I want to refer to a array where last row is dynamic and its value is saved in a cell.
Can anyone help me whats wrong in this code
Sub Drop_down()
'
' Drop_down Macro
'
'
Sheets("Raw Data").Select
Dim UC_count As Long
Application.ActiveSheet.UsedRange
UC_count = Worksheets("Raw Data").UsedRange.Rows.Count - 1
Sheets("Scoring Sheet").Select
Range("B4").Formula = "=VLOOKUP(R3C&"":""&INDEX(RA_Sheet!R4C1:R &(UC_count+4)C20,MATCH('Scoring Sheet'!RC1,RA_Sheet!R4C1:R&(UC_count+4)C1,0),MATCH('Scoring Sheet'!R3C,RA_Sheet!R4C1:R4C20,0)),'PV Lookup Table'!R1C6:R108C9,4,0)"
Instead of UC_count if I write number then the code works well.
Try this change:
Range("B4").Formula = "=VLOOKUP(R3C&"":""&INDEX(RA_Sheet!R4C1:R" & (UC_count+4 ) & "C20,MATCH('Scoring Sheet'!RC1,RA_Sheet!R4C1:R" & (UC_count+4 ) & "C1,0),MATCH('Scoring Sheet'!R3C,RA_Sheet!R4C1:R4C20,0)),'PV Lookup Table'!R1C6:R108C9,4,0)"
It inserts the values of UC Count properly into the formula. (i didn't test the rest of the formula though).
Another tip; perhaps use a named range; and just update the named range when needed?
Related
I'm new to this site, but I have already found some nice advice on how to solve problems in VBA. Now I'm here to ask help on a sub that gives me problems with Hyperlinks.
In particular, my problem is similar to the one described in this topic:
Excel VBA Get hyperlink address of specific cell
I have a worksheet full of hyperlink, but I need to extract only the addresses present in the "H" column, starting from "H6" and writing them into the "N" column, starting from "N6".
I put down this code:
Sub EstraiIndirizzoPut()
Dim IndirizzoInternet As Hyperlink
Dim ISINs As String
i = 6
For Each IndirizzoInternet In Sheets("XXX").Range("H" & i).Hyperlinks
IndirizzoInternet.Range.Offset(0, 6).Value = IndirizzoInternet.Address
ISINs = Mid(IndirizzoInternet.Address, 78, 12)
Range("N" & i).Value = ISINs
i = i + 1
Next
End Sub
It works fine only for the first "H6" cell, but at the "Next" point, when it should read the "H7" cell, it goes instead to "End Sub", terminating the routine, altough the "H7" cell, as well many others down the column, are filled with hyperlinks (it gives me "Nothing" value).
Could you please suggest me where I get this wrong? Many thanks.
Your loop isnt set up correctly. Try it like this instead:
For i = 6 to 100
Set IndirizzoInternet = Sheets("XXX").Range("H" & i).Hyperlinks
IndirizzoInternet.Range.Offset(0, 6).Value = IndirizzoInternet.Address
ISINs = Mid(IndirizzoInternet.Address, 78, 12)
Range("N" & i).Value = ISINs
Next
How do you know when to stop the loop? Is it a preset number of rows? If it not, you will want to have something determine the last row to process and replace the 100 with that variable.
I've started to use Macros this weekend (I tend to pick up quickly in regards to computers). So far I've been able to get by with searching for answers when I have questions, but my understanding is so limited I'm to a point where I'm no longer understanding the answers. I am writing a function using VBA for Excel. I'd like the function to result in a range, that can then be used as a variable for another function later. This is the code that I have:
Function StartingCell() As Range
Dim cNum As Integer
Dim R As Integer
Dim C As Variant
C = InputBox("Starting Column:")
R = InputBox("Starting Row:")
cNum = Range(C & 1).Column
Cells(R, cNum).Select
The code up to here works. It selects the cell and all is well in the world.
