This question already has answers here:
Add leading zeroes/0's to existing Excel values to certain length
(8 answers)
Closed 3 years ago.
I have values of 120, 136, 7120, 72136. The maximum string length should be 5, how do I go about making 120 "00120" and 136 "00136" etc?
Single line will work
VBA
Range("A1").Value = "'" & Format(Range("A1").Value, "00000")
Excel
=TEXT(A1,"00000")
In your simple case you can try something simple like this:
Sub FiveCharString()
Dim myStr As String
myStr = "136"
If Len(myStr) = 2 Then
myStr = "000" & myStr
ElseIf Len(myStr) = 3 Then
myStr = "00" & myStr
ElseIf Len(myStr) = 4 Then
myStr = "0" & myStr
End If
Debug.Print myStr
End Sub
Returns 00136.
Function FillWithZero(number as long, digitCount as long) as string
FillWithZero = Right(String(digitCount , "0") & number , digitCount)
End Function
Use a custom number format in your cells.
See Using a custom number format to display leading zeros
or Keeping leading zeros and large numbers
Or .NumberFormat = "00000" on your range.
I don't recommend to convert it into a string (unless it is something like a serial number that is not treated as an actual number).
A simpler version compared to Dean's
Sub StrLength()
Dim i As Long, str As String
str = "136"
i = Len(str)
StrLength = String(ExpectedLength - Len(str), "0") & str
End Sub
Small subroutines as these can easily be used as Functions, where you call the function in a regular sub. For example, when you are looping through a range of cells:
Function StrLength(str As String, ExpectedLength As Long) As String
Dim i As Long
i = Len(str)
StrLength = String(ExpectedLength - Len(str), "0") & str
End Function
Sub Test()
Dim c As Range
For each c In ThisWorkbook.Sheets(1).Range("A1:B200")
If Len(c.Value) < 5 Then c.Value = StrLength(Str:=c.Value, ExpectedLength:=5)
Next c
End Sub
Related
I have created a VBA code to remove all special characters available in a column. As an example I have a Alphanumeric character with some special characters in every cells of a column:
Suppose in a cell I have a value: abc#123!-245
After executing my code I got output abc 123 245
Here my code is working fine to remove all the special characters. My code is given below:
Sub ReplaceSpecial()
Dim cel As Range
Dim strVal As String
Dim i As Long
Application.ScreenUpdating = False
For Each cel In Selection
strVal = cel.Value
For i = 1 To Len(strVal)
Select Case Asc(Mid(strVal, i, 1))
Case 32, 48 To 57, 65 To 90, 97 To 122
' Leave ordinary characters alone
Case Else
Mid(strVal, i, 1) = " "
End Select
Next i
cel.Value = strVal
Next cel
Application.ScreenUpdating = True
End Sub
Now if I want to remove the space for my output so that output should look like abc123245, how to do that in VBA?
Input: abc#123!-245
Current Output: abc 123 245
Required Output: abc123245
You could construct a new string with just the permitted characters.
Sub ReplaceSpecial()
Dim cel As Range
Dim strVal As String, temp As String
Dim i As Long
Application.ScreenUpdating = False
For Each cel In Selection
strVal = cel.Value
temp = vbNullString
For i = 1 To Len(strVal)
Select Case Asc(Mid(strVal, i, 1))
Case 32, 48 To 57, 65 To 90, 97 To 122
temp = temp & Mid(strVal, i, 1)
End Select
Next i
cel.Value = temp
Next cel
Application.ScreenUpdating = True
End Sub
My sole intention for this late post was to
test some features of the â–ºApplication.Match() function (comparing a string input against valid characters) and to
demonstrate a nice way to "split" a string into single characters as alternative and possibly instructive solution (see help function String2Arr()).
I don't intend, however to show better or faster code here.
Application.Match() allows not only to execute 1 character searches in an array, but to compare even two arrays in one go,
i.e. a character array (based on an atomized string input) against an array of valid characters (blanks, all digits and chars from A to Z).
As Application.Match is case insensitive, it suffices to take e.g. lower case characters.
All findings of input chars return their position in the valid characters array (otherwise resulting in Error 2042).
Furthermore it was necessary to exclude the wild cards "*" and "?", which would have been considered as findings otherwise.
