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#11
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pulling out Numbers
hi Mike,
the problem is , it removes the zero at the begining of a numbers after pulling out them,like: aab0125 turns to 125 instead of 0125 "Mike H" wrote: Another way. I'm a bit cofused why the previous method only removed th dash but try this Sub anotherway() For Each cell In Selection For a = 1 To Len(cell) Select Case Mid(cell, a, 1) Case "0" To "9" newstring = newstring & Mid(cell, a, 1) End Select Next a cell.Value = newstring ActiveCell.NumberFormat = "General" newstring = "" Next cell End Sub Mike "peyman" wrote: hi Mike, it just removes dash "-" from string!! not letters. "Mike H" wrote: With a macro:- Sub removeletters() For Each cell In Selection For a = 1 To Len(cell) If Asc(Mid(cell, a, 1)) 47 And Asc(Mid(cell, a, 1)) 59 Or _ Asc(Mid(cell, a, 1)) 64 And Asc(Mid(cell, a, 1)) 91 Then _ newstring = newstring & Mid(cell, a, 1) Next cell.Value = newstring ActiveCell.NumberFormat = "General" newstring = "" Next End Sub Select the cells to convert and run this Mike "peyman" wrote: hi, how can I pull out numbers from a string?like: aa012985 to 012985 12ab-059 to 12059 the letters or characters might be either at the first, middle or at the end of a string. thanx in advance. |
#12
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pulling out Numbers
hi Ron,
the problem in Mikes macro is , it removes the zero at the begining of a numbers after pulling out them,like: aab0125 turns to 125 instead of 0125 "Ron Rosenfeld" wrote: On Tue, 21 Aug 2007 11:18:02 -0700, Mike H wrote: With a macro:- Sub removeletters() For Each cell In Selection For a = 1 To Len(cell) If Asc(Mid(cell, a, 1)) 47 And Asc(Mid(cell, a, 1)) 59 Or _ Asc(Mid(cell, a, 1)) 64 And Asc(Mid(cell, a, 1)) 91 Then _ newstring = newstring & Mid(cell, a, 1) Next cell.Value = newstring ActiveCell.NumberFormat = "General" newstring = "" Next End Sub Your For/Next loop can be simplified: For a = 1 To Len(cell) If Mid(cell, a, 1) Like "#" Then newstring = newstring & Mid(cell, a, 1) End If Next --ron |
#13
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pulling out Numbers
Hi,
It doesn't I suspect the zero is a letter "o" or "O" Mike "peyman" wrote: hi Mike, the problem is , it removes the zero at the begining of a numbers after pulling out them,like: aab0125 turns to 125 instead of 0125 "Mike H" wrote: Another way. I'm a bit cofused why the previous method only removed th dash but try this Sub anotherway() For Each cell In Selection For a = 1 To Len(cell) Select Case Mid(cell, a, 1) Case "0" To "9" newstring = newstring & Mid(cell, a, 1) End Select Next a cell.Value = newstring ActiveCell.NumberFormat = "General" newstring = "" Next cell End Sub Mike "peyman" wrote: hi Mike, it just removes dash "-" from string!! not letters. "Mike H" wrote: With a macro:- Sub removeletters() For Each cell In Selection For a = 1 To Len(cell) If Asc(Mid(cell, a, 1)) 47 And Asc(Mid(cell, a, 1)) 59 Or _ Asc(Mid(cell, a, 1)) 64 And Asc(Mid(cell, a, 1)) 91 Then _ newstring = newstring & Mid(cell, a, 1) Next cell.Value = newstring ActiveCell.NumberFormat = "General" newstring = "" Next End Sub Select the cells to convert and run this Mike "peyman" wrote: hi, how can I pull out numbers from a string?like: aa012985 to 012985 12ab-059 to 12059 the letters or characters might be either at the first, middle or at the end of a string. thanx in advance. |
#14
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pulling out Numbers
Mike,
It uses "Regular Expressions" which is a feature included with VBA. The CreateObject method sets up the reference to Microsoft VBScript Regular Expressions 5.5. Instead of that method, one can also select Tools/References (from the VBEditor top menu) and select that as a reference. This latter method has the advantage that properties will pop up when you are writing the macro, as they do for other objects. The Pattern "\D" means "match every character that is not a digit (not 0..9). The replace method then looks at "str" and replaces every match with nothing (""), thereby removing all the non-digits. http://support.microsoft.com/default...02&Product=vbb http://msdn2.microsoft.com/en-us/library/1400241x.aspx On Tue, 21 Aug 2007 11:36:00 -0700, Mike H wrote: Ron, Excellent. any chance of a quick rundown on how it works or a web reference please? Mike H "Ron Rosenfeld" wrote: On Tue, 21 Aug 2007 10:28:00 -0700, peyman wrote: hi, how can I pull out numbers from a string?like: aa012985 to 012985 12ab-059 to 12059 the letters or characters might be either at the first, middle or at the end of a string. thanx in advance. You can use a UDF. To enter the UDF, alt-F11 opens the VB Editor. Ensure your project is highlighted in the Project Explorer window, then Insert/Module and paste the code below into the window that opens. To use this UDF, enter =ExtrNums(cell_ref) into some cell. =============================== Option Explicit Function ExtrNums(str As String) Dim re As Object Set re = CreateObject("VBScript.RegExp") re.Global = True re.Pattern = "\D" ExtrNums = re.Replace(str, "") End Function ================================ --ron --ron |
#16
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pulling out Numbers
Thanks Ron,
