For starters, there are LOTS of questions that have been asked with this topic. However all the ones I kept clicking on were in languages other than VBA and I did not understand the syntax of those languages.
When I did a google search I found this answer which seemed promising. AH FIDDLE STICKS! I just realized that answer for VB and probably explains why its not working in my VBA
Situation
I have a variable called DimScale that is an integer. I want to create a string called DimName that will start with "mm-" and be following by the integer from DimScale with leading 0s such that there are a minimum of characters after "mm-".
IF DimScale = 25
Then DimName = "mm-0025"
IF DimScale = 235
Then DimName = "mm-0235"
Note Dimscale >=1 and <= 9999
What I have tried
Dim Dimscale as Integer
Dim Dimension_Style_Name as String
String.Format("{0:0000}", DimScale)
Dimension_Style_Name = DimScale$
Dimension_Style_Name.Format("{0:0000}", DimScale)
I have read the gist too that Dimscale get converted to a string and then is sent through a loop of adding a leading zero until the length of the string equals the 4 characters in my case for the integer part.
I have also seen the case with IF statments where IF Dimscale <10 then "000"& If Dimscale <100 then "00"& etc.
Is there a way to do it like like the VB method in VBA?
maybe:
DimName = "mm-" & format(DimScale,"0000")
As per #MathieuGuindon valuable (as usual) contribution:
Format (fully-qualified VBA.Strings.Format) takes a Variant parameter, and returns a Variant - you can also use its little brother Format$, which takes a String and returns a String, eliminating implicit conversions along the way
I had a similar need to apply leading zeros ( 12 to 00012 ) to a specified range. But everything I'd found thus-far used an iterative cell-by-cell approach. I found an older but still valuable posting from SiddHarth Rout. His posting pertains to case conversion ( lower to upper case ) but I found it adapted nicely to applying leading zeros.
Here is link to SiddHarth's posting:
Convert an entire range to uppercase without looping through all the cells
Here is the adaptation for applying leading zeros to a specified range:
Sub rngLeadingZeros(rng As Range, nbrZeros As Integer)
' Add leading zeros to a specified range.
Dim strZeros As String
Dim x As Integer
'build string as required for text() function:
For x = 1 To nbrZeros
strZeros = strZeros & "0"
Next
'make sure the range is formatted as text:
rng.NumberFormat = "#"
'apply the format to the range:
rng = Evaluate("index(text(" & rng.Address & ", """ & strZeros & """),)")
End Sub
Sub testZ()
With ActiveSheet
rngLeadingZeros .Range("e3:e9"), 5
End With
End Sub
Related
I have a row with values something like this, How to extract first two digits that come after the text 'ABCD' to another cell, any formula or vba? There may be a few chars in between or sometimes none.
ABCD 10 sadkf sdfas
ABCD-20sdf asdf
ABCD 40
ABCD50 asdf
You can do this with a worksheet formula. No need for VBA.
Assuming you do not need to test for the presence of two digits:
=MID(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890")),2)
If you need to test for the presence of two digits, you can try:
=IF(ISNUMBER(-RIGHT(MID(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890")),2),1)),MID(A1,MIN(FIND({1,2,3,4,5,6,7,8,9,0},A1&"1234567890")),2),"Invalid")
In general, it is always a good idea to show some code in StackOverflow. Thus, you show that you have tried something and you give some directions for the answer.
Concerning the first two digits extract, there are many ways to do this. Starting from RegEx and finishing with a simple looping of the chars and checking each one of them.
This is the loop option:
Public Function ExtractTwoDigits(inputString As String) As Long
Application.Volatile
Dim cnt As Long
Dim curChar As String
For cnt = 1 To Len(inputString)
curChar = Mid(inputString, cnt, 1)
If IsNumeric(curChar) Then
If Len(ExtractTwoDigits) Then
ExtractTwoDigits = ExtractTwoDigits & curChar
Exit Function
Else
ExtractTwoDigits = curChar
End If
End If
Next cnt
ExtractTwoDigits = -1
End Function
Application.Volatile makes sure that the formula recalculates every time;
-1 is the answer if no two digits exist in the inputString;
IsNumeric checks whether the string inside is numeric;
As a further step, you may try to make the function a bit robust, extracting the first 1, 3, 4 or 5 digits, depending on a parameter that you put. Something like this =ExtractTwoDigits("tarato123ra2",4), returning 1232.
RegEx Version:
Public Function GetFirstTwoNumbers(ByVal strInput As String) As Integer
Dim reg As New RegExp, matches As MatchCollection
With reg
.Global = True
.Pattern = "(\d{2})"
End With
Set matches = reg.Execute(strInput)
If matches.Count > 0 Then
GetFirstTwoNumbers = matches(0)
Else
GetFirstTwoNumbers = -1
End If
End Function
You have to enable Microsoft Regular Expressions 5.5 under extras->references. The pattern (\d{2}) matches 2 digits, return value is the number, if not existing -1.
