I have tried putting the addrow inside the loop...
at the beginning and end of the loop statement...
and nothing...

AARGH....

<smile>
> 
> From: "Walter Purvis" <[EMAIL PROTECTED]>
> Date: 2006/08/14 Mon PM 09:30:29 EST
> To: "'REALbasic NUG'" <[email protected]>
> Subject: RE: my silly listbox
> 
>  
> 
> >  -----Original Message-----
> >  From: [EMAIL PROTECTED] 
> >  [mailto:[EMAIL PROTECTED] On 
> >  Behalf Of Bill Johnson
> >  Sent: Monday, August 14, 2006 8:17 PM
> >  To: REALbasic NUG
> >  Subject: Re: my silly listbox
> >  
> >  
> >  On 14-Aug-06, at 5:03 PM, [EMAIL PROTECTED] wrote:
> >  >
> >  > Here is my code....
> >  >
> >  > after it runs I am left with the number 20 in what I would 
> >  call cell 
> >  > 0,1
> >  >
> >  > Regards
> >  >
> >  > ========================
> >  >   dim i as integer
> >  >   me.HasHeading=true
> >  >
> >  >   me.heading(0)="Heading1"
> >  >   me.Heading(1)="heading2"
> >  >
> >  >   me.addrow""
> >  >
> >  >   for i = 1 to 20
> >  >     me.Cell(0,0)=str(i)
> >  >   next i
> >  
> >  The first parameter of the cell property is its row number 
> >  and the second is its column number. When you say
> >  
> >  > for i = 1 to 20
> >  >     me.Cell(0,0)=str(i)
> >  >   next i
> >  
> >  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
> 
> 
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