I have a situation where I want to have an input box where a string like "C123q23C456a45" is entered. The pattern will remain the same, so it will always be C,3 numbers,A-z,number( this number can be either 2 or 3 digits).
I want to split this string with the C's, so the example above will be split into:
C456a45 and C123q23. this output is showed in a message box in the form of C123 has 23 of q and c456 has 45 of a.
if that makes sense.
how do i approach this?
thanks appreciate it
If you can live without VBA to do this, you can use these formulas if you have the string you want to split in A1:
=LEFT(A1;FIND("C";A1;FIND("C";A1)+2)-1) for C123q23
=RIGHT(A1;LEN(A1)-LEN(LEFT(A1;FIND("C";A1;FIND("C";A1)+2)-1))) for C456a45
** UPDATED **
Sub Macro1()
Dim strFirstSplitA As String, str1A As String, str2A As String
Dim strA As String, strAA As String, strAAA As String
Dim str1B As String, strB As String, strBB As String, strBBB As String
Dim strMsgBoxA As String, strMsgBoxB As String
str1A = InStr(1, Range("A1"), "C") + 2
str2A = InStr(str1A, Range("A1"), "C") ' FIND("C";A1)+2)-1)
strFirstSplitA = Left(Range("A1"), str2A - 1)
strA = Left(strFirstSplitA, 4)
strAA = Right(strFirstSplitA, Len(strFirstSplitA) - 4)
strAAA = Right(strAA, Len(strAA) - Len(Left(strAA, 1)))
strMsgBoxA = strA & " has " & strAAA & " of " & Left(strAA, 1)
' C123 has 23 of q
str1B = Right(Range("A1"), Len(Range("A1")) - Len(strFirstSplitA))
strB = Left(str1B, 4)
strBB = Right(str1B, Len(str1B) - 4)
strBBB = Right(strBB, Len(strBB) - Len(Left(strBB, 1)))
strMsgBoxB = strB & " has " & strBBB & " of " & Left(strBB, 1)
' c456 has 45 of a
MsgBox strMsgBoxA & " and " & strMsgBoxB
End Sub
Not tested, but something like this:
a = Split("C123q23C456a45", "C")
For i = 1 To Ubound(a)
Debug.Print Left$(a(i), 3), Mid$(a(i), 5)
Next
Related
I have a comma separated lists in cells. All numbers are positive and between 1 and 10.
Example:
if I have in A1: (2,3,5,6), I would like to have missing numbers in B1:(1,4,7,8,9,10).
If A2: (1,10), then I would have in B2:(2,3,4,5,6,7,8,9)
If A3: (7), then I would have in B2:(1,2,3,4,5,6,8,9,10)
I searched for a solution online, but I couldn't find anything similar with comma separated numbers.
I'd be glad if I can have a solution here. Thanks.
Here is a user-defined function that should accomplish this... probably can be optimized.
Public Function MissingNumbers(ByVal numberList As String) As String
Dim temp As String
temp = Replace(numberList, "(", "")
temp = Replace(temp, ")", "")
Dim arr As Variant
arr = Split(temp, ",")
Dim newNumbers As String
newNumbers = "1,2,3,4,5,6,7,8,9,10,"
Dim i As Long
For i = LBound(arr) To UBound(arr)
newNumbers = Replace(newNumbers, arr(i) & ",", "")
Next
newNumbers = "(" & Left$(newNumbers, Len(newNumbers) - 1) & ")"
MissingNumbers = newNumbers
End Function
Just for fun demonstrating how to use negative filtering:
Function MissingList(ByVal numberList As String) As String
Dim given: given = Split(Mid(numberList, 2, Len(numberList) - 2), ",")
Dim series: series = GetSeries() ' i.e. numbers 1..10
Dim i As Long
For i = 0 To UBound(given)
series = Filter(series, given(i), False) ' << negative filtering
Next
MissingList = "(" & Replace(Join(series, ","), "0", "10") & ")"
End Function
As Filter executes a partial search in the 1..10 series, 10 has to be replaced temporarily by a unique 0.
