Asking to user the letter referencing a column ; then using the answer to get the range of this a cell with in this column at some row.
But the "Range" refuses to recognize the letter input.
What am I missing in the following simple two lines ?
UserCol = Application.InputBox(" Please enter the column...", Type:=0)
Set Z = Range(UserCol & 5)
You need to use Type:=2.
Using Type:=0 will return ="<input string>", rather than just <input string>.
So, after:
Dim UserCol As String
UserCol = Application.InputBox(" Please enter the column...", Type:=2)
You can either do:
Set Z = Cells(5, UserCol)
OR:
Set Z = Range(UserCol & "5")
I would also suggest that use Option Explicit and also fully qualify range references. e.g. instead of Set Z = Range(UserCol & "5"), use Set Z = Thisworkbook.sheets("MySheetName").Range(UserCol & "5")
Try this:
UserCol = Application.InputBox(" Please enter the column...", Type:=2)
Set Z = Range((UserCol & 5))
I set Type to 2 to return a string from your user's input (see her for more)
Additionally, I added a parenthesis to the Range, because (UserCol & 5) becomes e.g. A5, and you need Range(A5).
This way is easier:
Dim UserCol As String
UserCol = Application.InputBox(" Please enter the column...")
Set Z = Cells(5, UserCol)
I don't know how did you declare your UserCol or if you even declared it. If you didn't and to avoid problems always use Option Explicit on the top of your module.
This is a really simple bug to fix: Set Z = Range(UserCol & "5")
Why? Because when you use implicit conversion, by typing UserCol & 5, VBA includes a space between UserCol and the 5 (and also after the 5).
Here is a test:
MsgBox "A" & 5 'Outputs "A 5 "
MsgBox "A" & "5" 'Outputs "A5"
(As Gravitate points out, Application.InputBox(" Please enter the column...", Type:=0) is a Formula, so an input of "A" would give you "=""A"", and not "A" -since "=""A""5" is not a valid cell reference either, use Type:=2 for Text or InputBox(" Please enter the column...") without the Application. or the type filtering)
Related
Sub Botão34_Clique()
Dim x As String
Dim z As Integer
InputBox "FIND PRODUCT", "TYPE" = x
For z = 2 To 143
If Range("B" & z).Value = x Then
MsgBox Range("B" & z)
End If
Next z
End Sub
Hi guys,
I don't understand why the above code is not working.
The user's input will be compared with an existing product list.
When the input is equal to an existing product, the msgbox should indicate the cell in which it is present.
Unfortunately, I don't understand why the statement Range("B" & z).Value = x doesn't work...
Any ideas?
You don't write the result of the InputBox into the variable x.
What you do is to call the InputBox with 2 parameters.
The first parameter ("FIND PRODUCT") is the text that is shown in the input box as label.
The second parameter is used as form caption. You pass the expression "TYPE" = x as second parameter, and VBA will happily compare the string "TYPE" with the content of the variable x, resulting in either True or (most likely) False. So if you carefully look at the input box window, you will see the string "False" as caption.
Now the user enters something into the box and presses the "OK"-button. The entered value is returned, but you don't assign it to anything and the value is send to the big Computer Nirvana.
What you want is to assign the value to x. The correct syntax for this is
x = InputBox("FIND PRODUCT", "TYPE")
Note that in VBA, you have to put parenthesis around the list of parameters when you are calling a function and use the return value.
I want to separate a character string using the special characters in that string as cutting lines. After each division the next group of strings should be copied in the next column. The picture below shows how it should work.
My first approach doesn't work and maybe it's too complicated. Is there a simple solution to this task?
Sub SeparateString()
Dim i, j, k, counterA, counterB As Integer
Dim str1, str2 As String
Const Sonderz As String = "^!§$%&/()=?`*'_:;°,.-#+´ß}][{³²"
For i = 1 To Worksheets("Tabelle1").Range("A1").End(xlDown).Row
counterA = 0
For j = 1 To Len(Worksheets("Tabelle1").Range("A" & i))
counterB = 0
For k = 1 To Len(Sonderz)
If Mid(Worksheets("Tabelle1").Range("A" & i), j, 1) = Mid(Sonderz, k, 1) Then
counterA = counterA + 1
End If
If Mid(Worksheets("Tabelle1").Range("A" & i), j, 1) <> Mid(Sonderz, k, 1) And counterA = 0 And counterB = 0 Then
Worksheets("Tabelle1").Range("B" & i) = Worksheets("Tabelle1").Range("B" & i) & Mid(Worksheets("Tabelle1").Range("A" & i), j, 1)
counterB = counterB + 1
End If
Next k
Next j
Next i
End Sub
If you are interested and you do have access to Microsoft 365's dynamic arrays:
Formula in B1:
=LET(X,MID(A1,SEQUENCE(LEN(A1)),1),TRANSPOSE(FILTERXML(CONCAT("<t><s>",IF(ISNUMBER(FIND(X,"^!§$%&/()=?`*'_:;°,.-#+´ß}][{³²")),"</s><s>",X),"</s></t>"),"//s")))
Or nest a SUBSTITUTE() if you need to return string variables:
=LET(X,MID(A1,SEQUENCE(LEN(A1)),1),TRANSPOSE(SUBSTITUTE(FILTERXML(CONCAT("<t><s>'",IF(ISNUMBER(FIND(X,"^!§$%&/()=?`*'_:;°,.-#+´ß}][{³²")),"</s><s>'",X),"</s></t>"),"//s"),"'","")))
If VBA is a must, you could think about a regular expression to replace all the characters from a certain class with a uniform delimiter to use Split() on:
Sub Test()
Dim s As String: s = "CD!02?WX12EF"
Dim arr() As String
With CreateObject("vbscript.regexp")
.Global = True
.Pattern = "[!§$%&\/()=?`*'_:;°,.#+´ß}\][{³²^-]"
arr = Split(.Replace(s, "!"), "!")