Note: it only extracts 2 successive numbers.
If you place this function into a module, you can use it like normal formula.
Here a great site to to get into regEx.
I am trying to create a macro for Word 2013 that does the following: the macro should capture the value of a cell of a word table and then add another value and paste the result in another cell of the same table.
My code so far is:
Sub prueba()
Dim a As String, b As String, c As String
Dim entero1 As Double, entero2 As Double
Dim resultado As Double
Dim tabla1 As Table
Set tabla1 = ActiveDocument.Tables(1)
a = tabla1.Cell(Row:=1, Column:=3).Range
entero1 = CDbl(a)
End Sub
But when I run it I get an error 13
To evaluate the error add the following two lines to validate if the data type obtained in "a" was a string
MsgBox (TypeName(a))
MsgBox (a)
And I got the following
I believe that the CDbl function does not finish converting the string to double because as they see the chain has a small square, what is not like to erase it so that the conversion is achieved.
Thank you very much for your help.
One way of extracting just the numeric portion of the Range would be to use the Val function, e.g.
entero1 = Val(a)
If the string a contained, for instance, 123.23XYZ4567 then Val(a) would return the number 123.23.
That should ensure that the non-numeric character that you are getting at the end of your Range is removed.
The answer provided by YowE3K is elegant and has my vote. For further information:
That 'small square' is the end of cell marker which is part of Cell.Range.Text (.Text is the default property returned when returning a range object is inappropriate).
To actually remove the end of cell marker (Chr(13) & Chr(7)) you can use something like this:
?CDbl(Replace$(Selection.Range.Cells(1).Range.Text, Chr(13) & Chr(7), vbNullString))
A possible advantage of this approach is that it may provide better opportunity to trap errors if you are only expecting numeric characters.
Perhaps this is a simple solution for most, but I can't get this to work like it should according to syntax.
I have this line of text "Part Number123456Price$50.00"
I want to pull the part number out of it, so I use this function...
str = Mid(str, str.IndexOf("Part Number") + 12, str.IndexOf("Price"))
My results are str = "123456Price$50.0" every time. I know the part number can vary in length so I need a solid solution of pulling this out.
It can be confusing to mix the legacy VB string methods (such as Mid) with the .Net string methods (like IndexOf). The VB methods use 1 as the index of the first character while the .Net methods use 0.
The following code will extract the part number from a string
Dim str As String = "Part Number123456Price$50.00"
Dim iPart As Integer = str.IndexOf("Part Number") + 11
Dim iPrice As Integer = str.IndexOf("Price")
str = str.Substring(iPart, iPrice - iPart).Trim
The Mid() function of Visual Basic is documented as having three arguments: (1) a string, (2) the beginning location in the string, and (3) the number of characters to copy.
So if your string is "Part Number123456Price$50.00" and you want to pull the part number as a series of digits, the "123456" part of the string, using the Mid() function then you need to find the beginning of the part number digit string and to then know the number of digits.
If your string is in the variable str then you can find the offset by something like str.IndexOf("Number") + len("Number") which will provide the offset to after the string "Number".
Next you need to find the number of digits so you would do something like str.IndexOf("Price") to find where the text "Price" begins and then subtract from that offset the offset of where the digits begin.
The result of all of this is you need a bit of code something like the following. I have not tested this source as I am not a VB programmer so it may need a tweak and you might want to put some checks on data validity as well.
Dim TextNumber as String = "Number"
Dim TextPrice as String = "Price"
iOffset = str.IndexOf(TextNumber) + len(TextNumber)
str = Mid(str, iOffset, str.IndexOf(TextPrice) - iOffset)
Alternatively, if Price is always the format $00.00, this will also work.
Dim str as String = "Part Number123456Price$50.00"
str = str.Remove(str.IndexOf("Price"))
How do I get the length of character between beginning with space and ending with * Here is the image. Column B shows the total len before dasher(-) and my code
Sub xn()
Dim x As Integer
x = 1
If Worksheet("Sheet1").Range("A"& x).len(Right," ") Or _
Worksheet("Sheet1").Range("A"&x)len(Left,"-") Then
len(totallen)
End If
x = x + 1
End Sub
The code posted has multiple issues:
Worksheet is not a valid object - you need to use Worksheets.
.len is not a property of a Range object.
Even in .len was a property of a Range, you would need a
de-reference operator (aka '.') in here: Range("A"&x)len(Left,"-")
If you intend to use the function Len(), it only takes one argument.
You apparently are trying to loop, but you need to use either a For
or For Each loop - it won't loop automatically when you increment x
at the bottom of the sub.
Right is a function, but you're calling it without arguments and they are not optional.