Function ValidChars(ByVal s, Optional JoinResult As Boolean = True)
'Purp: return only valid characters if space,digits,"A-Z" or "a-z"
'compare all string characters against valid characters
Dim tmp: tmp = foundCharAt(s) ' get array with found positions in chars
'overwrite tmp array
Dim i As Long, ii As Long
For i = 1 To UBound(tmp)
If IsNumeric(tmp(i)) Then ' found in valid positions
If Not Mid(s, i, 1) Like "[?*]" Then ' exclude wild cards
ii = ii + 1
tmp(ii) = Mid(s, i, 1) ' get char from original string
End If
End If
Next
ReDim Preserve tmp(1 To ii) ' reduce to new size
'join tmp elements to resulting string (if argument JoinResult = True)
ValidChars = IIf(JoinResult, Join(tmp, ""), tmp)
End Function
Help function foundCharAt()
Returns an array of found character positions in the valid chars array:
Function foundCharAt(ByVal s As String) As Variant
'Purp: return array of found character positions in chars string
'Note: (non-findings show Error 2042; can be identified by IsError + Not IsNumeric)
Dim chars: chars = String2Arr(" 0123456789abcdefghijklmnopqrstuvwxyz")
foundCharAt = Application.Match(String2Arr(s), chars, 0)
End Function
Help function String2Arr()
Assigns an array of single characters after atomizing a string input:
Function String2Arr(ByVal s As String) As Variant
'Purp: return array of all single characters in a string
'Idea: https://stackoverflow.com/questions/13195583/split-string-into-array-of-characters
s = StrConv(s, vbUnicode)
String2Arr = Split(s, vbNullChar, Len(s) \ 2)
End Function
Use a regular expression's object and replace all unwanted characters by using a negated character class. For demonstration purposes:
Sub Test()
Dim str As String: str = "abc#123!-245"
With CreateObject("vbscript.regexp")
.Global = True
.Pattern = "[^0-9A-Za-z ]"
str = .Replace(str, "")
End With
Debug.Print str
End Sub
The pattern [^0-9A-Za-z ] is a negated character class and captured everything that is not a alphanumeric or a space character. You'll find a more in-depth explaination in this online demo.
At time of writing I'm unsure if you want to leave out the space characters or not. If so, just remove the space from the pattern.
Thought I'd chuck in another alternative using the Like() operator:
For i = Len(str) To 1 Step -1
If Mid(str, i, 1) Like "[!0-9A-Za-z ]" Then
str= Application.Replace(str, i, 1, "")
End If
Next
Or with a 2nd string-type variable (as per #BigBen's answer):
For i = 1 to Len(str)
If Mid(str, i, 1) Like "[0-9A-Za-z ]" Then
temp = temp & Mid(str, i, 1)
End If
Next
If you want to build on your current effort, replace:
cel.Value = strVal
with:
cel.Value = Replace(strVal, " ", "")
Consider:
Sub ReplaceSpecial()
Dim cel As Range
Dim strVal As String
Dim i As Long
Application.ScreenUpdating = False
For Each cel In Selection
strVal = cel.Value
For i = 1 To Len(strVal)
Select Case Asc(Mid(strVal, i, 1))
Case 32, 48 To 57, 65 To 90, 97 To 122
' Leave ordinary characters alone
Case Else
Mid(strVal, i, 1) = " "
End Select
Next i
cel.Value = Replace(strVal, " ", "")
Next cel
Application.ScreenUpdating = True
End Sub
A B C
1 numbers signs **Result**
2 *001* *alpha* 001-alpha
3 *001*111*221*104* *alpha*kappa*epislon*ETA* 001-alpha, 111-kappa, 221-epislon, 104-ETA
4 *001*085* *alpha*delta* 001-alpha, 085-delta
I'm trying to concatenate the values in columns A and B into the following format under the result section. Anything helps, thanks.