Help function GetSeries()
Function GetSeries()
' Purpose: get numbers 1..10
Const LAST As Long = 10: Const FIRST = 1
Dim tmp: tmp = Application.Transpose(Evaluate("row(" & FIRST & ":" & LAST & ")"))
tmp(LAST) = 0 ' replace 10 by 0 as search item 1 would filter out value 10, too
GetSeries = tmp
End Function
I am working on some software that cleans up data before sending it into another system. The data comes from all around the world and contains a variety of characters that have to be replaced. For example ‘, : ; #
The system that accepts the parsed data has very strict character set. It allows
the letters A to Z (upper case only)
the numerals 0 to 9
the special characters / -. Space < =
The data arrives in Excel spreadsheets so I have written the following code in a visual basic macro.
fhl_str contains the data to be cleansed
fhl_str = Replace(fhl_str, ",", " ")
fhl_str = Replace(fhl_str, "'", " ")
fhl_str = Replace(fhl_str, ":", " ")
fhl_str = Replace(fhl_str, ";", " ")
fhl_str = ucase(fhl_str)
Now, each time a new unwanted character arrives we have to add a new line of code. e.g. fhl_str = Replace(fhl_str, "#", " ")
My question is
Could I reverse the logic so that the macro looks for A to Z and 0 to 9 and deletes anything else. That way my code would be future proof for new unwanted characters.
Thanks
If you want to replace bad characters with a single space:
Sub KeepOnlyTheGood()
Dim i As Long, L As Long, v As String, CH As String
Dim r As Range
For Each r In Selection
t = ""
v = r.Value
L = Len(v)
For i = 1 To L
CH = Mid(v, i, 1)
If CH Like "[0-9A-Z]" Or CH = "/" Or CH = "-" Or CH = "." Or CH = " " Or CH = "<" Or CH = "=" Then
t = t & CH
Else
t = t & " "
End If
Next i
r.Value = t
Next r
End Sub
Here's some VBA that will do it if you find regex difficult to understand. It uses the ASCII code to determine the only characters to allow. If your scope changes you can modify the ASCII numbers in the Case statement.
Public Function RemoveSpecial(s As String) As String
Dim sResult As String
Dim nIndex As Integer
s = UCase$(s)
For nIndex = 1 To Len(s)
Select Case Asc(Mid$(s, nIndex, 1))
Case 65 To 90, 45 To 57, 32, 60 To 61
sResult = sResult & Mid$(s, nIndex, 1)
Case Else
sResult = sResult & " "
End Select
Next
RemoveSpecial = sResult
End Function
Usage:
Debug.Print RemoveSpecial("TeSt<>=.##")
or something like:
Range("A1") = RemoveSpecial("TeSt<>=.##")
ASCII Codes
Very simple I believe just can't get it myself
I'm taking a string example "88888<>88888<>88888<>" and all I want to do is remove the "<>" take the part after and place on 2nd line of a list so it would become
88888
88888
88888
This code does the work for you using .IndexOf() and .SubString(). There are other shorter ways too, a lot more actually but I wrote this so you can know what's happening in every step and know the exact solution.
Dim FInalString As String
Dim String_ As String = "88888<>8888888<>88888<>"
'now find the position of first <>
Dim firstsign_ As Integer = String_.IndexOf("<", 0)
'add string upto that point in the textbox
FInalString = "1. " & String_.Substring(0, firstsign_)
'now find the position of second <>
Dim secondSign_ As Integer = String_.IndexOf("<", firstsign_ + 2) 'here, not 0 because we have to skip the first one
' MsgBox(secondSign_)
'add the string upto that point in the textbox
FInalString = FInalString & vbCrLf & "2. " & String_.Substring(firstsign_ + 2, secondSign_ - firstsign_ - 2) 'vbCrLf means go to next line now
'now find the third sign, last
Dim lastsign_ As Integer = String_.IndexOf("<", secondSign_ + 2)
'add to the string
FInalString = FInalString & vbCrLf & "3. " & String_.Substring(secondSign_ + 2, lastsign_ - secondSign_ - 2)
MsgBox(FInalString)
Output:
88888
8888888
88888
I have no idea what this has to do with a WebClient. But this should work as you need.
