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#1
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Finding a string of 7 numbers in cell contents
Hi,
I have a list of cells, each containing address, tel. no. and customer ID for a specific customer. I'm looking to find the customer ID from a each cell and copy to another cell using vb. The customer ID could appear at the beginning, middle or end of the cell contents, however is always 7 digits long with a space at each side of it (unless it is the first part of the cell contents in which case there would be no space at the beginning). The complexity is that the cell also contains telephone numbers, however these are longer than 7 digits. Your help / ideas on how I could start tackling this would be much appreciated. Thanks, |
#2
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Finding a string of 7 numbers in cell contents
Try something like this:
Sub test() Dim i As Long Dim c As Long Dim n As Long Dim arr Dim arr2 Dim arr3 'test data range arr = Range(Cells(1), Cells(5, 1)) ReDim arr3(1 To UBound(arr), 1 To 1) For i = 1 To UBound(arr) arr2 = Split(arr(i, 1), Chr(32)) For c = 0 To UBound(arr2) If Len(arr2(c)) = 7 Then If IsNumeric(arr2(c)) Then n = n + 1 arr3(n, 1) = arr2(c) End If End If Next c Next i Range(Cells(3), Cells(UBound(arr), 3)) = arr3 End Sub RBS "anon" wrote in message ... Hi, I have a list of cells, each containing address, tel. no. and customer ID for a specific customer. I'm looking to find the customer ID from a each cell and copy to another cell using vb. The customer ID could appear at the beginning, middle or end of the cell contents, however is always 7 digits long with a space at each side of it (unless it is the first part of the cell contents in which case there would be no space at the beginning). The complexity is that the cell also contains telephone numbers, however these are longer than 7 digits. Your help / ideas on how I could start tackling this would be much appreciated. Thanks, |
#3
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Finding a string of 7 numbers in cell contents
=regexpreplace(" "&A1&" ","^.* (\d{7}) .*$","$1")
The function regexpreplace you will find he http://www.sulprobil.com/html/regexp.html Regards, Bernd |
#4
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Finding a string of 7 numbers in cell contents
Your use of this...
If IsNumeric(arr2(c)) Then could false hit in any number of ways (see my standard warning about IsNumeric after my signature); for example, one way being if this 123.456 came before this 1234567 in the text. I would do the test this way... If arr2(c) Like "#######" Then Rick From a previous post of mine... 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 "RB Smissaert" wrote in message ... Try something like this: Sub test() Dim i As Long Dim c As Long Dim n As Long Dim arr Dim arr2 Dim arr3 'test data range arr = Range(Cells(1), Cells(5, 1)) ReDim arr3(1 To UBound(arr), 1 To 1) For i = 1 To UBound(arr) arr2 = Split(arr(i, 1), Chr(32)) For c = 0 To UBound(arr2) If Len(arr2(c)) = 7 Then If IsNumeric(arr2(c)) Then n = n + 1 arr3(n, 1) = arr2(c) End If End If Next c Next i Range(Cells(3), Cells(UBound(arr), 3)) = arr3 End Sub RBS "anon" wrote in message ... Hi, I have a list of cells, each containing address, tel. no. and customer ID for a specific customer. I'm looking to find the customer ID from a each cell and copy to another cell using vb. The customer ID could appear at the beginning, middle or end of the cell contents, however is always 7 digits long with a space at each side of it (unless it is the first part of the cell contents in which case there would be no space at the beginning). The complexity is that the cell also contains telephone numbers, however these are longer than 7 digits. Your help / ideas on how I could start tackling this would be much appreciated. Thanks, |
#5
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Finding a string of 7 numbers in cell contents
Yes, I am aware of all that, but thought it might be OK with the data of the
OP. You could always do something like this if needed: If IsNumeric(arr2(c)) Then If CLng(arr2(c)) = arr2(c) Then n = n + 1 arr3(n, 1) = arr2(c) End If End If I am sure he will come back if more coding is needed. RBS "Rick Rothstein (MVP - VB)" wrote in message ... Your use of this... If IsNumeric(arr2(c)) Then could false hit in any number of ways (see my standard warning about IsNumeric after my signature); for example, one way being if this 123.456 came before this 1234567 in the text. I would do the test this way... If arr2(c) Like "#######" Then Rick From a previous post of mine... 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 "RB Smissaert" wrote in message ... Try something like this: Sub test() Dim i As Long Dim c As Long Dim n As Long Dim arr Dim arr2 Dim arr3 'test data range arr = Range(Cells(1), Cells(5, 1)) ReDim arr3(1 To UBound(arr), 1 To 1) For i = 1 To UBound(arr) arr2 = Split(arr(i, 1), Chr(32)) For c = 0 To UBound(arr2) If Len(arr2(c)) = 7 Then If IsNumeric(arr2(c)) Then n = n + 1 arr3(n, 1) = arr2(c) End If End If Next c Next i Range(Cells(3), Cells(UBound(arr), 3)) = arr3 End Sub RBS "anon" wrote in message ... Hi, I have a list of cells, each containing address, tel. no. and customer ID for a specific customer. I'm looking to find the customer ID from a each cell and copy to another cell using vb. The customer ID could appear at the beginning, middle or end of the cell contents, however is always 7 digits long with a space at each side of it (unless it is the first part of the cell contents in which case there would be no space at the beginning). The complexity is that the cell also contains telephone numbers, however these are longer than 7 digits. Your help / ideas on how I could start tackling this would be much appreciated. Thanks, |
#6
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Finding a string of 7 numbers in cell contents
Yes, I am aware of all that, but thought it might be OK with the data
of the OP. It probably would be, but why take the chance when the Like comparison is so easy and foolproof to perform. You could always do something like this if needed: If IsNumeric(arr2(c)) Then If CLng(arr2(c)) = arr2(c) Then False hits are still possible with this. Off the top of my head... arr2(c) = "(12345)" arr2(c) = "$12345$" {substitute regional currency symbol or $} arr2(c) = "12,3,45" arr2(c) = "&123456" Rick |
#7
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Finding a string of 7 numbers in cell contents
Yes, you are right there. Have tested and all your examples passed the test.