End With
For Each el In arr
Debug.Print el
Next
End Sub
The caret has been moved from being the 1st character to any but the first to prevent a negated-character class; also the hyphen has been moved to the back to prevent an array-construct of characters. Funny enough, if you actually want to be less verbose you could throw these character in a more condense class [!#-/:;?[]-`{}§=°´ß³²].
Assuming the first data is in cell A2,
I would go with the simple use of find() with left() mid() and right()
=left(A2,find("!",A2,1)-1)
then:
=mid(A2,find("!",A2,1)+1,find("?",A2,1)-find("!",A2,1)-1)
and:
=right(A2,len(A2)-find("?",A2,1))
Tested and working with one correction done.
You can also do this in Power Query which has a command to split by ANY delimiter.
In the code below, I generate a list of all possible special characters defined as characters NOT in the set of A..Za..z0..9 and you can easily add to that list by editing the code if you want to include other characters in the permitted list.
Edit: If you only have a few special characters, you can just hard-code the list, eg {"!","?"} instead of using List.Generate, but in your question you did not necessarily restrict the list of special characters, even though those are the only two showing in your examples
To use Power Query:
Select some cell in your Data Table
Data => Get&Transform => from Table/Range
When the PQ Editor opens: Home => Advanced Editor
Make note of the Table Name in Line 2
Paste the M Code below in place of what you see
Change the Table name in line 2 back to what was generated originally.
Read the comments and explore the Applied Steps to understand the algorithm
let
//change Table name in next line to reflect actual table name
Source = Excel.CurrentWorkbook(){[Name="Table15"]}[Content],
#"Changed Type" = Table.TransformColumnTypes(Source,{{"Source", type text}}),
//Generate list of "special characters" for splitting
//the List.Contains function contains the non-special characters
splitterList = List.RemoveNulls(
List.Generate(()=>0,
each _ <= 255,
each _ +1,
each if List.Contains({"A".."Z","a".."z","0".."9"}, Character.FromNumber(_)) then null else Character.FromNumber(_))),
splitIt = Table.SplitColumn(#"Changed Type", "Source",
Splitter.SplitTextByAnyDelimiter(splitterList))
in
splitIt
I have a dataset of names in a column in Excel.
However, only some but not all of the names have a letter attached to the end of it (e.g. John Doe A, Kai Jin, Johnny Desplat Lang B, etc).
Can anyone think of a method to remove the letter from the end of the name from each row, if it is there? Such that, using the example above, I will be left with: John Doe, Kai Jin, Johnny Desplat Lang, etc.
I am fairly familiar with VBA and Excel and would be open to trying anything at all.
Thank you for your help with this question! Apologies beforehand if this seems like an elementary question but I have no idea how to begin to solve it.
"I am fairly familiar with VBA and Excel and would be open to trying anything at all."
If so, then this can be done with a simple formula if you wish to avoid VBA. With your value in A1:
=IF(MID(A1,LEN(A1)-1,1)=" ",LEFT(A1,LEN(A1)-2),A1)
If you must use VBA, I think the Like operator comes in handy:
Sub Test()
Dim arr As Variant: arr = Array("John Doe A", "Kai Jin", "Johnny Desplat Lang B")
For Each el In arr
If el Like "* ?" Then 'Or "* [A-Z]" if you must check for uppercase alpha.
Debug.Print Left(el, Len(el) - 2)
Else
Debug.Print el
End If
Next
End Sub
Just for fun and in order to demonstrate another approach via the Filter() function:
Function ShortenName(ByVal FullName As Variant) As String
'Purpose: remove a single last letter
Dim n: n = Split(FullName, " "): n = Len(n(UBound(n)))
ShortenName = Left(FullName, Len(FullName) + 2 * (n = 1))
End Function
Explanation
Applying the Split() function upon the full name and isolating the last name token (via UBound()) allows to check for a single letter length (variable n).
