Here is an example from the Internet that should help you solve this:

To find the tracks whose names contain the Wild literal string, 
you use % wildcard at the beginning and end of the pattern:

SELECT
    trackid,
    name    
FROM
    tracks
WHERE
    name LIKE '%Wild%';

Thanks. Just got it.

Try providing answers to the matching pattern question but seems not matching at all .can some help? In the e-commerce database we have a products table. The columns are id, name, description and price.

Find all the products where the pattern 't-shirt' can be found anywhere in the product name.

Hey Bright so for this challenge, you want to use the LIKE keyword

Kris is on the right track with the formatting as well the example solution