Array from Range in Excel VBA
Asked Answered
Y

5

7

Well I've been struggling with the little bit of code and can't seem to get around it ... I'm trying to get an array from a range of cells, the array however is showing up to be 1 element wide.
Well here's the code:

Dim item As Variant
MsgBox Range("D19:H19").Count    
item = Range("D19:H19").Value
MsgBox LBound(item) & " " & UBound(item)   

as per my understanding item should contain a 2D array... however I'm getting the following result 1st MsgBox prints 5 2nd MsgBox prints 1 1

What's going wrong?

Yesteryear answered 10/10, 2009 at 22:1 Comment(4)
I'm no vba expert, but taking the value of a range (or array) seems odd. Is it returning the first value of the range?Distill
Generally VBA would return a 2D array with row values being the 1st dimension and column values being the 2nd dimension , however here I am seeking a range which has 5 cells all in one row, so I'm not quite sure whether a 1d or 2d array will be returned...Yesteryear
@dave: and yes its returning only the first value of the range... any ideas why it's doing so?!Yesteryear
It does not actually return the first value of the range. Select a 2D-range of cells and press Tab several times. You'll see a selected cell that is being highlighted. That's the cell Value returns the content of.Volnak
F
11

The problem is in LBound and UBound

jtolle was correct about the LBound and UBound.

LBound(item, 2)

UBound(item, 2)

However, item must not be dimmed as an array (you'll get an error).

I think this is what you want

Dim item As Variant
MsgBox Range("D19:H19").Count
item = Range("D19:H19").Value

MsgBox LBound(item, 2) & " " & UBound(item, 2)

For i = LBound(item, 2) To UBound(item, 2)
  MsgBox item(1, i)
Next
Frontpage answered 10/10, 2009 at 22:1 Comment(2)
That's correct. My edit was meant to describe the form of the array returned by getting the Value property of the Range object and putting it in item. I wasn't intending to suggest that item should be declared as an array, only that once assigned, it would look the same as a variable that had been declared as such.Vest
@imfrancisd: Works like a charm! I still didn't understand what was I doing wrong, I don't seem to understand the array concepts in VBA..Yesteryear
V
3

Your item should contain a 2-D array as expected. If you stick a breakpoint in your code and look at the little "Locals" window in the VBA editor, you should see that. Your calls to LBound and UBound are getting the bounds in the first dimension. If you call Lbound(item,2) and UBound(item,2), you should get 1 and 5 as you expect.

EDIT: That is, once you've made the assignment, item would look like something you could have declared as such:

Dim item(1 to 1, 1 to 5)

One of the banes of VBA programming is that arrays can have arbitrary lower bounds. So all of your code needs to be aware of that.

Vest answered 11/10, 2009 at 0:19 Comment(0)
V
1

That's correct as is. Even if you select an array of cells, you still have the option to select one single cell out of the array (and step for example with tab through the items of this array)

.Value

only gets you the content of the currently single-selected cell.

if you want the enumeration of the array, you may call the .Cells()-method of the Range-object

Assuming that D19 until H19 contain "a" through "e" respectively, calling

Range("D19:H19").Cells(2)

returns you "b". Note that this is a one-based array and can be 2-dimensional. Cells() takes at most 2 parameters to specify the inner offset from the selection's origin.

hope that clarifies... regards

Volnak answered 10/10, 2009 at 22:14 Comment(6)
The code snippet here support.microsoft.com/kb/213798 in the "To take values from a worksheet and fill the array" section shows a similar code, so I wonder why my code is not functioning!Yesteryear
Well I tested this on Excel 2k3. Possibly, this is due to the fact that Value might be 2-dimensional... lbound takes a second parameter which expresses as what we call dimension. I'm not sure anymore which parameter corresponds to which value. What does VarType(Range("D19:H19").Value) return? It should be vbArray... You might try Ubound(item,1) instead of Ubound(1)Volnak
well I tried Ubound(item, 1) but to no avail.. I still get the 1 1 from the 2nd MsgBox output.. And regarding what (Range("D19:H19").Value) should return... I think i'm not really sure..I'm just working on code samples..Yesteryear
hmm strange then... well the article refers to all the versions before 2007, but not 2007 itself. Therefore it's possible that it works differently now. How about the solution with .Cells(number) ? ... Yet another point: Try declaring item with item() instead of itemVolnak
Using .Cells(number) I will have to iterate through all the cells in the range in a for-next loop, right?... well if there is no other choice I won't mind going for this solution...as long as it works consistently...Yesteryear
yes you would have to, unfortunately. if Range("D19:H19").Value = yourArray doesn't work, I cannot see any other solution. In the end, Excel doesn't do anything else except that it runs way much faster because it is compiled into Excel's basic functionsVolnak
K
1

Try this:

Dim item As Variant
MsgBox Range("D19:H19").Count
item = Application.Transpose(Range("D19:H19").Value)
MsgBox LBound(item) & " " & UBound(item)
Klusek answered 15/10, 2009 at 19:14 Comment(2)
I thought "Transpose" was meant for columns... will check this out anyway though...Yesteryear
FYI, transpose is not available in Excel 2003.Capote
H
0

if you want a 1D array, to join it for an IN clause, for example, you should transpose your range. I've found you have to transpose twice for a row, once for a column of data like this:

Dim rngRow As Range, rngColumn As Range

Set rngRow = Sheets(1).Range("A1", "Z1")
Set rngColumn = Sheets(1).Range("A1", "A20")

Dim arrRowValues, arrColValues
arrRowValues = WorksheetFunction.Transpose(WorksheetFunction.Transpose(rngRow))
arrColValues = WorksheetFunction.Transpose(rngColumn)

Dim numList As String, stringList As String
numList = Join(arrRowValues, ",")
stringList = "'" & Join(arrColValues, "','") & "'"

worth a play.

Harness answered 14/10, 2009 at 13:51 Comment(0)

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