The function result returns the entire string length minus 2 (last letter plus preceding space) in case of a single letter (the the condition n = 1 then results in True equalling -1). - Alternatively you could have coded: ShortenName = Left(FullName, Len(FullName) - IIf(n = 1, 2, 0))
I have a basic programming background and have been self sufficient for many years but this problem I can't seem to solve. I have a program in VBA and I need to compare two strings. I have tried using the following methods to compare my strings below but to no avail:
//Assume Cells(1, 1).Value = "Cat"
Dim A As String, B As String
A="Cat"
B=Cell(1, 1).Value
If A=B Then...
If A Like B Then...
If StrCmp(A=B, 1)=0 Then...
I've even tried inputting the Strings straight into the code to see if it would work:
If "Cat" = "Cat" Then...
If "Cat" Like "Cat" Then...
If StrCmp("Cat" = "Cat", 1) Then...
VBA for some reason does not recognize these strings as equals. When going through Debugger it shows that StrComp returns 1. Do my strings have different Char lengths? Thanks for any help.
Posting as answer because it doesn't fit in the comments:
I find it hard to believe that something like:
MsgBox "Cat" = "Cat"
Would not display True on your machine. Please verify.
However, I do observe that you are most certainly using StrComp function incorrectly.
The proper use is StrComp(string, anotherstring, [comparison type optional])
When you do StrComp(A=B, 1) you are essentially asking it to compare whether a boolean (A=B will either evaluate to True or False) is equivalent to the integer 1. It is not, nor will it ever be.
When I run the following code, all four message boxes confirm that each statement evaluates to True.
Sub CompareStrings()
Dim A As String, B As String
A = "Cat"
B = Cells(1, 1).Value
MsgBox A = B
MsgBox A Like B
MsgBox StrComp(A, B) = 0
MsgBox "Cat" = "Cat"
End Sub
Update from comments
I don't see anything odd happening if I use an array, just FYI. Example data used in the array:
Modified routine to use an array:
Sub CompareStrings()
Dim A As String, B() As Variant
A = "Cat"
B = Application.Transpose(Range("A1:A8").Value)
For i = 1 To 8
MsgBox A = B(i)
MsgBox A Like B(i)
MsgBox StrComp(A, B(i)) = 0
MsgBox "Cat" = B(i)
Next
End Sub
What I would check is how you're instantiating the array. Range arrays (as per my example) are base 1. If it assigned some other way, it is most likely base 0, so check to make sure that you're comparing the correct array index.
I am currently encountering a problem which doesn't seem that hard to fix but, yet, I can't find a clean way of doing it on my own.
I am using the "Replace" function to change some expressions in a sentence typed by an user. For example, if the user types "va", I want it to be turned into "V. A." instead so it will match more easily with my database for further operations.
Here is my simple code to do it :
sMain.Range("J3").Replace "VA", "V. A."
It works well.
Problem is, it's not only spotting "VA" as an individual expression, but also as a part of words.
So if my user types "Vatican", it's gonna turn it into : "V. A.tican"... which of course I don't want.
Do you know how to easily specify my code to make it ONLY consider replacing the whole words matching the expression? (I have dozens of lines of these replacement so ideally, it would be better to act directly on the "replace" functions - if possible).
Thanks in advance !
Do this:
sMain.Range("J3").Replace " VA ", "V. A."
then handle the cases where the original string starts or ends with VA
also, handle all cases of separators which could be (for example) tab, space or comma.
To do that:
const nSep As Integer = 3
Dim sep(nSep) As String
sep(1) = " "
sep(2) = vbTab
sep(3) = ","
for i=1 to nSep
for j=1 to nSep
sMain.Range("J3").Replace sep(i) & "VA" & sep(j), "V. A."
next
next
Can split it up and check each word. I have put it into a function for easy of use and flexibility.
Function ReplaceWordOnly(sText As String, sFind As String, sReplace As String) As String
On Error Resume Next
Dim aText As Variant, oText As Variant, i As Long
aText = Split(sText, " ")
For i = 0 To UBound(aText)
oText = aText(i)
' Check if starting with sFind
If LCase(Left(oText, 2)) = LCase(sFind) Then
Select Case Asc(Mid(oText, 3, 1))
Case 65 To 90, 97 To 122
' obmit if third character is alphabet (checked by ascii code)
Case Else
aText(i) = Replace(oText, sFind, sReplace, 1, -1, vbTextCompare)
End Select
End If
Next
ReplaceWordOnly = Join(aText, " ")
End Function
Example output:
?ReplaceWordOnly("there is a vatican in vA.","Va","V. A.")
there is a vatican in V. A..