从数据库获取每日销售额

Question

With the help of this great community I get this solution to group the daily sales by 2 hours interval.

SELECT   SUBSTRING(DATE(o.date_purchased) + 
         INTERVAL (HOUR(o.date_purchased) -
         HOUR(o.date_purchased) MOD 2) HOUR , 12) AS time_purchased,
         ROUND(SUM(ot.value), 2) AS sales_today,
         COUNT(*) AS orders_today
FROM     orders AS o
JOIN     orders_total AS ot ON ot.orders_id = o.orders_id
WHERE    ot.class = 'ot_total'
AND      o.date_purchased >= CURRENT_DATE()
AND      o.date_purchased <  CURRENT_DATE() + INTERVAL 1 DAY
GROUP BY DATE(o.date_purchased) +
         INTERVAL (HOUR(o.date_purchased) -
         HOUR(o.date_purchased) MOD 2) HOUR

How do I change this query to get the daily sales of every day of a month. Starting automatically with the first day and ending with the last day of the actual month?

Answer 1

To get daily sales of current month you can write your query as

SELECT   DATE(o.date_purchased)  AS date_purchased,
         ROUND(SUM(ot.value), 2) AS sales_per_day,
         COUNT(*) AS orders_today
FROM     orders AS o
JOIN     orders_total AS ot ON ot.orders_id = o.orders_id
WHERE    ot.class = 'ot_total'
AND      o.date_purchased >= LAST_DAY(NOW() - INTERVAL 1 MONTH) + INTERVAL 1 DAY
AND      o.date_purchased <= LAST_DAY(NOW())
GROUP BY DATE(o.date_purchased)