How To Generate URL Slug From Metaobject?

Technical Q&A
1

I have a metaobject called “brand” within a metafield “created_by” and I want to use the slug of the “name” key of the brand to create a link to the brand page template. Example link… https://www.beforeourtime.com/pages/brand/colors-greys

“colors & greys” is the name of a brand. This is the metaobject definition… https://snipboard.io

This is how I reference the brand name on my product page template…

{{ product.metafields.custom.created_by.value.name }}

2

Hi,

  1. Create a Metafield–>Assign a Value(This value could be a string that you want to use as a part of the URL slug.)

  2. Generate the slug using the metafield value, you can use liquid code. Here’s an example for a product:

{% assign product_metafield = product.metafields.namespace.key %}
{% assign slug = product_metafield | handle %}

In this code:

product.metafields.namespace.key should be replaced with the actual namespace and key of your metafield.

handle is a filter in liquid that converts a string into a URL-friendly slug. It removes spaces, converts to lowercase, and replaces special characters.

Use the generated Slug in Your URLs

Example

custom product URL

{{ product.title }}

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2 Likes
3

Thanx Bro!