Dyamic range - selecting the starting value

Hi Guys -

So I have this dynamic range, I'm sure you can see what it does:

=OFFSET(Visits!$A$4, 0, COUNTA(Visits!$4:$4)-1, 1,-MIN(chtLen,COUNTA(Visits!$4:$4))-1)

chtLen is the number of months, as selected by the user.

So this formula currently selects cell A4, goes right to the end of the last entry in that row, goes back one cell, and for the width of the range goes back the specified in the number of months they have selected.

However, I want to do this differently. Rather than the above, I want to achieve this:

Depending on a month and year selected, the range starts at a relevant column, then goes back for the number of months selected. Is this possible?

So for example:

The user selects August 2012, the value of which is in column 5 (E) of the source data. This column number is returned using a MATCH formula:

=MATCH(Traffic!F3,Visits!$1:$1,0)

I can specifiy a fixed row number here (4).

So, in this case I want the range to start from cell E4, then go back the length selected (chtLen).

So I guess instead of the fixed cell (A4) in the OFFSET formula above, it would need to reference the starting cell (which depends on the month/year selected): this starting cell would be found using the MATCH formula for the column number, and the row number can be specified as a number.

If this doesn't make sense I will mock up a spreadsheet, as the one I have built contains sensitive data.

Thanks,
John

Actually, I've just realised there was an error or two:

1) If chtLen is higher that the number of columns left of the starting column, this causes no cells to be selected.
2) Column a contains row name (Visits, Page Views, etc). This is sometimes included in the range. Can this be excluded?

Thanks,
John

Awesome - that was perfect, just needed a slight modification (when the last column is selected, no need to move to right any columns).

=OFFSET(INDEX(Visits!$4:$4,MATCH(Traffic!F3,Visits!$1:$1,0)), 0, 0, 1,-MIN(chtLen,COUNTA(Visits!$4:$4))-1)

Thanks so Much!!!!

John