I'm only on my 4th read of John Walkenbach's 2003 bible so perhaps some way off this yet. Mike "Ron Rosenfeld" wrote: Mike, It uses "Regular Expressions" which is a feature included with VBA. The CreateObject method sets up the reference to Microsoft VBScript Regular Expressions 5.5. Instead of that method, one can also select Tools/References (from the VBEditor top menu) and select that as a reference. This latter method has the advantage that properties will pop up when you are writing the macro, as they do for other objects. The Pattern "\D" means "match every character that is not a digit (not 0..9). The replace method then looks at "str" and replaces every match with nothing (""), thereby removing all the non-digits. http://support.microsoft.com/default...02&Product=vbb http://msdn2.microsoft.com/en-us/library/1400241x.aspx On Tue, 21 Aug 2007 11:36:00 -0700, Mike H wrote: Ron, Excellent. any chance of a quick rundown on how it works or a web reference please? Mike H "Ron Rosenfeld" wrote: On Tue, 21 Aug 2007 10:28:00 -0700, peyman wrote: hi, how can I pull out numbers from a string?like: aa012985 to 012985 12ab-059 to 12059 the letters or characters might be either at the first, middle or at the end of a string. thanx in advance. You can use a UDF. To enter the UDF, alt-F11 opens the VB Editor. Ensure your project is highlighted in the Project Explorer window, then Insert/Module and paste the code below into the window that opens. To use this UDF, enter =ExtrNums(cell_ref) into some cell. =============================== Option Explicit Function ExtrNums(str As String) Dim re As Object Set re = CreateObject("VBScript.RegExp") re.Global = True re.Pattern = "\D" ExtrNums = re.Replace(str, "") End Function ================================ --ron --ron |
#17
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pulling out Numbers
On Tue, 21 Aug 2007 12:16:02 -0700, Mike H
wrote: Thanks Ron, I'm only on my 4th read of John Walkenbach's 2003 bible so perhaps some way off this yet. Mike Regular expressions are extremely powerful tools for string manipulation. Harlan Grove turned me on to them, and I've found them quite worthwhile, although I'm still a novice in their use. --ron |
#18
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pulling out Numbers
hi Ron, Can we do it reverse.pulling out only letters. aa125df36 to aadf???? thank you "Ron Rosenfeld" wrote: On Tue, 21 Aug 2007 12:16:02 -0700, Mike H wrote: Thanks Ron, I'm only on my 4th read of John Walkenbach's 2003 bible so perhaps some way off this yet. Mike Regular expressions are extremely powerful tools for string manipulation. Harlan Grove turned me on to them, and I've found them quite worthwhile, although I'm still a novice in their use. --ron |
#19
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pulling out Numbers
On Tue, 21 Aug 2007 15:12:01 -0700, peyman
wrote: hi Ron, Can we do it reverse.pulling out only letters. aa125df36 to aadf???? thank you It's pretty simple. In the code I posted, you just need to change the "\D", which selects all non-digits for removal, to a "\d" which selects all digits for removal. But for more flexibility, you could include the arguments for what to find, and what to replace it with, in the function code. For example, with this UDF: ===================================== Option Explicit Function ReSub(str As String, FindText As String, ReplaceWith As String) Dim re As Object Set re = CreateObject("VBScript.RegExp") re.Global = True re.Pattern = FindText ReSub = re.Replace(str, ReplaceWith) End Function ====================================== You could use this formula to replace all digits with a null string, resulting in pulling out only letters: =resub(A1,"\d","") If you wanted to pull out only digits, then: =resub(A1,"\D","") If you wanted to replace the digits with a tilde, you could use: =resub(A1,"\d","~") If you wanted to replace all of the letters with the word Peyman, with leading and ending spaces, then: =resub(A6,"\D"," Peyman ") And many other solutions --ron |
#20
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pulling out Numbers
thanx Ron.It's excellent
"Ron Rosenfeld" wrote: On Tue, 21 Aug 2007 15:12:01 -0700, peyman wrote: hi Ron, Can we do it reverse.pulling out only letters. aa125df36 to aadf???? thank you It's pretty simple. In the code I posted, you just need to change the "\D", which selects all non-digits for removal, to a "\d" which selects all digits for removal. But for more flexibility, you could include the arguments for what to find, and what to replace it with, in the function code. For example, with this UDF: ===================================== Option Explicit Function ReSub(str As String, FindText As String, ReplaceWith As String) Dim re As Object Set re = CreateObject("VBScript.RegExp") re.Global = True re.Pattern = FindText ReSub = re.Replace(str, ReplaceWith) End Function ====================================== You could use this formula to replace all digits with a null string, resulting in pulling out only letters: =resub(A1,"\d","") If you wanted to pull out only digits, then: =resub(A1,"\D","") If you wanted to replace the digits with a tilde, you could use: =resub(A1,"\d","~") If you wanted to replace all of the letters with the word Peyman, with leading and ending spaces, then: =resub(A6,"\D"," Peyman ") And many other solutions --ron |
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