PHP & MySQL SELECT Substring - php

Question

$content is a variable with a 'detailed description'.
product_id is column which might contain a substring of the detailed description ($content) in a MySQL table called products
I am trying to create a select statement that would find a record if the product_id is CONTAINED in the $content variable. Then I want to update another table called receive_sms with the url field from the SELECT staement
Researching on the website I have come up with the following.... But it doesn't work
$mysqlic = mysqli_connect("testsms.cloudaccess.net", "username", "password", "testsms2");
$prod_res=mysqli_query($mysqlic,"SELECT url from products
WHERE %product_id% LIKE %$content%");
mysqli_query($mysqlic,"INSERT INTO recieve_sms (comments) VALUES ('$prod_res')");
Any Ideas??

Answer 1

It should be:
$prod_res = mysqli_query($mysqlic, "SELECT url from products
WHERE '$content' LIKE CONCAT('%', product_id, '%')");
or:
$prod_res = mysqli_query($mysqlic, "SELECT url from products
WHERE LOCATE(product_id, '$content') != 0");
You need to put $content in quotes.
Actually, it would be better if you used a prepared query, then '$content' would become a placeholder ?.
After you query, you need to call mysqli_fetch_assoc() to get the column value:
$row = mysqli_fetch_assoc($prod_res);
$url = $row['url'];

Answer 2

Well, first of all, I don't understand why you're using wildcards on you field name. You want to check if a value is contained within the value of that field, consider changing this:
... WHERE %product_id% LIKE %$content%);
to
... WHERE product_id LIKE '%$content%');
Also, please, use prepared statements to avoid SQL injection. You're using MySQLi, which supports them.
EDIT
Also, the return of a mysqli_query is a MySQLi Resource. You'll have to fetch the results from the resource to gain access to the value you're looking for.