[allcalParticipant]

I added this to create a “made by link” to shop products:

function show_designer () {
echo ‘<p class=”showdesigner”>By: Taylor and Tessier</p>’;
}
add_action(‘woocommerce_shop_loop_item_title’,’show_designer’, 20);

It works but I want to make that function only show for a single woocommerce category “Taylor and Tessier” slug: taylor-and-tessier

How do I wrap that function to only show for that specific woocommerce category?

http://aperfectcomplement.com/new/shop/

[NicoModerator]

Hi There,

Thanks for writing in. You could add this code below

if ( is_category() || x_is_portfolio_category() || x_is_product_category() )

For example:

function show_designer () {
<code>if ( is_category('categoryhere'))</code>
echo ‘<p class=”showdesigner”>By: Taylor and Tessier</p>’;
}
add_action(‘woocommerce_shop_loop_item_title’,’show_designer’, 20);

Let us know how it goes.

Feel free to ask us again.

Thanks.

[allcalParticipant]

Couldn’t get it to work. Can you give the correct code instead of an example?

[LelyModerator]

Hi There,

Please use this code:

function show_designer () {
if ( is_category('taylor-and-tessier')){
echo '<p class="showdesigner">By: Taylor and Tessier</p>';
}
add_action('woocommerce_shop_loop_item_title','show_designer', 20);

Hope this helps.

[allcalParticipant]

That code is giving: Parse error: syntax error, unexpected $end in /home/content/37/13168137/html/new/wp-content/themes/x/functions.php

[DarshanaModerator]

Hi there,

Could you please try updating the code as follows.

function show_designer () {
  if ( is_category('taylor-and-tessier')){
    echo '<p class="showdesigner">By: Taylor and Tessier</p>';
  }
}
add_action('woocommerce_shop_loop_item_title','show_designer', 20);

Hope that helps.

[Author]

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