Sending sql query and sql logs, how to insert everything?

Question

MySQL

Good afternoon, I am studying and doing parsing for myself (training).
I want to make it send a request to the database, but it doesn't work.
When adding a product to the site, there are requests:

FROM s_products AS p
            LEFT JOIN s_lang_products l ON l.product_id=p.id AND l.lang_id = 1
            WHERE 
                1 
                AND p.url = 'nazvanie'
            GROUP BY p.id
            LIMIT 1
        	SHOW TABLES LIKE '%s_languages%'
        	SELECT DISTINCT
                p.id,
                p.url,
                p.brand_id,
                p.position,
                p.created as created,
                p.visible,
                p.featured,
                p.rating,
                p.votes,
                p.last_modify,
                l.name, l.meta_title, l.meta_keywords, l.meta_description, l.annotation, l.body, l.special
            
      FROM s_products AS p
            LEFT JOIN s_lang_products l ON l.product_id=p.id AND l.lang_id = 1
            WHERE 
                1 
                AND p.url = 'nazvanie'
            GROUP BY p.id
            LIMIT 1
        	INSERT INTO s_products SET id='0', name='Название', visible='1', featured=NULL, brand_id='0', url='nazvanie', meta_title='Title', meta_keywords='Keywords', meta_description='Description', annotation='', body='', rating='0', votes='0', last_modify='2018-02-20 19:20:01'
        	UPDATE s_products SET position=id WHERE id=2888
        	SELECT count(*) as count FROM s_lang_products WHERE lang_id='1' AND product_id=2888 LIMIT 1
        	INSERT INTO s_lang_products SET name='Название', meta_title='Title', meta_keywords='Keywords', meta_description='Description', annotation='', body='', lang_id='1', product_id='2888'	    	
        	SHOW TABLES LIKE '%s_languages%'
        	SELECT DISTINCT
                p.id,
                p.url,
                p.brand_id,
                p.position,
                p.created as created,
                p.visible,
                p.featured,
                p.rating,
                p.votes,
                p.last_modify,
                l.name, l.meta_title, l.meta_keywords, l.meta_description, l.annotation, l.body, l.special
            
      FROM s_products AS p
            LEFT JOIN s_lang_products l ON l.product_id=p.id AND l.lang_id = 1
            WHERE 1 AND p.id = 2888 GROUP BY p.id
            LIMIT 1
        	select category_id from s_products_categories where product_id='2888'
        	DELETE FROM s_products_categories WHERE product_id='2888'
        	update s_categories set last_modify=now() where id=24
        	INSERT IGNORE INTO s_products_categories SET product_id='2888', category_id='24', position=0
        	INSERT INTO s_variants SET id='', name='', sku='', price='', currency_id='2', compare_price='', stock=NULL, yandex='0', product_id='2888'
        	SELECT count(*) as count FROM s_lang_variants WHERE lang_id='1' AND variant_id=2858 LIMIT 1
        	INSERT INTO s_lang_variants SET name='', lang_id='1', variant_id='2858'
        	SELECT count(*) as count FROM s_lang_variants WHERE lang_id='2' AND variant_id=2858 LIMIT 1
        	INSERT INTO s_lang_variants SET name='', lang_id='2', variant_id='2858'
        	SELECT count(*) as count FROM s_lang_variants WHERE lang_id='3' AND variant_id=2858 LIMIT 1
        	INSERT INTO s_lang_variants SET name='', lang_id='3', variant_id='2858'
        	SHOW TABLES LIKE '%s_languages%'
        	SELECT 
                v.id, 
                v.product_id, 
                v.price, 
                NULLIF(v.compare_price, 0) as compare_price, 
                v.sku, 
                IFNULL(v.stock, '10') as stock, 
                (v.stock IS NULL) as infinity, 
                v.attachment, 
                v.currency_id, 
                v.yandex, 
                c.rate_from, 
                c.rate_to, 
                l.name

            FROM s_options po
            LEFT JOIN s_features f ON f.id=po.feature_id
            LEFT JOIN s_lang_features l ON l.feature_id=f.id AND l.lang_id = 1
            WHERE 
                1 
                AND po.product_id in('2888') 
                 
                AND po.lang_id=1 
            ORDER BY f.position
        	SHOW TABLES LIKE '%s_languages%'
        	SELECT 
                f.id, 
                f.position, 
                f.in_filter, 
                f.yandex, 
                f.auto_name_id, 
                f.auto_value_id, 
                f.url, 
                l.name

Provided:
in the table s_products
url - is taken from the name of the product in Cyrillic immediately. The title is naming .
ID - dynamic.
The same goes and write immediately to another table

Where ID and PRODUCT_ID must be the same and a dynamic one is generated (1,2,3,4.......).
I did it on snot as I understood it, I have this code:

INSERT INTO s_products (url,name,brand_id,meta_title,meta_keywords,meta_description) 
VALUES ((SELECT MAX(product_id) FROM s_lang_products),'%Название%','0','%Название%','%Название%','%Название%')
ON DUPLICATE KEY UPDATE `url` = 
(SELECT MAX(product_id) FROM s_lang_products),
 `name` = '%Название%',
 `brand_id` = '0',
 `meta_title` = '%Название%',
 `meta_keywords` = '%Название%',
 `meta_description` = '%Название%';
 
 INSERT INTO s_lang_products (lang_id,product_id,name,meta_title,meta_keywords,meta_description) 
VALUES ('1',(SELECT MAX(product_id) FROM s_products),%Название%,'%Название%','%Название%','%Название%')
ON DUPLICATE KEY UPDATE `lang_id` = '1', 
`product_id` = (SELECT MAX(product_id) FROM s_products), 
`name` = '%Название%', 
`meta_title` = '%Название%', 
`meta_keywords` = '%Название%', 
`meta_description` = '%Название%';

2 databases
s_lang_products
s_products
Where:
(SELECT MAX(product_id) FROM s_lang_products) - creates the following numbers from the s_lang_products table
1 query, it adds everything.
But the second request does not want
How to connect it all ((brains have burst already ...
Please tell me how to make it so that when adding a product through the database, the product is added immediately to 2 tables with the same (ID and PRODUCT_ID), and when it is added The next item was checked.If there is a item, so as not to create a new one.

Answer 1

little text,
see the picture in my signature