как сформировать запрос чтоб было равенство 2021-10-30 23:59:59 = 10.21
SELECT *
FROM `table`
JOIN ( SELECT '10.21' AS parameter) criteria
ON `date` >= CONCAT('20', SUBSTRING_INDEX(parameter, '.', -1), '-', SUBSTRING_INDEX(parameter, '.', 1), '-', '01' AS DATE)
AND `date` < CONCAT('20', SUBSTRING_INDEX(parameter, '.', -1), '-', SUBSTRING_INDEX(parameter, '.', 1), '-', '01' AS DATE) + INTERVAL 1 MONTH
WITH RECURSIVE
cte AS ( SELECT CAST(DATE_FORMAT(@range_from, '%Y-%m-01') AS DATE) month_start,
LAST_DAY(@range_from) month_end
UNION ALL
SELECT month_start + INTERVAL 1 MONTH,
LAST_DAY(month_start + INTERVAL 1 MONTH)
FROM cte
WHERE month_start < DATE_FORMAT(@range_till, '%Y-%m-01')
)
SELECT cte.month_start, COUNT(employee.id) employees_amount
FROM cte
LEFT JOIN employee ON employee.date_employment <= cte.month_end
AND ( employee.date_dismissal >= cte.month_start
OR employee.date_dismissal IS NULL )
GROUP BY 1;
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY list_name ORDER BY date DESC) rn
FROM table
)
SELECT *
FROM cte
WHERE rn = 1
SELECT t1.*
FROM table t1
NATURAL JOIN (
SELECT list_name, MAX(date) date
FROM table
GROUP BY 1
)
SELECT *
FROM table
ORDER BY user_id = @user_id DESC, id DESC LIMIT 1
SELECT *
FROM table
ORDER BY user_id IS NULL, id DESC LIMIT 1
SELECT *
FROM products
CROSS JOIN categories
LEFT JOIN products_category USING (product_id, category_id)
DELETE t1
FROM review_detail t1
JOIN review_detail t2 USING (title, nickname)
WHERE t1.detail_id < t2.detail_id;
INSERT INTO people (last_name, first_name, sex, birthday)
VALUES ('Иванов', 'Иван', 'm', TO_DATE('1999-04-04', 'YYYY-MM-DD'));
SELECT DISTINCT
client_id,
FIRST_VALUE(massage)
OVER (PARTITION BY client_id
ORDER BY massage IS NULL, dttm) massage
FROM test
WITH cte AS ( SELECT *,
LAG(`datetime`) OVER (PARTITION BY login ORDER BY `datetime`) lag_datetime,
LAG(event) OVER (PARTITION BY login ORDER BY `datetime`) lag_event
FROM history )
SELECT login, SUM(TIMESTAMPDIFF(MINUTE, lag_datetime, `datetime`)) duration
FROM cte
WHERE (event, lag_event) = (2,1)
GROUP BY login;