If I want to set a field as a css value.

I try to make a templete,and I want to set a select field as a css style selector.

In the templete I want to make field in some tags.
like
<ul class="field1value field2value field3value">

I write
<ul class="iteminfo <?php foreach ($items[$i]->positions as $field) : ?> <?php echo $field->display; ?><?php endforeach; ?>">
But it error. :oops:


Sorry my english is poor. :oops:
Thanks for everyone!

What error message do you get?

<br />

<b>Warning</b>: Invalid argument supplied for foreach() in <b>..\www\TIWC\components\com_flexicontent\templates\profile\category_items.php</b> on line <b>236</b><br />

The line236 just what I add in the templete. :

Have you created a position called color and put your fields here? I'm not that good with FC templates, but I guess you have to do that to use positions.

Do that and try the code below

EDIT: Is it an item view or category view you're working with?

yes, I add a position in category.xml
<fieldgroups>
<group>images</group>
<group>otherinfo</group>
<group>color</group>
</fieldgroups>
than add a field also named color in the color position.

I really happy to find out Flexicontent is so powerfull.
I want to design a a category templete as school table.
But I'm not good in PHP.
So I want to set fields as css class with jQuery to control every category items.

I tried to build my first template last day. :oops:
140.122.88.193/TIWC/index.php?op ... e&Itemid=4

Many Thanks!!

Ok, so the template you want to create is a category template, where all the items are listed and when you click the item title you see the detailed view of the template. Correct?
And every item should have its own row color?

I guess you have duplicated the default template or something?

Do I understand correctly?