Set StartingCell = Range(Cell.Address)
End Function
I suppose I have no idea how to save this location as the StartingCell(). I used the same code as I had seen in another very similar situation with the "= Range(Cell.Address)." But that's not working here. Any ideas? Do I need to give more information for help? Thanks for your input!
Edit: I forgot to add that I'm using the InputBox to select the starting cell because I will be reusing this code with multiple data sets and will need to put each data set in a different location, each time this will follow the same population pattern.
Thank you A.S.H & Shai Rado
I've updated the code to:
Function selectQuadrant() As Range
Dim myRange As Range
Set myRange = Application.InputBox(Prompt:="Enter a range: ", Type:=8)
Set selectQuadrant = myRange
End Function
This is working well. (It appears that text is supposed to show "Enter a range:" but it only showed "Input" for the InputBox. Possibly this could be because I'm on a Mac?
Anyhow. I was able to call the function and set it to a new variable in my other code. But I'm doing something similar to set a long (for a color) so I can select cells of a certain color within a range but I'm getting all kinds of Object errors here as well. I really don't understand it. (And I think I'm dealing with more issues because, being on a mac, I don't have the typical window to edit my macros. Just me, basically a text box and the internet.
So. Here also is the Function for the Color and the Sub that is using the functions. (I've edited both so much I'm not sure where I started or where the error is.)
I'm using the functions and setting the variables to equal the function results.
Sub SelectQuadrantAndPlanets()
Dim quadrant As Range
Dim planetColor As Long
Set quadrant = selectQuadrant()
Set planetColor = selectPlanetColor() '<This is the row that highlights as an error
Call selectAllPlanets(quadrant, planetColor)
End Sub
This is the function I'm using to select the color that I want to highlight within my range
I would alternately be ok with using the interior color from a range that I select, but I didn't know how to set the interior color as the variable so instead I went with the 1, 2 or 3 in the input box.
Function selectPlanetColor() As Long
Dim Color As Integer
Color = InputBox("What Color" _
& vbNewLine & "1 = Large Planets" _
& vbNewLine & "2 = Medium Planets" _
& vbNewLine & "3 = Small Planets")
Dim LargePlanet As Long
Dim MediumPLanet As Long
Dim smallPlanet As Long
LargePlanet = 5475797
MediumPlanet = 9620956
smallPlanet = 12893591
If Color = 1 Then
selectPlanetColor = LargePlanet
Else
If Color = 2 Then
selectPlanetColor = MediumPlanet
Else
If Color = 3 Then
selectPlanetColor = smallPlanet
End If
End If
End If
End Function
Any help would be amazing. I've been able to do the pieces individually but now drawing them all together into one sub that calls on them is not working out well for me. Thank you VBA community :)
It's much simpler. Just
Set StartingCell = Cells(R, C)
after getting the inputs, then End Function.
The magic of the Cells method is it accepts, for its second parameter, both a number or a character. That is:
Cells(3, 4) <=> Cells(3, "D")
and
Cells(1, 28) <=> Cells(3, "AB")
One more thing, you can prompt the user directly to enter a range, with just one input box, like this:
Dim myRange as Range
Set myRange = Application.InputBox(Prompt:="Enter a range: ", Type:=8)
The Type:=8 specifies the input prompted for is a Range.
Last thing, since you are in the learning process of VBA, avoid as much as possible:
using the Select and Activate stuff
using unqualified ranges. This refers to anywhere the methods Cells(..) or Range(..) appear without a dot . before them. That usually leads to some random issues, because they refer to the ActiveSheet, which means the behavior of the routine will depend on what is the active worksheet at the moment they run. Avoid this and always refer explicitly from which sheet you define the range.
Continuing your line of thought of selecting the Range bu Selecting the Column and Row using the InputBox, use the Application.InputBox and add the Type at the end to restrict the options of the user to the type you want (Type:= 1 >> String, Type:= 2 >> Number).
