"how many more optional values will you have ?"

Arnon

Arnon, hi;

Based on your suggestion I'm using this code:

thisform.transGrid1.SetAll("dynamicbackcolor", ";
IIF(col1=0, RGB(255,255,255),; && white
IIF(col1=1, RGB(255,0,0),; &&
IIF(col1=2, RGB(0,255,0),;
IIF(col1=3, RGB(0,0,255),;
IIF(col1=4, RGB(0,255,255),; && green
IIF(col1=5, RGB(255,255,0),; && yellow
IIF(col1=6, RGB(0,128,0),;
RGB(0,0,0))))))))", "Column")

It works, but as you see it does change the whole row's color depending on the value of col1 as first column; now my grid should end up looking like this:

Record col1 col2 col3 col4 col5... coln
1 1 (blue) 2(red) 1(blue) ...
2 3 (grn) 1(blue) 2(red) ...
3 4 (yel) 5(orn) 1(blue) ...

So, basically, each cell has to be painted in a different color, because the have a different value, even if they happen to be located in the same row.

What do you think?