On Wed, 2004-05-26 at 11:12, James Earl wrote: > On Wed, 2004-05-26 at 10:08, James Earl wrote: > > On Tue, 2004-05-25 at 18:14, Andreas L. Delmelle wrote: > > > > > Say you have a max record width (sum of all respective max field widths) > > > of > > > 65, then each column gets its width according to a calculation like > > > > > > [( proportional-width / 65 ) * remaining-table-width ] > > > > I'll post any success that I have! > > I'm not sure if I'm doing this right... but the following gave > acceptable results: > > Columns (max field length): > > 1. 20 > 2. 10 > 3. 5 > 4. 1 > 5. 11 > 6. 6 > > Total: 53 > > Calculations: > > 1. (20 / 53) * 33 = 12.5 (e.g. proportional-column-width(12.5)) > 2. (10 / 53) * 43 = 8.1 > 3. (5 / 53) * 48 = 4.5 > 4. (1 / 53) * 52 = 0.9 > 5. (11 / 53) * 42 = 8.7 > 6. (6 / 53) * 47 = 5.3 > > I've just tried this once so far, but it worked great. I have two > templates set up. Once exports xsl-fo, while the other one exports > html. The PDF generated by FOP looks almost identical in terms of the > column widths. > > I just need to take into account the max length of the column heading > when it is larger than the column data. > > James
Okay, I see now that a string length WILL WORK! Man do I feel stupid. :) Well, at least now I understand why! Man, this is easy, who needs auto table layout!!! --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