Formula solution
Using Textjoin and Filterxml function, of which Textjoin available in Office 365 or Excel 2019 and Filterxml available in Excel 2013 & later versions of Excel
In C2, array formula (confirm by pressing Ctrl+Shift+Enter) copied down :
=TEXTJOIN(", ",1,IFERROR(TEXT(FILTERXML("<a><b>"&SUBSTITUTE(A2,"*","</b><b>")&"</b></a>","//b"),"000")&FILTERXML("<a><b>"&SUBSTITUTE(B2,"*","</b><b>-")&"</b></a>","//b"),""))
I'm assuming this is doable with formulas but it might get unwieldy, so perhaps a UDF like this:
Public Function JoinNumbersAndSigns(ByVal numbersRng As Range, ByVal signsRng As Range) As String
Dim nums As String
nums = numbersRng.Cells(1).Value
nums = Mid$(nums, 2, Len(nums) - 2) ' remove leading and trailing *
Dim signs As String
signs = signsRng.Cells(1).Value
signs = Mid$(signs, 2, Len(signs) - 2) ' remove leading and trailing *
Dim tempNums As Variant
tempNums = Split(nums, "*")
Dim tempSigns As Variant
tempSigns = Split(signs, "*")
Dim i As Long
For i = LBound(tempNums) To UBound(tempNums)
Dim tempString As String
Dim sep As String
tempString = tempString & sep & tempNums(i) & "-" & tempSigns(i)
sep = ", "
Next i
JoinNumbersAndSigns = tempString
End Function
In Action:
The nums = Mid$(nums, 2, Len(nums) - 2) and similar line for signs could probably be made more robust, but should work given your current data.
Here's another approach using regular expressions ...
Option Explicit
Public Function Link(vNumbers As Range, vSigns As Range) As Variant
' ADD REFERENCE TO "Microsoft VBScript Regular Expressions 5.5"
Dim vRegEx As New RegExp
Dim vNumbersMatches As MatchCollection
Dim vSignsMatches As MatchCollection
Dim vCounter As Long
' The two parameters must only reference a single cell
If vNumbers.Cells.Count <> 1 Or vSigns.Cells.Count <> 1 Then
Link = CVErr(xlErrRef)
Exit Function
End If
' use regular expression to get the numbers
vRegEx.Pattern = "([0-9]+)"
vRegEx.Global = True
vRegEx.MultiLine = True
Set vNumbersMatches = vRegEx.Execute(vNumbers.Text)
' Use regular expression to get the signs
vRegEx.Pattern = "([^\*]+)"
vRegEx.Global = True
vRegEx.MultiLine = True
Set vSignsMatches = vRegEx.Execute(vSigns.Text)
' If the number of Numbers and Signs differs, then return an error
If vNumbersMatches.Count <> vSignsMatches.Count Then
Link = CVErr(xlErrValue)
Exit Function
End If
' Loop through the Numbers and Signs, appending each set
For vCounter = 0 To vNumbersMatches.Count - 1
Link = Link & vNumbersMatches.Item(vCounter) & "-" & vSignsMatches.Item(vCounter) & IIf(vCounter < vNumbersMatches.Count - 1, " ,", "")
Next
End Function
And the output ...
As long as there will always be a correlation between the number of elements in A & B this will work
Sub SplitandConcat()
' Declare working vars
Dim lRow As Long: lRow = 2
Dim sOutputString As String
Dim iWorkIndex As Integer
Dim CommaSpace As String
While ActiveSheet.Cells(lRow, 1) <> ""
CommaSpace = ""
'Split the incoming string on delimiter
arInput1 = Split(ActiveSheet.Cells(lRow, 1), "*")
arInput2 = Split(ActiveSheet.Cells(lRow, 2), "*")
' For each non blank item in the 1st array join the corresponding item int the second
For iWorkIndex = 0 To UBound(arInput1)
If arInput1(iWorkIndex) <> "" Then
ActiveSheet.Cells(lRow, 3) = ActiveSheet.Cells(lRow, 3) & CommaSpace & arInput1(iWorkIndex) & "-" & arInput2(iWorkIndex)
CommaSpace = ", "
End If
Next iWorkIndex
' check next row
lRow = lRow + 1
Wend
End Sub
I'm facing a problem as a non dev. I have a column in Excel that contains info as such:
46843 xxxx xxx x
xxxx 65483 xxxx
xxxx xxx 65432 xxxxx 4 xx
"x" being normal caracters.
What I want is to be able to extract only the numbers of five digits only.
