vba, flip it over the pattern

[montecarlo2012]

Hello people
with this code,

Sub rur()
      For i = 1 To 5
            For j = 1 To 5
                  Cells(i, j) = i
            Next
      Next
End Sub

I got this

how can I get

Thank for reading.
Please, some help in this.

 

[Marc L]

Hi,​

​

according to Excel basics it could be achieved without looping, anyway :​

Sub Demo1()
    For R& = 1 To 5:  [A:E].Rows(R).Value2 = 6 - R:  Next
End Sub

 

[montecarlo2012]

Your respond is appreciated,
would be real helpful for me
IF---in case is possible for you
be able to see the translation from

'for(i=5;i>=1,i--)
'for(j=5;j>=1,j--)
'print("%d",i)

to the VBA language. Please
IF, you don't mind

 

[Marc L]

​

Something like For i = 5 To 1 Step -1 - to see in VBA help - but this logic obviously not well fits your initial post …​

 

[mikerickson]

Dim myArray() as Variant
With SomeRange
    Redim myArray(1 to .Rows.Count, 1 To .Columns.Count)
    For i = 1 to .Rows.Count
        For j = 1 to .Columns.Count
            myArray(.Rows.Count - i + 1, j) = .Cells(i, j).Value
        Next j
    Next i

    .Value = myArray
End With

There is no in-place looping that won't require a temporary storage location.

 

[Marc L]

Or according to the initial post obviously :​

      For i = 1 To 5
      For j = 1 To 5
          Cells(i, j) = 6 - i
      Next j, i

 

[montecarlo2012]

Thank you Mike.

 

[montecarlo2012]

Thank you "Marc L" .

 

[Rick Rothstein]

I'm guessing your goal is to understand looping, but if your goal is simply to produce the output you showed no matter the method, then you could consider this...

Sub MonteCarlo2012()
  [A1:E5] = [{5;4;3;2;1}]
End Sub

 

[navic]

Another way using a formula (copy across)

=INDEX(A$1:A$5,COUNTA(A1:A$5))