use MOD 3  in output

Dan Haley <[EMAIL PROTECTED]> wrote:

<cf_untested confidence_level="medium" but="you get the idea">

<cfset variables.num_cols = 3>
<cfset variables.col_len = int((query.recordcount - 1) / variables.num_cols)
+ 1>

<cfloop from="1" to="#variables.col_len#" index="variables.i">
        <cfloop from="1" to="#variables.num_cols#" index="variables.j">
        <cfoutput>
        #query.field[((variables.i - 1) * variables.num_cols) +
variables.j]#
        </cfoutput>
        </cfloop>
</cfloop>

</cf_untested>

Dan

-----Original Message-----
From: Michael Kear [mailto:[EMAIL PROTECTED]]
Sent: Thursday, August 31, 2000 11:12 PM
To: [EMAIL PROTECTED]
Subject: How to display across a row instead of down?



I want to display the results of my query in 3 columns instead of one.  I
know the method is something to do with mod3, but I'm dashed if i can
figure it out.

What I want to get is:

[record1] [record2] [record3]
[record4] [record5] [record6]
etc

Can anyone help me please with how to code it?

Cheers,
Mike Kear
AFP Web Development
Windsor, NSW, Australia


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