Similarly, Left is a function, but you're also calling it without
the required arguments.
totallen is not declared anywhere, so Len(totallen) will assume
that totallen is a Variant (default for undeclared variables), then
cast it to a String, and then always return 0 because it has never
been given a value.
Anything else I may have missed.
The solution is to use the InStr function. It returns the location in a string of a given sub-string.
Sub xn()
Dim x As Long
Dim sheet As Worksheet
Set sheet = ActiveWorkbook.Worksheets("Sheet1")
For x = 1 To sheet.Range("A" & sheet.Rows.Count).End(xlUp).Row
sheet.Cells(x, 2) = InStr(1, sheet.Cells(x, 1), "-") - 1
Next x
End Sub
I'd also recommend taking a look at the MSDN article on Looping Through a Range of Cells (2003 vintage, but still valid), and Error Finding Last Used cell In VBA.
I need help with extracting 5-digit numbers only from one column to another in Excel 2010. These numbers can be in any position of the string (beginning of the string, anywhere in the middle, or at the end). They can be within brackets or quotes like:
(15478) or "15478" or '15478' or [15478]
I need to ignore any numbers that are less than 5 digits and include numbers that start with 1 or more leading zeros (like 00052, 00278, etc.) and ensure that leading zeros are copied over to the next column. Could someone help me with either creating a formula or UDF?
Here is a formula-based alternative that will extract the first 5 digit number found in cell A1. I tend to prefer reasonably simple formula solutions over VBA in most situations as formulas are more portable. This formula is an array formula and thus must be entered with Ctrl+Shift+Enter. The idea is to split the string up into every possible 5 character chunk and test each one and return the first match.
=MID(A1,MIN(IF(NOT(ISERROR(("1"&MID(A1,ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE)),5)&".1")*1))*ISERROR(MID(A1,ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE))+5,1)*1)*ISERROR(MID(A1,ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE))-1,1)*1),ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE)),9999999999)),5)
Let's break this down. First we have an expression I used twice to return an array of numbers from 1 up to 4 less than the length of your initial text. So if you have a string of length 10 the following will return {1,2,3,4,5,6}. Hereafter the below formula will be referred to as rowlist. I used R1C1 notation to avoid potential circular references.
ROW(INDIRECT("R1C[1]:R"&(LEN(A1)-4)&"C[1]",FALSE))
Next we will use that array to split the text into an array of 5 letter chunks and test each chunk. The test being performed is to prepend a "1" and append ".1" then verify the chunk is numeric. The prepend and append eliminate the possibility of white space or decimals. We can then check the character before and the character after to make sure they are not numbers. Hereafter the below formula will be referred to as isnumarray.
NOT(ISERROR(("1"&MID(A1,rowlist,5)&".1")*1))
*ISERROR(MID(A1,rowlist+5,1)*1)
*ISERROR(MID(A1,rowlist-1,1)*1)
Next we need to find the first valid 5 digit number in the string by returning the current index from a duplicate of the rowlist formula and returning a large number for non-matches. Then we can use the MIN function to grab that first match. Hereafter the below will be referred to as minindex.
MIN(IF(isnumarray,rowlist,9999999999))
Finally we need to grab the numeric string that started at the index returned by the MIN function.
MID(A1,minindex,5)
The following UDF will return the first five digit number in the string, including any leading zero's. If you need to detect if there is more than one five digit number, the modifications are trivial. It will return a #VALUE! error if there are no five-digit numbers.
Option Explicit
Function FiveDigit(S As String, Optional index As Long = 0) As String
Dim RE As Object
Set RE = CreateObject("vbscript.regexp")
With RE
.Pattern = "(?:\b|\D)(\d{5})(?:\b|\D)"
.Global = True
FiveDigit = .Execute(S)(index).submatches(0)
End With
End Function
As you may see from the discussion between Mark and myself, some of your specifications are unclear. But if you would want to exclude decimal numbers, when the decimal portion has five digits, then the regex pattern in my code above should be changed:
.Pattern = "(?:\d+\.\d+)|(?:\b|\D)(\d{5})(?:\b|\D)"
I just wrote this UDF for you , basic but will do it...
It will find the first 5 consecutive numbers in a string, very crude error checking so it just says Error if anything isn't right
Public Function GET5DIGITS(value As String) As String
Dim sResult As String
Dim iLen As Integer
sResult = ""
iLen = 0
For i = 1 To Len(value)
If IsNumeric(Mid(value, i, 1)) Then
sResult = sResult & Mid(value, i, 1)
iLen = iLen + 1
Else
sResult = ""
iLen = 0
End If
If iLen = 5 Then Exit For
Next
If iLen = 5 Then
GET5DIGITS = Format(sResult, "00000")
Else
GET5DIGITS = "Error"
End If
End Function