毛网商城：牛客网SQL刷题四-电商场景（某东商城） 2024-04-16 07:13:16 0 0 SQL13 计算商城中2021年每月的GMV数据DROP TABLE IF EXISTS tb_order_overall;CREATE TABLE tb_order_overall ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', uid INT NOT NULL COMMENT '用户ID', event_time datetime COMMENT '下单时间', total_amount DECIMAL NOT NULL COMMENT '订单总金额', total_cnt INT NOT NULL COMMENT '订单商品总件数', `status` TINYINT NOT NULL COMMENT '订单状态') CHARACTER SET utf8 COLLATE utf8_bin;INSERT INTO tb_order_overall(order_id, uid, event_time, total_amount, total_cnt, `status`) VALUES (301001, 101, '2021-10-01 10:00:00', 15900, 2, 1), (301002, 101, '2021-10-01 11:00:00', 15900, 2, 1), (301003, 102, '2021-10-02 10:00:00', 34500, 8, 0), (301004, 103, '2021-10-12 10:00:00', 43500, 9, 1), (301005, 105, '2021-11-01 10:00:00', 31900, 7, 1), (301006, 102, '2021-11-02 10:00:00', 24500, 6, 1), (391007, 102, '2021-11-03 10:00:00', -24500, 6, 2), (301008, 104, '2021-11-04 10:00:00', 55500, 12, 0);题目场景逻辑说明： 用户将购物车中多件商品一起下单时，订单总表会生成一个订单（但此时未付款，status-订单状态为0，表示待付款）； 当用户支付完成时，在订单总表修改对应订单记录的status-订单状态为1，表示已付款； 若用户退货退款，在订单总表生成一条交易总金额为负值的记录（表示退款金额，订单号为退款单号，status-订单状态为2表示已退款）。 问题：请计算商城中2021年每月的GMV，输出GMV大于10w的每月GMV，值保留到整数。注：GMV为已付款订单和未付款订单两者之和。结果按GMV升序排序。输出示例：示例数据输出如下：month GMV2021-10 1098002021-11 111900解释：2021年10月有3笔已付款的订单，1笔未付款订单，总交易金额为109800；2021年11月有2笔已付款订单，1笔未付款订单，总交易金额为111900（还有1笔退款订单由于已计算了付款的订单金额，无需计算在GMV中）。SQLselect date_format(event_time,'%Y-%m') month,round(sum(total_amount)) gmv from tb_order_overall where year(event_time)='2021' and status in(0,1) group by date_format(event_time,'%Y-%m') having GMV>100000 order by gmv asc; SQL14 统计2021年10月每个退货率不大于0.5的商品各项指标数据DROP TABLE IF EXISTS tb_user_event;CREATE TABLE tb_user_event ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', uid INT NOT NULL COMMENT '用户ID', product_id INT NOT NULL COMMENT '商品ID', event_time datetime COMMENT '行为时间', if_click TINYINT COMMENT '是否点击', if_cart TINYINT COMMENT '是否加购物车', if_payment TINYINT COMMENT '是否付款', if_refund TINYINT COMMENT '是否退货退款') CHARACTER SET utf8 COLLATE utf8_bin;INSERT INTO tb_user_event(uid, product_id, event_time, if_click, if_cart, if_payment, if_refund) VALUES (101, 8001, '2021-10-01 10:00:00', 0, 0, 0, 0), (102, 8001, '2021-10-01 10:00:00', 1, 0, 0, 0), (103, 8001, '2021-10-01 10:00:00', 1, 1, 0, 0), (104, 8001, '2021-10-02 10:00:00', 1, 1, 1, 0), (105, 8001, '2021-10-02 10:00:00', 1, 1, 1, 0), (101, 8002, '2021-10-03 10:00:00', 1, 1, 1, 0), (109, 8001, '2021-10-04 10:00:00', 1, 1, 1, 1);题目问题：请统计2021年10月每个有展示记录的退货率不大于0.5的商品各项指标，注： 商品点展比=点击数÷展示数； 加购率=加购数÷点击数； 成单率=付款数÷加购数；退货率=退款数÷付款数， 当分母为0时整体结果记为0，结果中各项指标保留3位小数，并按商品ID升序排序。 