优化代码SELECT SUM(IF(order_date BETWEEN '2021-11-01' AND '2021-11-15', 1, 0)) AS new_customer_count, SUM(IF(order_date BETWEEN '2021-11-01' AND '2021-11-15', amount, 0)) AS new_customer_amount, SUM(IF(order_date BETWEEN '2021-11-01' AND '2021-11-15' AND o1.customer_id IS NOT NULL, 1, 0)) AS old_customer_count, SUM(IF(order_date BETWEEN '2021-11-01' AND '2021-11-15' AND o1.customer_id IS NOT NULL, amount, 0)) AS old_customer_amount FROM orders o LEFT JOIN ( SELECT customer_id FROM orders WHERE order_date < '2020-11-01' ) o1 ON o.customer_id = o1.customer_id WHERE order_date BETWEEN '2021-11-01' AND '2021-11-15'

时间: 2024-01-24 12:18:03 浏览: 81
RAR

SQL-define-sum.rar_SUM

这段 SQL 代码看起来已经很优化了，但是如果要进一步优化，可以考虑以下几点： 1. 使用索引：为了提高查询效率，可以在 order_date 和 customer_id 字段上创建索引。 2. 简化 IF 语句：可以将 IF(order_date BETWEEN '2021-11-01' AND '2021-11-15', 1, 0) 简化为 (order_date BETWEEN '2021-11-01' AND '2021-11-15')，同样的方法也可以用于其他 IF 语句。 3. 使用子查询的优化：可以使用 EXISTS 或 IN 替换掉子查询，以提高查询效率。 优化后的 SQL 代码如下： SELECT SUM(order_date BETWEEN '2021-11-01' AND '2021-11-15') AS new_customer_count, SUM(IF(order_date BETWEEN '2021-11-01' AND '2021-11-15', amount, 0)) AS new_customer_amount, SUM(order_date BETWEEN '2021-11-01' AND '2021-11-15' AND o1.customer_id IS NOT NULL) AS old_customer_count, SUM(IF(order_date BETWEEN '2021-11-01' AND '2021-11-15' AND o1.customer_id IS NOT NULL, amount, 0)) AS old_customer_amount FROM orders o LEFT JOIN (SELECT DISTINCT customer_id FROM orders WHERE order_date < '2020-11-01') o1 ON o.customer_id = o1.customer_id WHERE order_date BETWEEN '2021-11-01' AND '2021-11-15'; 其中，使用了 DISTINCT 关键字来去重，以替换掉子查询。
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优化代码SELECT SUM(CASE WHEN order_date BETWEEN '2021-11-01' AND '2021-11-15' THEN 1 ELSE 0 END) AS new_customer_count, SUM(CASE WHEN order_date BETWEEN '2021-11-01' AND '2021-11-15' THEN amount ELSE 0 END) AS new_customer_amount, SUM(CASE WHEN order_date BETWEEN '2021-11-01' AND '2021-11-15' AND customer_id IN ( SELECT customer_id FROM orders WHERE order_date < '2020-11-01' ) THEN 1 ELSE 0 END) AS old_customer_count, SUM(CASE WHEN order_date BETWEEN '2021-11-01' AND '2021-11-15' AND customer_id IN ( SELECT customer_id FROM orders WHERE order_date < '2020-11-01' ) THEN amount ELSE 0 END) AS old_customer_amount FROM orders WHERE order_date BETWEEN '2021-11-01' AND '2021-11-15'

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优化代码SELECT SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 1, 1, 0)) AS new_customer_count, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 0, 1, 0)) AS old_customer_count, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 1, payment_amount, 0)) AS new_customer_payment_amount, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 0, payment_amount, 0)) AS old_customer_payment_amount, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND category = 'A', 1, 0)) AS category_A_customer_count, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND category = 'A', payment_amount, 0)) AS category_A_payment_amount, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND category = 'B', 1, 0)) AS category_B_customer_count, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND category = 'B', payment_amount, 0)) AS category_B_payment_amount, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND product_id = 'P1', 1, 0)) AS product_P1_customer_count, SUM(IF(order_date BETWEEN '2022-10-31' AND '2022-11-11' AND product_id = 'P1', payment_amount, 0)) AS product_P1_payment_amount FROM orders o LEFT JOIN (SELECT DISTINCT order_date FROM orders WHERE order_date BETWEEN '2022-10-31' AND '2022-11-11') d1 ON o.order_date = d1.order_date LEFT JOIN (SELECT DISTINCT order_date FROM orders WHERE order_date BETWEEN '2021-10-31' AND '2022-10-30' AND order_date NOT IN (SELECT order_date FROM orders WHERE order_date BETWEEN '2022-10-31' AND '2022-11-11')) d2 ON o.order_date = d2.order_date LEFT JOIN (SELECT DISTINCT order_date FROM orders WHERE order_date BETWEEN '2021-11-12' AND '2022-10-30' AND order_date IN (SELECT order_date FROM orders WHERE order_date BETWEEN '2022-10-31' AND '2022-11-11')) d3 ON o.order_date = d3.order_date WHERE d1.order_date IS NOT NULL OR d2.order_date IS NOT NULL OR d3.order_date IS NOT NULL;