Dim str As String = "88888<>88888<>88888<>"
Dim strarr As String() = str.Split("<>")
Dim result As String = ""
For i As Integer = 0 To strarr.Length
result &= (i + 1) & ". " & strarr(i)
Next
I need to add brackets around the numbers in a string found in cells on my Excel worksheet.
For example, say I am given:
913/(300+525)
I need to get this in return:
[913]/([300]+[525])
The equations are fairly simple, should only have to deal with + - * / ( ) characters.
I attempted looping through the string character by character using the MID function but I can't get the loop(s) working correctly and end up getting a jumbled mess of random brackets and numbers back. I also considered using regular expressions but I've never used them before and have no idea if this would be a good application.
Please let me know if you need anything else. Thank you for your time!
They can be decently long. Here is another example:
I have:
(544+(1667+1668+1669+1670+1671+1672+1673)-1674)
But I need:
([544]+([1667]+[1668]+[1669]+[1670]+[1671]+[1672]+[1673])-[1674])
I just threw this together but it should work
Function generateBrackets(Equation As String) As String
Dim temp As String
Dim brackets As Boolean
Dim x 'If we're using Option Explicit, or just to be safe
For x = 1 To Len(Equation)
If Not IsNumeric(Mid(Equation, x, 1)) And brackets = False Then
temp = temp & Mid(Equation, x, 1)
ElseIf Not IsNumeric(Mid(Equation, x, 1)) And brackets = True Then
temp = temp & "]" & Mid(Equation, x, 1)
brackets = False
ElseIf IsNumeric(Mid(Equation, x, 1)) And brackets = False Then
temp = temp & "[" & Mid(Equation, x, 1)
brackets = True
ElseIf IsNumeric(Mid(Equation, x, 1)) And brackets = True Then
temp = temp & Mid(Equation, x, 1)
End If
Next x
generateBrackets = temp
End Function
Here is a way which caters for Decimal numbers.
'~~> Add here whatever operators your equation
'~~> is likely to have
Const delim As String = "+()-/"
Sub Sample()
Dim MyAr
Dim sSamp As String
sSamp = "(5.44+(16.67+1668+1669+1670+1671+1672+1673)-1674)"
MyAr = Split(GetNewString(sSamp))
For i = 0 To UBound(MyAr)
sSamp = Replace(sSamp, MyAr(i), "[" & MyAr(i) & "]")
Next i
Debug.Print sSamp
End Sub
Function GetNewString(s As String) As String
Dim sTemp As String
sTemp = s
For i = 1 To Len(delim)
sTemp = Replace(sTemp, Mid(delim, i, 1), " ")
Next i
Do While InStr(1, sTemp, " ")
sTemp = Replace(sTemp, " ", " ")
Loop
GetNewString = Trim(sTemp)
End Function
Input
"(5.44+(16.67+1668+1669+1670+1671+1672+1673)-1674)"
Output
([5.44]+([16.67]+[1668]+[1669]+[1670]+[1671]+[1672]+[1673])-[1674])
I am trying to soft code the output variables, so that I don't have to modify the VBA code each time i need to modify the outputs.
This is the code that works
Sub Working()
Dim cat(1 To 10)
Dim bat(1 To 10)
For i = 1 To 10
cat(i) = i * 10
bat(i) = i * 5
Next i
Sheet2.Range("A2:A11") = Application.Transpose(cat())
Sheet2.Range("B2:B11") = Application.Transpose(bat())
End Sub
This is the ideal way i would want to write, but doesnt work
Sub not_working()
Dim cat(1 To 10)
Dim bat(1 To 10)
For i = 1 To 10
cat(i) = i * 10
bat(i) = i * 5
Next i
a = 3
Do While Sheet1.Cells(a, 1) <> ""
OutVar = Sheet1.cells(a, 1) & "()"
Sheet3.Range( _
Cells(2, a - 2).Address, Cells(11, a - 2).Address _
) = Application.Transpose(Outvar)
a = a + 1
Loop
End Sub
' Sheet1.cells(3,1) = cat - these cells contain the variable names
' Sheet1.cells(4,1) = bat - these cells contain the variable names
Can someone please suggest if it is possible to do so?
