Ok
thought I would put my 2cents in here. I figured if you wanted to adapt it
for more than 3 columns or less than 3 columns it would be silly to be hard
coding in cells so here you go.
<cfset query = queryNew("") />
<cfset queryAddColumn(query, "column",
listToArray("1,2,3,4,5,6,7,8,9,10,11")) />
<cfset columns = 3 />
<cfset rows = CEILING(query.recordCount / columns) />
<cfoutput>
<table border="1">
<cfloop from="1" to="#rows#" index="row">
<tr>
<cfloop from="0" to="#columns-1#" index="cell">
<td>
<cfif row + ((rows * cell)) LTE query.recordCount>
#query["column"][row + ((rows * cell))]#
<cfelse>
</cfif>
</td>
</cfloop>
</tr>
</cfloop>
</table>
</cfoutput>
This will also fill any empty cells automatically
Have fun
Steve Onnis
________________________________
From: cfaussie@googlegroups.com [mailto:[EMAIL PROTECTED] On Behalf
Of Andrew Scott
Sent: Thursday, 5 April 2007 6:12 PM
To: cfaussie@googlegroups.com
Subject: [cfaussie] Re: outputting recordset
Yes they are very quick and it wouldn't make any perfomace issue one way or
the other.
On 4/5/07, Dale Fraser <[EMAIL PROTECTED]> wrote:
Ahh,
Picky, in my defence, array functions are very quick :P
Regards
Dale Fraser
http://dale.fraser.id.au/blog
-----Original Message-----
From: cfaussie@googlegroups.com [mailto: cfaussie@googlegroups.com
<mailto:cfaussie@googlegroups.com> ] On Behalf
Of Scott Thornton
Sent: Thursday, 5 April 2007 4:26 PM
To: cfaussie@googlegroups.com
Subject: [cfaussie] outputting recordset
Hi,
You probably don't want to calculate the length of the array within
every
iteration of the loop if it does not change.
<cfset arraylength = arrayLen(data)>
>>> "Dale Fraser" < [EMAIL PROTECTED]> 05/04/2007 3:29 pm >>>
Working example.
<cfset data = listToArray('1,2,3,4,5,6,7,8,9,10,11,12,13') />
<cfset cols = 3 />
<cfset rows = ceiling(arrayLen(data) / cols) />
<cfoutput>
<table>
<cfloop index="i" from="1" to="#rows#">
<tr>
<cfif i lte
arrayLen(data)><td>#data[i]#</td></cfif>
<cfif i+rows*1 lte
arrayLen(data)><td>#data[i+rows*1]#</td></cfif>
<cfif i+rows*2 lte
arrayLen(data)><td>#data[i+rows*2]#</td></cfif>
</tr>
</cfloop>
</table>
</cfoutput>
Regards
Dale Fraser
http://dale.fraser.id.au/blog
_____
From: cfaussie@googlegroups.com <mailto:cfaussie@googlegroups.com>
[mailto:[EMAIL PROTECTED] On Behalf
Of Taco Fleur
Sent: Thursday, 5 April 2007 3:22 PM
To: cfaussie@googlegroups.com <mailto:cfaussie@googlegroups.com>
Subject: [cfaussie] Re: outputting recordset
Hi Seona,
that is it! Thanks.
* count
* count + divisor
* count + (divisor * 2)
On 4/5/07, Seona Bellamy < [EMAIL PROTECTED]> wrote:
On 05/04/07, Dale Fraser < [EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED] > >
wrote:
So are you saying a fixed number of rows 6
Or a fixed number of columns 3?
I believe the line in the original email was:
I need to output a recordset in a table, infinite rows, and maximum
3
columns in width.
I'm more interested in knowing if there was a reason for the number
of empty
cells in the third column. Because I'm pretty sure that the columns
could
have been closer to even in length, and that would be easier to do I
think.
The maths is making better sense in my head, anyway. Then again,
knowing my
history with maths, I'm not sure that's saying much...
Basically, what I'm thinking is something along these lines:
* Divide your number of records by 3 and round it off to an integer
(I'll
call this divisor). This is the number of records you will have in
each of
the first two (complete) columns. The third column may or may not
have this
many records.
* Loop over your recordset this many times, with a counter of some
sort to
count the iterations.
* In the loop, get the three cells as follows:
* count
* count + divisor
* count + (divisor * 2)
I haven't tested this, so it's purely theoretical and may or may not
work.
Oh, and you'd probably want to test if the value exists in the third
column
because that one may not be full.
Cheers,
Seona.
Web Design, Web development, Graphic Design and Complete Internet
Solutions
an industry leader with commercial IT experience since 1994 .
www.aegeon.com.au
Phone: +613 8676 4223
Mobile: 0404 998 273
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