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优化代码SELECT SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 1 THEN 1 ELSE 0 END) AS new_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 0 THEN 1 ELSE 0 END) AS old_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 1 THEN payment_amount ELSE 0 END) AS new_customer_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 0 THEN payment_amount ELSE 0 END) AS old_customer_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'A' THEN 1 ELSE 0 END ELSE 0 END) AS category_A_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'A' THEN payment_amount ELSE 0 END ELSE 0 END) AS category_A_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'B' THEN 1 ELSE 0 END ELSE 0 END) AS category_B_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'B' THEN payment_amount ELSE 0 END ELSE 0 END) AS category_B_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN product_id = 'P1' THEN 1 ELSE 0 END ELSE 0 END) AS product_P1_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN product_id = 'P1' THEN payment_amount ELSE 0 END ELSE 0 END) AS product_P1_payment_amount FROM orders WHERE (order_date BETWEEN '2022-10-31' AND '2022-11-11') OR (order_date BETWEEN '2021-10-31' AND '2022-10-30' AND order_date NOT IN (SELECT order_date FROM orders WHERE order_date BETWEEN '2022-10-31' AND '2022-11-11')) OR (order_date BETWEEN '2021-11-12' AND '2022-10-30' AND order_date IN (SELECT order_date FROM orders WHERE order_date BETWEEN '2022-10-31' AND '2022-11-11'))

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优化代码，并提供新脚本SELECT SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 1 THEN 1 ELSE 0 END) AS new_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 0 THEN 1 ELSE 0 END) AS old_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 1 THEN payment_amount ELSE 0 END) AS new_customer_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' AND is_new_customer = 0 THEN payment_amount ELSE 0 END) AS old_customer_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'A' THEN 1 ELSE 0 END ELSE 0 END) AS category_A_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'A' THEN payment_amount ELSE 0 END ELSE 0 END) AS category_A_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'B' THEN 1 ELSE 0 END ELSE 0 END) AS category_B_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN category = 'B' THEN payment_amount ELSE 0 END ELSE 0 END) AS category_B_payment_amount, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN product_id = 'P1' THEN 1 ELSE 0 END ELSE 0 END) AS product_P1_customer_count, SUM(CASE WHEN order_date BETWEEN '2022-10-31' AND '2022-11-11' THEN CASE WHEN product_id = 'P1' THEN payment_amount ELSE 0 END ELSE 0 END) AS product_P1_payment_amount FROM orders WHERE (order_date BETWEEN '2022-10-31' AND '2022-11-11') OR (order_date BETWEEN '2021-10-31' AND '2022-10-30' AND order_date NOT IN (SELECT order_date FROM orders WHERE order_date BETWEEN '2022-10-31' AND '2022-11-11')) OR (order_date BETWEEN '2021-11-12' AND '2022-10-30' AND order_date IN (SELECT order_date FROM orders WHERE order_date BETWEEN '2022-10-31' AND '2022-11-11'))

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这段代码是一个SQL查询语句，用于计算用户在活动期间内的签到金币数。代码中使用了两个临时表t1和t2。 t1表根据条件筛选出活动期间内的数据，并使用DISTINCT关键字去重，得到每位用户在每天签到的第几次编号。这个...

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如果编号是0，则表示该用户当天是连续签到的第七天，应获得7个金币；其他情况下，用户当天应获得1个金币。 3. 最后，使用主查询从t2中获取每个用户每个月的总金币数，并按照日期和用户ID进行排序。 希望以上解释对...

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with t as( select t2.tag,CAST(t1.start_time as date) fdate,COUNT(IIF(t1.if_like=1,1,null)) flike, COUNT(IIF(t1.if_retweet=1,1,null)) fzf from tb_user_video_log t1 join tb_video_info t2 on t1.video_id=t2.video_id where DATEDIFF(DAY,start_time,'2021-10-03')<9 group by t2.tag,CAST(t1.start_time as date) ) select t.*, sum(t.flike) over(partition by tag order by fdate desc rows between current row and 6 preceding) from t “BETWEEN CURRENT ROW AND ... PRECEDING”不是有效的窗口框架，不能与 OVER 子句一起使用。

SUM(t.flike) OVER (PARTITION BY tag ORDER BY fdate DESC ROWS BETWEEN 6 PRECEDING AND CURRENT ROW) FROM t; 这将计算每个标签 (tag) 在按照日期 (fdate) 降序排列的窗口中，包括当前行以及前 6 行的 ...

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2. 使用 BETWEEN 代替两个 DATE_FORMAT：可以将 DATE_FORMAT(checkin_time, '%Y-%m-%d') >= DATE_FORMAT('2023-04-01', '%Y-%m-%d') 和 DATE_FORMAT(checkin_time, '%Y-%m-%d') <= DATE_FORMAT('2023-05-22', '%Y...

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SELECT SUM(MAX(TRUNCATE(CAST(t.disk_space_utilization AS DECIMAL(30,8)), 0))) OVER() * 1.0/100 AS calc FROM ( SELECT a.sync_time, a.equipment_monitor_id, a.equipment_id, a.cpu_load, a....

优化以下SQL语句使该语句可以在MySQL5.0中可以运行：select sum(max(TRUNCATE(cast(t0.disk_space_utilization as decimal(30,8)), 0))) over ( order by max(TRUNCATE(cast(t0.disk_space_utilization as decimal(30,8)), 0)))*1.0/100 as calc from (SELECT a.sync_time, a.equipment_monitor_id, a.equipment_id, a.cpu_load, a.memory_load, a.disk_space, a.disk_space_utilization, a.network_card_throughput, a.control_unit_temperature FROM nec_equipment_monitor_statistics a order by a.sync_time) t0 where (t0.equipment_id='IT-A01-32' AND str_to_date(t0.sync_time,'%Y-%m-%d %H:%i:%s')>=str_to_date('20230519','%Y%m%d') and str_to_date(t0.sync_time,'%Y-%m-%d %H:%i:%s')<str_to_date('20230520','%Y%m%d'))

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