If I understand your requirement correctly, a ragged array will meet it.
If you have a variable of type Variant, you can set that variable to, for example, an integer, a real, a string, a boolean or an array.
If you have an array of type Variant, you can set each element of that array to a different type of value.
In my code below, I have variant array Main. I set:
Main(0) to a 1D array,
Main(1) to a larger 1D array,
Main(2) to a 2D array,
Main(3) to a single integer,
Main(4) to the used range of a worksheet.
This is called a ragged array because each element is a different size.
Having loaded the array with values, I use a general routine to output each element of Main according to its nature.
Each of your 200-300 variables would become an element of Main.
Have a look at my code. This is only a brief introduction to what can be achieved with variant arrays. Come back with questions if you think I am heading in the correct direction but have not gone far enough.
Option Explicit
Sub DemoRaggedArray()
Dim InxDim As Long
Dim InxMain As Long
Dim InxWCol As Long
Dim InxWRow As Long
Dim Main() As Variant
Dim NumOfDim As Long
Dim Work() As Variant
ReDim Main(0 To 5)
Work = Array(1, "A", True)
Main(0) = Work
Main(1) = Array(2, "B", False, 1.2)
ReDim Work(1 To 2, 1 To 3)
Work(1, 1) = 1
Work(1, 2) = 2.5
Work(1, 3) = DateSerial(2012, 12, 27)
Work(2, 1) = True
Work(2, 2) = "String"
Main(2) = Work
Main(3) = 27
' Cells A1:C4 of the worksheet have been set to their addresses
Main(4) = WorksheetFunction.Transpose(Worksheets("Sheet2").UsedRange.Value)
For InxMain = LBound(Main) To UBound(Main)
Debug.Print "Type of Main(" & InxMain & ") is " & VarTypeName(Main(InxMain))
Select Case VarType(Main(InxMain))
Case vbEmpty, vbNull
' No value
Case Is >= vbArray
' Array
NumOfDim = NumDim(Main(InxMain))
Debug.Print " Main(" & InxMain & ") is dimensioned as: (";
For InxDim = 1 To NumOfDim
Debug.Print LBound(Main(InxMain), InxDim) & " To " & _
UBound(Main(InxMain), InxDim);
If InxDim < NumOfDim Then
Debug.Print ", ";
End If
Next
Debug.Print ")"
Select Case NumOfDim
Case 1
For InxWCol = LBound(Main(InxMain)) To UBound(Main(InxMain))
Debug.Print " (" & InxWCol & ")[" & _
VarTypeName(Main(InxMain)(InxWCol)) & "]";
Select Case VarType(Main(InxMain)(InxWCol))
Case vbEmpty, vbNull, vbArray
' No code to handle these types
Case Else
Debug.Print "=" & Main(InxMain)(InxWCol);
End Select
Next
Debug.Print
Case 2
For InxWRow = LBound(Main(InxMain), 2) To UBound(Main(InxMain), 2)
For InxWCol = LBound(Main(InxMain), 1) To UBound(Main(InxMain), 1)
Debug.Print " (" & InxWCol & "," & InxWRow & ")[" & _
VarTypeName(Main(InxMain)(InxWCol, InxWRow)) & "]";
Select Case VarType(Main(InxMain)(InxWCol, InxWRow))
Case vbEmpty, vbNull, vbArray
' No code to handle these types
Case Else
Debug.Print "=" & Main(InxMain)(InxWCol, InxWRow);
End Select
Next
Debug.Print
Next
Case Else
Debug.Print " There is no display code for this number of dimensions"
End Select
Case Else
' Single variable
Debug.Print " Value = " & Main(InxMain)
End Select
Next
End Sub
Public Function NumDim(ParamArray TestArray() As Variant) As Integer
' Returns the number of dimensions of TestArray.