Function StartingCell Code
Function StartingCell() As Range
Dim cNum As Integer
Dim R As Integer
Dim C As Variant
C = Application.InputBox(prompt:="Starting Column:", Type:=2) '<-- type 2 inidcates a String
R = Application.InputBox(prompt:="Starting Row:", Type:=1) '<-- type 1 inidcates a Number
Set StartingCell = Range(Cells(R, C), Cells(R, C))
End Function
Sub TestFunc Code (to test the function)
Sub TestFunc()
Dim StartCell As Range
Dim StartCellAddress As String
Set StartCell = StartingCell '<-- set the Range address to a variable (using the function)
StartCellAddress = StartCell.Address '<-- read the Range address to a String
End Sub
In Excel's VBA I want to create a formula which both takes the value from the source cell and the format.
Currently I have:
Function formEq(cellRefd As Range) As Variant
'thisBackCol = cellRefd.Interior.Color
'With Application.Caller
' .Interior.Color = thisBackCol
'End With
formEq = cellRefd.Value
End Function`
This returns the current value of the cell. The parts that I have commented out return a #VALUE error in the cell. When uncommented it seems the colour of the reference is saved however the Application.Caller returns a 2023 Error. Does this mean that this is not returning the required Range object?
If so how do I get the range object that refers to the cell that the function is used? [obviously in order to set the colour to the source value].
Here's one approach showing how you can still use ThisCell:
Function CopyFormat(rngFrom, rngTo)
rngTo.Interior.Color = rngFrom.Interior.Color
rngTo.Font.Color = rngFrom.Font.Color
End Function
Function formEq(cellRefd As Range) As Variant
cellRefd.Parent.Evaluate "CopyFormat(" & cellRefd.Address() & "," & _
Application.ThisCell.Address() & ")"
formEq = cellRefd.Value
End Function
This is the solution I found to the above question using Tim William's magic:
Function cq(thisCel As Range, srcCel As Range) As Variant
thisCel.Parent.Evaluate "colorEq(" & srcCel.Address(False, False) _
& "," & thisCel.Address(False, False) & ")"
cq = srcCel.Value
End Function
Sub colorEq(srcCell, destCell)
destCell.Interior.Color = srcCell.Interior.Color
End Sub
The destCell is just a cell reference to the cell in which the function is called.
The interior.color can be exchanged or added to with other formatting rules. Three extra points to note in this solution:
By keeping the value calculation in the formula this stops the possibility for circular referencing when it destCell refers to itself. If placed in the sub then it continually recalculates; and
If the format is only changed when the source value is changed, not the format as this is the only trigger for a UDF to run and thus change the format;
Application.Caller or Application.ThisCell cannot be integrated as when it refers to itself and returns a value for itself it triggers an infinite loop or "circular reference" error. If incorporated with an Address to create a string then this works though as per Tim William's answer.
I've created a macro variable active_search, which I'm trying to include in an evaluate function. However, the evaluate function is not outputting the correct value. Does anyone have suggestions on how to do this?
Code:
' this accurately provides the cell reference for the formula below
active_search = Cells(i, search_col).Address
active_match = Cells(i, match_col).Address
MsgBox active_search ' = &B$3 (or row i)
MsgBox active_match ' = $A$3
' formulas for the criteria below
' check_search = [IFERROR(MATCH(B3,list_keywords,0),0)] << Note: this works correctly
check_search = [IFERROR(MATCH(active_search,list_keywords,0),0)]
check_match = [IFERROR(SEARCH("exact",active_match,0)]
MsgBox check_search ' currently = 0 but should be 2250
MsgBox check_match
Thank you!
Let us expound what #simoco commented.
[] which is a shortcut for Evaluate function doesn't work on Variables.
To make it work, you have to explicitly use it as what simoco does in his comment.
To get the correct values use these instead (one already provided by simoco)
check_search = Evaluate("IFERROR(MATCH(" & active_search & "list_keywords,0),0)")
'~~> provided that list_keywords is named Range
check_match = Evaluate("IFERROR(SEARCH(""exact""," & active_match & ",0),0)")
Syntax: Evaluate(name) wherein the name argument is in a form of string.