I started something like this but struggle to put a loop so that it scans all the string:
Function test()
val_in = "rue 4 qsdqsd CURIE 38320 EYBENS"
Filte = Left(val_in, 5)
If IsNumeric(Filte) Then
test = Left(val_in, 5)
Else
sp1 = InStr(1, val_in, " ")
sp2 = InStr(sp1 + 1, val_in, " ")
spt = sp2 + sp1
If spt > 5 Then
extr = Mid(val_in, spt, 5)
End If
End If
End Function
How could I turn the part after "Else" into a loop so that it would scan every space of the string and extract only the numbers that contains 5 digits?
Using regex
Option Explicit
Public Function GetNumbers(ByVal rng As Range) As Variant
Dim arr() As String, i As Long, matches As Object, re As Object
Set re = CreateObject("VBScript.RegExp")
With re
.Global = True
.MultiLine = True
.IgnoreCase = False
.Pattern = "\b\d{5}\b"
If .test(rng.Value) Then
Set matches = .Execute(rng.Value)
ReDim arr(0 To matches.Count - 1)
For i = LBound(arr) To UBound(arr)
arr(i) = matches(i)
Next i
Else
arr(i) = rng.Value
End If
End With
GetNumbers = Join(arr, ",")
End Function
Data:
If there is more than one match a comma separated list is returned.
Sub TestMe()
Dim valIn As String
valIn = "rue 4 qsdqsd CURIE 38320 EYBENS 43443"
Dim i As Long
Dim splitted As Variant
splitted = Split(valIn)
For i = LBound(splitted) To UBound(splitted)
If IsNumeric(splitted(i)) And Len(splitted(i)) = 5 Then
Debug.Print splitted(i)
End If
Next i
End Sub
Considering that in your example you mean that the 5 digit numbers are splitted by space, the above works. It splits the string by space to an array and loops through the elements of the array. If the element is with 5 chars and is numeric, it prints it.
If the rule for the spaces is not something that one can count on, here is a different implementation:
Sub TestMe()
Dim valIn As String
valIn = "44244rue4qsdqsdCURIE383201EYBENS43443"
Dim i As Long
For i = 1 To Len(valIn) - 4
If IsNumeric(Mid(valIn, i, 5)) Then
Debug.Print Mid(valIn, i, 5)
End If
Next i
End Sub
It starts looping through the string, checking whether each 5 chars are numeric. When you have numeric 6 chars, it gives two results - 1 to 5 and 2 to 6. Thus 383201 is "translated" as the following 2:
38320
83201
If you have always space between words/numbers then this should do
Sub test()
Dim TestStr As String
Dim Temp As Variant
Dim i As Long, FoundVal As Long
TestStr = "rue 4 qsdqsd CURIE 38320 EYBENS"
Temp = Split(TestStr, " ")
For i = 0 To UBound(Temp)
If Len(Trim(Temp(i))) = 5 And IsNumeric(Temp(i)) Then
FoundVal = Temp(i)
MsgBox FoundVal
End If
Next i
End Sub
From the solution you are trying to apply (creating custom function in VBA) I understand that you actually need to use it in a formula.
To find number with five digits from cell A1 you can use the following formula without VBA:
=IF(ISERROR(FIND("0"&REPT("1",5)&"0",CONCAT(0+(ISNUMBER(-MID(" "&A1&" ",ROW(INDIRECT("1:"&LEN(A1)+2)),1))+0)))),"",MID(A1,FIND("0"&REPT("1",5)&"0",CONCAT(0+(ISNUMBER(-MID(" "&A1&" ",ROW(INDIRECT("1:"&LEN(A1)+2)),1))+0))),5))
To search for other number of digits change the three occurrences of number 5 to your desired digits count in the formula.
Example
Say I have a string:
"I say ""Hello world"" and she says ""Excuse me?"""
VBA will interpret this string as:
I say "Hello world" and she says "Excuse me?"
A more complex example:
I have a string:
"I say ""Did you know that she said """"Hi there!"""""""
VBA interprets this string as:
I say "Did you know that she said ""Hi there!"""
If we remove "I say "
"Did you know that she said ""Hi there!"""
we can continue parsing the string in vba:
Did you know that she said "Hi there!"
