Even / Odd row skipping

I am generating a two column table where we would like to populate left column first and then the right column.

One way I thought of how to do this is to put the odd numbered rows in the left column and then the even numbered rows in the right column. All of this is within an interate statement. Since there are 20 possible replicants, how can I build a loop to check for odd/even rows without having to code in a bunch of case checks. On top of that, there will be 12 sections with the similar replicants.

The code snippet below works to skip row two, now I'd like to do all even rows from 2 to 20.

<iw_perl>$n++;</iw_perl>
<iw_if expn='{iw_value name="$n" /} eq "2" '>
<iw_then>
<iw_next></iw_next>
</iw_then>
</iw_if>
... if not skipped, place the result into the table

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Comments

Adam Stoller

I'm not positive (haven't tried it out), but I think you can do something like:

...
<iw_if expn='{iw_value name="$n"/} % 2 == 0'>
...

Basicaly x % y returns the remainder from doing x / y - so: 2 % 2 is 0, 3 % 2 is 1, 4 % 2 = 0, ...

--fish
(Interwoven Senior Technical Consultant)

Migrateduser

Correction to my problem -
My apologies - was not thinking clearly in stating my problem.

Really, I need to figure out how to divide up the rows into two columns. With the left being populated first and then the 2nd half on the right.

For example:
A D
B E
C F

The problem is that the number of rows will vary, I guess that means I have to count the number of rows up front, and then divide by two and then place them into iterate loops with a counter until it matches the calculated stop point...Any better ideas on how to do this ?

Thanks in advance,

Adam Stoller

If you're willing to break out into a bit of Perl - you could read the replicant list into an array and use the scalar value of that array to control your loop. Something like (and again, this is off the top of my head, unverified as fully functional):

my $list_size = scalar(@list);
my $half = $list_size / 2;
$half++ if ($list_size % 2); # put the bigger half on the left
my $idx = 0;
while ($idx < $half){
    ... left-column $list[$idx] ...
    if ($idx + $half < $list_size){
       ... right-column $list[$idx+$half] ...
    }
    $idx++;
}

--fish
(Interwoven Senior Technical Consultant)

MattP

Or if you didn't want as much perl, based on the previous post, you could do something like...

<table>
<iw_if expn='{iw_value name="$n"/} % 2 == 0'>
<iw_then>
<tr> <td> iw value goes here </td>
</iw_then>
<iw_else>
<td> iw value goes here </td></tr>
</iw_else>
</iw_if>
</table>

Matt

Matthew Petitjean
BOC Group
Murray Hill, NJ 07974 USA

Adam Stoller

Matt - your solution works, I believe, for the initial description of the problem - in which you get something like:

1	2
3	4
5

but I believe it won't work for the revised description:

1	4
2	5
3

--fish
(Interwoven Senior Technical Consultant)