Q1: The routine below does not work.
Sub test()
TXT = Sheets("INDEX").Cells(2, 1).Value
AKO = Sheets("TOBESEEN").Cells(1, 1).Address
Sheets("INDEX").Hyperlinks.Add Anchor:=Sheets("INDEX").Cells(1, 2), Address:="",
SubAddress:=AKO, TextToDisplay:=TXT
End Sub
Q2. Is there a place where I can see ALL the properties of the cell? When I type the "dot" after cells, VBA DOES NOT GIVE any options.
Like
sheet1.cell(1,2).VALUE
sheet1.cell(1,2).ADDRESS
sheet1.cell(1,2).?
I suspect my problem is related to the definition of AKO, but I am not sure what the correct property is (if not ADDRESS)
Thank you
Q1: Try to use this code:
Sub test()
TXT = CStr(Sheets("INDEX").Cells(2, 1).Value)
AKO = Sheets("TOBESEEN").Cells(1, 1).Address
Sheets("INDEX").Hyperlinks.Add _
Anchor:=Sheets("INDEX").Cells(1, 2), _
Address:="", _
SubAddress:=Sheets("TOBESEEN").Name & "!" & AKO, _
TextToDisplay:=TXT
End Sub
Q2: you can use F2 help. There you can see a list of properties and methods of any object.
When you create a hyperlink in a document, you need to give the "full" address of the cell you are linking to (sheet name and cell address), not just the .Address (which is the absolute address on a given sheet, but doesn't include the sheet information). To get the full address (including the sheet) of a cell, you need to add an additional parameter to the Address():
.Address(External:=True)
which will give you the full address, including workbook and sheet name. Unfortunately, if you create this link and then change the name of the workbook (maybe because you had not saved it up to this point), the link will break.
It is therefore (slightly) more robust to create a link that doesn't include the workbook name. The following routine does this for you. Note that is usually a good idea to split your code into small self-contained functions that do one thing particularly well - you can re-use your code more easily, and the main program becomes easier to read and maintain. I suggest therefore that you create the following function:
Sub addLink(target As Range, location As Range, text As String)
' create a hyperlink with text "text"
' in the cell "location"
' pointing to the range "target"
Dim linkAddress As String
linkAddress = target.Address(external:=True) ' this is the "full" address
'remove everything between brackets - this is the workbook name:
bLeft = InStr(1, linkAddress, "[") ' location of left bracket
bRight = InStr(bLeft, linkAddress, "]") ' location of right bracket
linkAddress = Left(linkAddress, bLeft - 1) + Mid(linkAddress, bRight + 1)
' now create the link:
location.Parent.Hyperlinks.Add _
anchor:=location, _
Address:="", _
SubAddress:=linkAddress, _
TextToDisplay:=text
End Sub
You can call this from your code above as follows:
Dim TXT As String, AKO As Range, LOC As Range ' define types when possible
TXT = Sheets("INDEX").Cells(2, 1).Value ' I called this text
Set AKO = Sheets("TOBESEEN").Cells(1, 1) ' I called this target (note - need Set)
Set LOC = Sheets("INDEX").Cells(1, 2) ' I called this location
addLink AKO, LOC, TXT
As for the second part of your question - "I can't see the properties of Cells()". Yes, that is annoying. It happens that Cells() is a lot like a Range object; you can see in the above that I was able to do Set AKO = Sheets("TOBESEEN").Cells(1,1) which demonstrates this. You can declare a variable as being of the type Range, and then tooltips will work for you:
Dim r As Range
r.
and as you type r. you will see a list of the properties of Range (which are also the properties of Cells):
It is annoying that Cells doesn't just do this by itself. It is annoying that Address() doesn't have an option to include the sheet name and not the workbook name. Excel is full of annoyances. In fact, there is even a book called "Excel Annoyances". And more could be written, I'm sure…
At least there are usually VBA tricks to work around these things and get the job done.