' If there is an official way of determining the number of dimensions, I cannot find it.
' This routine tests for dimension 1, 2, 3 and so on until it get a failure.
' By trapping that failure it can determine the last test that did not fail.
' Coded June 2010. Documentation added July 2010.
' * TestArray() is a ParamArray because it allows the passing of arrays of any type.
' * The array to be tested in not TestArray but TestArray(LBound(TestArray)).
' * The routine does not validate that TestArray(LBound(TestArray)) is an array. If
' it is not an array, the routine return 0.
' * The routine does not check for more than one parameter. If the call was
' NumDim(MyArray1, MyArray2), it would ignore MyArray2.
Dim TestDim As Integer
Dim TestResult As Integer
On Error GoTo Finish
TestDim = 1
Do While True
TestResult = LBound(TestArray(LBound(TestArray)), TestDim)
TestDim = TestDim + 1
Loop
Finish:
NumDim = TestDim - 1
End Function
Function VarTypeName(Var As Variant)
Dim Name As String
Dim TypeOfVar As Long
TypeOfVar = VarType(Var)
If TypeOfVar >= vbArray Then
Name = "Array of type "
TypeOfVar = TypeOfVar - vbArray
Else
Name = ""
End If
Select Case TypeOfVar
Case vbEmpty
Name = Name & "Uninitialised"
Case vbNull
Name = Name & "Contains no valid data"
Case vbInteger
Name = Name & "Integer"
Case vbLong
Name = Name & "Long integer"
Case vbSingle
Name = Name & "Single-precision floating-point number"
Case vbDouble
Name = Name & "Double-precision floating-point number"
Case vbCurrency
Name = Name & "Currency"
Case vbDate
Name = Name & "Date"
Case vbString
Name = Name & "String"
Case vbObject
Name = Name & "Object"
Case vbError
Name = Name & "Error"
Case vbBoolean
Name = Name & "Boolean"
Case vbVariant
Name = Name & "Variant"
Case vbDataObject
Name = Name & "Data access object"
Case vbDecimal
Name = Name & "Decimal"
Case vbByte
Name = Name & "Byte"
Case vbUserDefinedType
Name = Name & "Variants that contain user-defined types"
Case Else
Name = Name & "Unknown type " & TypeOfVar
End Select
VarTypeName = Name
End Function
Output from DemoRaggedArray
Type of Main(0) is Array of type Variant
Main(0) is dimensioned as: (0 To 2)
(0)[Integer]=1 (1)[String]=A (2)[Boolean]=True
Type of Main(1) is Array of type Variant
Main(1) is dimensioned as: (0 To 3)
(0)[Integer]=2 (1)[String]=B (2)[Boolean]=False (3)[Double-precision floating-point number]=1.2
Type of Main(2) is Array of type Variant
Main(2) is dimensioned as: (1 To 2, 1 To 3)
(1,1)[Integer]=1 (2,1)[Boolean]=True
(1,2)[Double-precision floating-point number]=2.5 (2,2)[String]=String
(1,3)[Date]=27/12/2012 (2,3)[Uninitialised]
Type of Main(3) is Integer
Value = 27
Type of Main(4) is Array of type Variant
Main(4) is dimensioned as: (1 To 3, 1 To 4)
(1,1)[String]=A1 (2,1)[String]=B1 (3,1)[String]=C1
(1,2)[String]=A2 (2,2)[String]=B2 (3,2)[String]=C2
(1,3)[String]=A3 (2,3)[String]=B3 (3,3)[String]=C3
(1,4)[String]=A4 (2,4)[String]=B4 (3,4)[String]=C4
Type of Main(5) is Uninitialised
Note the date is displayed as "27/12/2012" because that is the default date format for my country. If you run this code, it will display in your country's default format.