 1 7 8 22 88 6 4

--
Garry

Free usenet access at http://www.eternal-september.org
Classic VB Users Regroup!
comp.lang.basic.visual.misc
microsoft.public.vb.general.discussion

For example...


Function FilterString$(ByVal TextIn$, Optional IncludeChars$, _
Optional IncludeLetters As Boolean = True, _
Optional IncludeNumbers As Boolean = True)
' Filters out all unwanted characters in a string.
' Arguments: TextIn The string being filtered.
' IncludeChars [Optional] Any non alpha-numeric characters to
keep.
' IncludeLetters [Optional] Keeps any letters.
' IncludeNumbers [Optional] Keeps any numbers.
'
' Returns: String containing only wanted characters.
' Comments: Works very fast using the Mid$() function over other methods.

Const sSource As String = "FilterString()"

'The basic characters to always keep by default
Const sLetters As String = "abcdefghijklmnopqrstuvwxyz"
Const sNumbers As String = "0123456789"

Dim i&, sKeepers$

sKeepers = IncludeChars
If IncludeLetters Then _
sKeepers = sKeepers & sLetters & UCase(sLetters)
If IncludeNumbers Then sKeepers = sKeepers & sNumbers

For i = 1 To Len(TextIn)
If InStr(sKeepers, Mid$(TextIn, i, 1)) Then _
FilterString = FilterString & Mid$(TextIn, i, 1)
Next
End Function 'FilterString()

In the IW:
?nopad_zeros(filterstring("Part# 0000006004",,false))
Returns 6004

?nopad_zeros(filterstring("Part# 0000060040",,false))
Returns 60040

--
Garry

Free usenet access at http://www.eternal-september.org
Classic VB Users Regroup!
comp.lang.basic.visual.misc
microsoft.public.vb.general.discussion