In that case Like "#######" may be the best test. Not sure how fast it is and that could be an issue. How would you test if the number of figures was variable? RBS "Rick Rothstein (MVP - VB)" wrote in message ... Yes, I am aware of all that, but thought it might be OK with the data of the OP. It probably would be, but why take the chance when the Like comparison is so easy and foolproof to perform. You could always do something like this if needed: If IsNumeric(arr2(c)) Then If CLng(arr2(c)) = arr2(c) Then False hits are still possible with this. Off the top of my head... arr2(c) = "(12345)" arr2(c) = "$12345$" {substitute regional currency symbol or $} arr2(c) = "12,3,45" arr2(c) = "&123456" Rick |
#8
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Finding a string of 7 numbers in cell contents
Yes, you are right there. Have tested and all your examples passed the
test. In that case Like "#######" may be the best test. Not sure how fast it is and that could be an issue. I'm not sure how Like compares (no pun intended<g) to IsNumeric speedwise; but I know it is slower than InStr, so it should not be use in place of that when testing for substring inclusion within a larger text string. How would you test if the number of figures was variable? This will test a number string to see if it is composed of all digits, no matter how many digits (literally, no matter how many digits; well, up to 2 billion or so digits at least)... If NumberString Like String(Len(NumberString), "#") Then Rick |
#9
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Finding a string of 7 numbers in cell contents
Have compared Like "#######" with IsNumeric etc. and depending on the
supplied value Like can be a lot slower, I would say about 5 times slower: Option Explicit Private lStartTime As Long Private Declare Function timeGetTime Lib "winmm.dll" () As Long Sub test() Dim i As Long Dim v Dim b As Boolean 'v = 1234567 'v = "1234567" v = Cells(1) StartSW For i = 0 To 100000 b = IsDigitsOnly(v) Next i StopSW MsgBox b End Sub Function IsDigitsOnly(v As Variant) As Boolean IsDigitsOnly = v Like String(Len(v), "#") End Function Function IsDigitsOnly2(v As Variant) As Boolean If IsNumeric(v) Then If CLng(v) = v Then IsDigitsOnly2 = True End If End If End Function Sub StartSW() lStartTime = timeGetTime() End Sub Function StopSW(Optional bMsgBox As Boolean = True, _ Optional vMessage As Variant, _ Optional lMinimumTimeToShow As Long = -1) As Variant Dim lTime As Long lTime = timeGetTime() - lStartTime If lTime lMinimumTimeToShow Then If IsMissing(vMessage) Then StopSW = lTime Else StopSW = lTime & " - " & vMessage End If End If If bMsgBox Then If lTime lMinimumTimeToShow Then MsgBox "Done in " & lTime & " msecs", , vMessage End If End If End Function But if the test value is supplied like this: "1234567" then Like is faster. Probably speed doesn't come into it for the OP, so I take Like will be best. RBS "Rick Rothstein (MVP - VB)" wrote in message ... Yes, you are right there. Have tested and all your examples passed the test. In that case Like "#######" may be the best test. Not sure how fast it is and that could be an issue. I'm not sure how Like compares (no pun intended<g) to IsNumeric speedwise; but I know it is slower than InStr, so it should not be use in place of that when testing for substring inclusion within a larger text string. How would you test if the number of figures was variable? This will test a number string to see if it is composed of all digits, no matter how many digits (literally, no matter how many digits; well, up to 2 billion or so digits at least)... If NumberString Like String(Len(NumberString), "#") Then Rick |
#10
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Finding a string of 7 numbers in cell contents
On Wed, 30 Jul 2008 05:08:53 -0700 (PDT), anon
wrote: Hi, I have a list of cells, each containing address, tel. no. and customer ID for a specific customer. I'm looking to find the customer ID from a each cell and copy to another cell using vb. The customer ID could appear at the beginning, middle or end of the cell contents, however is always 7 digits long with a space at each side of it (unless it is the first part of the cell contents in which case there would be no space at the beginning). The complexity is that the cell also contains telephone numbers, however these are longer than 7 digits. Your help / ideas on how I could start tackling this would be much appreciated. Thanks, Place the UDF =CustomerID(cell_ref) in the destination cell. Place the code below in a regular module: ================================ Option Explicit Function CustomerID(str) As String Dim re As Object, mc As Object Set re = CreateObject("vbscript.regexp") re.Pattern = "\b\d{7}\b" If re.test(str) = True Then Set mc = re.Execute(str) CustomerID = mc(0).Value End If End Function =============================== --ron |
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