Problem
Ultimately I want some function, sBasicQuote(quotedStringHierarchy as string), which returns a string containing the next level up in the string hierarchy.
E.G.
dim s as string
s = "I say ""Did you know that she said """"Hi there!"""""""
s = sBasicQuote(s) ' returns 'I say "Did you know that she said ""Hi there!"""'
s = sBasicQuote(s) ' returns 'Did you know that she said "Hi there!"'
s = sBasicQuote(s) ' returns 'Hi there!'
I just can't figure out an algorithm that would work with this... You almost need to replace all double quotes, but when you've replaced the nth double quote you have to skip to the n+1th douple quote?
How does one implement this in VBA?
You could do something like this
Public Sub test()
Dim s As String
s = "I say ""Did you know that she said """"Hi there!"""""""
Debug.Print DoubleQuote(s, 0)
Debug.Print DoubleQuote(s, 1)
Debug.Print DoubleQuote(s, 2)
End Sub
Public Function DoubleQuote(strInput As String, intElement As Integer) As String
Dim a() As String
strInput = Replace(strInput, String(2, Chr(34)), String(1, Chr(34)))
a = Split(strInput, chr(34))
DoubleQuote = a(intElement)
End Function
Another slightly modified version is a little more accurate
`Public Function DoubleQuote(strInput As String, intElement As Integer) As String
Dim a() As String
Dim b() As String
Dim i As Integer
ReDim b(0)
a = Split(strInput, Chr(34))
' ***** See comments re using -1 *******
For i = 0 To UBound(a) - 1
If Len(a(i)) = 0 Then
b(UBound(b)) = Chr(34) & a(i + 1) & Chr(34)
i = i + 1
Else
b(UBound(b)) = a(i)
End If
ReDim Preserve b(UBound(b) + 1)
Next i
DoubleQuote = b(intElement)
End Function`
I think the following will return what you are looking for in your nested quote example. Your first example is not really a situation of nested quotes.
Option Explicit
Sub NestedQuotes()
Const s As String = "I say ""Did you know that she said """"Hi there!"""""""
Dim COL As Collection
Dim Start As Long, Length As Long, sTemp As String, V As Variant
Set COL = New Collection
sTemp = s
COL.Add sTemp
Do Until InStr(sTemp, Chr(34)) = 0
sTemp = COL(COL.Count)
sTemp = Replace(sTemp, String(2, Chr(34)), String(1, Chr(34)))
Start = InStr(sTemp, Chr(34)) + 1
Length = InStrRev(sTemp, Chr(34)) - Start
sTemp = Mid(sTemp, Start, Length)
COL.Add sTemp
Loop
For Each V In COL
Debug.Print V
Next V
End Sub
My Solution
I spent some more time thinking and came up with this solution.
Function sMineDoubleQuoteHierarchy(s As String) As String
'Check the number of quotes in the string are even - sanity check
If (Len(s) - Len(Replace(s, """", ""))) Mod 2 <> 0 Then sMineDoubleQuoteHierarchy = "Error - Odd number of quotes found in sMineDoubleQuoteHierarchy() function": Exit Function
'First thing to do is find the first and last *single* quote in the string
Dim lStart, lEnd, i As Long, fs As String
lStart = InStr(1, s, """")
lEnd = InStrRev(s, """")
'After these have been found we need to remove them.
s = Mid(s, lStart + 1, lEnd - lStart - 1)
'Start at the first character
i = 1
Do While True
'Find where the next double quote is
i = InStr(1, s, """""")
'if no double quote is found then concatenate with fs with the remainder of s
If i = 0 Then Exit Do
'Else add on the string up to the char before the ith quote
fs = fs & Left(s, i - 1)
'Replace the ith double quote with a single quote
s = Left(s, i - 1) & Replace(s, """""", """", i, 1)
'Increment by 1 (ensuring the recently converted double quote is no longer a single quote
i = i + 1
Loop
'Return fs
sMineDoubleQuoteHierarchy = s
End Function
What's going on in this solution?
The first part of the process is removing the first and last single quote from the string and returning the text between them. Then we loop through the string replacing each instance of "" and replacing it with ". Each time we do this we skip to the next character to unsure strings like """" go to "" instead of ".
