Excel in Office XP- Range/Selection Object

[drkestrel]

This is what I am trying to do

1) User manually select a random range in Excel for Office XP (e.g. A8:M10, A20:M20, A30:M:32)

2) User click a VBA Button
2.1 For each row in time
2.1.1) Code should modify the first "column" of the row
2.1.2) Code should set the "Pattern" of the row to grey
2.1.3) On error, stop processing subsequent rows (rows processed could stay as is)

How could I find out the Range "co-ordinates", I think I need the absolute range (because after processing each row, I want to highlight that row Grey (which would lose my selection). How could this be done?

 

[xlbo]

You can use syntax like this to pick up the 4 corners of your selection - then use the cells syntax ( cells(row,col) ) to reference the rows and columns

stRw = selection.rows(1).row
enRw = selection.rows(selection.rows.count).row
lCol = selection.columns(1).column
rCol = selection.columns(selection.columns.count).column

Rgds, Geoff
[blue]Si hoc signum legere potes, operis boni in rebus Latinus alacribus et fructuosis potiri potes![/blue]
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[drkestrel]

Given objCurWorksheet as the Worksheet object that I am working on, how could I find out obtain the Range object of what the user has selected?

 

[xlbo]

give a man a fish and he'll eat for a day, teach him how to fish etc etc etc. I'm quite disapointed here that you cannot breach the small gap between what I have given you and what you need.

dim SelRange as range, startRow as long, endRow as long, startColumn as integer, endColumn as integer
with objCurWorksheet.Selection
set selRange = range(.address)
startRow = .rows(1).row
endRow = .rows(.rows.count).row
startColumn = .columns(1).column
endColumn = .columns(.columns.count).column
msgbox selRange.address
msgbox "Start Row is:" & startRow & vbcrlf & "End Row is:" & endRow & vbcrlf & "Start Column is:" & startcolumn & vbcrlf & "End column is:" & endColumn
end with

Rgds, Geoff
[blue]Si hoc signum legere potes, operis boni in rebus Latinus alacribus et fructuosis potiri potes![/blue]
Want the [red]best[/red] answers to your questions ? faq222-2244