On 14-Aug-06, at 6:30 PM, Walter Purvis wrote:
you're addressing the cell at row 0 and column 0 which is
the first cell in the listbox no matter how many rows you add.
To fill in the cells of the row you've just added with the
cell's column index increment the column parameter.
dim i as integer
me.HasHeading=true
me.heading(0)="Heading1"
me.Heading(1)="heading2"
me.addrow""
for i = 0 to 20
me.Cell(me.lastIndex,i)=str(i)
next
LastIndex is the index of the last row added to the listbox
which is the one you've just added with me.addrow"".
Well, sure, but me.addrow is not inside the loop, so at the end of
the day
he's still only going to have one row in the listbox and at some
point he's
going to be addressing columns which don't exist, unless he has 20+
column;
i.e., at the end, he's setting column #20 -> me.cell(0,20) = 20
Yes, that gives you 1 row with 21 columns filled in a listbox with 21
columns. This will give you 21 rows(listbox is zero based) with the
first 2 columns filled in.
dim i as integer
me.HasHeading=true
me.ColumnCount = 2
me.heading(0)="Heading1"
me.Heading(1)="heading2"
for i = 0 to 20
me.addrow""
me.Cell(me.lastIndex,0)= "Row " + str(i) + " in column 0"
me.Cell(me.lastIndex,1)= "Row " + str(i) + " in column 1"
next
See, and it's not even the end of the day yet. :D
Bill Johnson
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