Does anyone else have a better/more compact solution?
Edit
After all the suggestions in this forum I settled with this. It's got some extra error trapping to find validate nested strings.
Public Function DoubleQuoteExtract(ByVal s As String, Optional ByRef ErrorLevel As Boolean) As String
'This effectively parses the string like BASIC does by removing incidents of "" and replacing them with "
'SANITY CHECK - Check even number of quotes
Dim countQuote As Double
countQuote = Len(s) - Len(Replace(s, """", ""))
'Calculate whether or not quote hierarchy is correct:
'"..." - Is okay - Count Quotes = 2 - Count Quotes / 2 = 1
'""..."" - Is not okay - Count Quotes = 4 - Count Quotes / 2 = 2
'"""...""" - Is okay - Count Quotes = 6 - Count Quotes / 2 = 3
'""""..."""" - Is not okay - Count Quotes = 8 - Count Quotes / 2 = 4
'etc.
'Ultimately: IF CountQuotes/2 = Odd The string hierarchy is setup fine
' IF CountQuotes/2 = Even, The string Hierarchy is setup incorrectly.
Dim X As Double: X = countQuote / 2
Dim ceil As Long: ceil = Int(X) - (X - Int(X) > 0)
If ceil Mod 2 <> 0 Then sDoubleQuoteExtract = "#Error - Incorrect number of double quotes forming an incomplete hierarchy.": GoTo ErrorOccurred
'If an odd number of quotes are found then they cannot be paired correctly, thus throw error
If countQuote Mod 2 <> 0 Then sDoubleQuoteExtract = "#Error - Odd number of quotes found in sMineDoubleQuoteHierarchy() function": GoTo ErrorOccurred
'Find the next incident of single quote. Trim the string to this
s = Mid(s, InStr(1, s, String(1, Chr(34))))
'replace all instances of "" with "
s = Replace(s, String(2, Chr(34)), String(1, Chr(34)))
'Finally trim off the first and last quotes
DoubleQuoteExtract = Mid(s, 2, Len(s) - 2)
ErrorLevel = False
Exit Function
ErrorOccurred:
ErrorLevel = True
End Function
I'd like to create a function in vba to extract the first nth words from a string and to look like this
ExtractWords(affected_text, delimiter, number_of_words_to_extract)
I tried a solution but it only extracts the first two words.
Function FirstWords(myStr As Variant, delimiter,words_to_extract) As Variant
FirstWords = Left(myStr, InStr(InStr(1, myStr, delimiter) + 1, myStr, delimiter, vbTextCompare) - 1)
End Function
Any ideas? Thanks
Use Split() function. It returns array of String, split using the delimiter and limit of words you specify.
Dim Result As Variant
Result = Split("Alice,Bob,Chuck,Dave", ",") 'Result: {"Alice,"Bob","Chuck","Dave"}
Result = Split("Alice,Bob,Chuck,Dave", ",", 2) 'Result: {"Alice,"Bob"}
#Taosique's answer using Split is excellent, but if you want the result returned as a string you can do the following:
Function FirstWords(myStr As String, delimiter As String, words_to_extract As Long) As Variant
Dim i As Long, k As Long
For i = 1 To Len(myStr)
If Mid(myStr, i, 1) = delimiter Then
k = k + 1
If k = words_to_extract Then
FirstWords = Mid(myStr, 1, i)
Exit Function
End If
End If
Next I
'if you get to here -- trouble
'unless the delimiter count is words_to_extract - 1
If k = words_to_extract - 1 Then
FirstWords = myStr
Else
FirstWords = CVErr(xlErrValue)
End If End Function
Sub test()
Debug.Print FirstWords("This is a test. I hope it works", " ", 4)
Debug.Print FirstWords("This is a test. I hope it works", " ", 10)
End Sub
When test is run it first displays the string "This is a test." then prints an error condition.
Much the same effect as the above can be achieved by first splitting the string using Split and then rejoining it using Join. A subtle difference is the behavior if there are less than words_to_extract words. The Split then Join approach will return the whole string. The above code treats this as an error condition and, if used as a UDF worksheet function, will display #VALUE! in any cell that contains it.