variable column selection

[nigelandrewfoster]

Hi. I want to replace
Columns("a:g") with a variable, so I can select a variable range of columns. Can this easily be done without looping, please?
Thanks

 

[MickG]

Do you mean:-

Dim rng As Range
Set rng = Columns("A:G")
rng.Select

Mick

 

[nigelandrewfoster]

Hi MIck. No, the number of columns I want to hide varies, so I need the equivalent of
x=1
y=7
Columns(x:y).Hideentire columns (or whatever)=true

 

[nigelandrewfoster]

Hi MIck. No, the number of columns I want to hide varies, so I need the equivalent of
x=1
y=7
Columns(x:y).Hideentire columns (or whatever)=true

 

[MickG]

Possibly:-

Dim x, y
x = 3
y = 5
Columns.Hidden = False
'Columns (C to G)
Columns(Chr(64 + x)).Resize(, y).Hidden = True

Mick

 

[nigelandrewfoster]

Displays 'Compile error: Expected: list separator or ) at the first period before 'resize'

 

[MickG]

Try stepping through this code a line at a time, See if any of these lines work:-
It works in xl2002

Dim x, y
x = 3
y = 5
Columns.Hidden = False
'Columns (C to G)
MsgBox Chr(64 + x)
MsgBox Columns(Chr(64 + x)).Resize(, y).Address
MsgBox Columns(Chr(64 + x) & ":" & Chr(64 + 3)).Address
'Columns(Chr(64 + x)).Resize(, y).Hidden = True
Columns(Chr(64 + x) & ":" & Chr(64 + x)).Resize(, y).Hidden = True

also

Dim x, y
x = 3
y = 5
MsgBox Range(Chr(64 + x) & ":" & Chr(64 + x)).Resize(, y).Address

Mick

 

[jim may]

Possibly:-

Dim x, y
x = 3
y = 5
Columns.Hidden = False
'Columns (C to G)
Columns(Chr(64 + x)).Resize(, y).Hidden = True

Mick

The above worked for me WITHOUT ERROR, in testing. Jim

 

[Weaver]

Something along lines of:

Sub test()
    a = 3
    b = 7
    Range(Columns(a), Columns(b)).Select
End Sub

 

[nigelandrewfoster]

Thankd guys. IT worked when I typed it in again. Using phone for internet so couldn't copy and paste. No idea why it came up with that error before. Much appreciated. Your efforts are not wasted. I have learned a lot from these examples.