输出示例：示例数据的输出结果如下product_id ctr cart_rate payment_rate refund_rate8001 0.833 0.800 0.750 0.3338002 1.000 1.000 1.000 0.000解释：在2021年10月商品8001被展示了6次，点击了5次，加购了4次，付款了3次，退款了1次，因此点击率为5/6=0.833，加购率为4/5=0.800，成单率为3/4=0.750，退货率为1/3=0.333（保留3位小数）；SQLselect product_id, round(sum(if_click)/count(product_id),3) ctr, if(sum(if_click)=0,0,round(sum(if_cart)/sum(if_click),3)) cart_rate, if(sum(if_cart)=0,0,round(sum(if_payment)/sum(if_cart),3)) payment_rate, if(sum(if_payment)=0,0,round(sum(if_refund)/sum(if_payment),3)) refund_rate from tb_user_event where date_format(event_time,'%Y-%m')='2021-10' group by product_id having refund_rate<=0.5 order by product_id asc SQL15 某店铺的各商品毛利率及店铺整体毛利率数据DROP TABLE IF EXISTS tb_order_overall;CREATE TABLE tb_order_overall ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', uid INT NOT NULL COMMENT '用户ID', event_time datetime COMMENT '下单时间', total_amount DECIMAL NOT NULL COMMENT '订单总金额', total_cnt INT NOT NULL COMMENT '订单商品总件数', `status` TINYINT NOT NULL COMMENT '订单状态') CHARACTER SET utf8 COLLATE utf8_bin;INSERT INTO tb_order_overall(order_id, uid, event_time, total_amount, total_cnt, `status`) VALUES (301001, 101, '2021-10-01 10:00:00', 30000, 3, 1), (301002, 102, '2021-10-01 11:00:00', 23900, 2, 1), (301003, 103, '2021-10-02 10:00:00', 31000, 2, 1);DROP TABLE IF EXISTS tb_product_info;CREATE TABLE tb_product_info ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', product_id INT NOT NULL COMMENT '商品ID', shop_id INT NOT NULL COMMENT '店铺ID', tag VARCHAR(12) COMMENT '商品类别标签', in_price DECIMAL NOT NULL COMMENT '进货价格', quantity INT NOT NULL COMMENT '进货数量', release_time datetime COMMENT '上架时间') CHARACTER SET utf8 COLLATE utf8_bin;DROP TABLE IF EXISTS tb_order_detail;CREATE TABLE tb_order_detail ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', product_id INT NOT NULL COMMENT '商品ID', price DECIMAL NOT NULL COMMENT '商品单价', cnt INT NOT NULL COMMENT '下单数量') CHARACTER SET utf8 COLLATE utf8_bin;INSERT INTO tb_product_info(product_id, shop_id, tag, in_price, quantity, release_time) VALUES (8001, 901, '家电', 6000, 100, '2020-01-01 10:00:00'), (8002, 902, '家电', 12000, 50, '2020-01-01 10:00:00'), (8003, 901, '3C数码', 12000, 50, '2020-01-01 10:00:00');INSERT INTO tb_order_detail(order_id, product_id, price, cnt) VALUES (301001, 8001, 8500, 2), (301001, 8002, 15000, 1), (301002, 8001, 8500, 1), (301002, 8002, 16000, 1), (301003, 8002, 14000, 1), (301003, 8003, 18000, 1);题目场景逻辑说明： 用户将购物车中多件商品一起下单时，订单总表会生成一个订单（但此时未付款，status-订单状态为0表示待付款），在订单明细表生成该订单中每个商品的信息； 当用户支付完成时，在订单总表修改对应订单记录的status-订单状态为1表示已付款； 若用户退货退款，在订单总表生成一条交易总金额为负值的记录（表示退款金额，订单号为退款单号，status-订单状态为2表示已退款）。问题：请计算2021年10月以来店铺901中商品毛利率大于24.9%的商品信息及店铺整体毛利率。注：商品毛利率=(1-进价/平均单件售价)*100%；店铺毛利率=(1-总进价成本/总销售收入)*100%。结果先输出店铺毛利率，再按商品ID升序输出各商品毛利率，均保留1位小数。输出示例：示例数据的输出结果如下：product_id profit_rate店铺汇总 31.0%8001 29.4%8003 33.3%解释：店铺901有两件商品8001和8003；8001售出了3件，销售总额为25500，进价总额为18000，毛利率为1-18000/25500=29.4%，8003售出了1件，售价为18000，进价为12000，毛利率为33.3%；店铺卖出的这4件商品总销售额为43500，总进价为30000，毛利率为1-30000/43500=31.0%SQLselect if(product_id is null ,'店铺汇总',product_id),concat(round(profit_rate*100,1),'%') from ( select product_id,(1-sum(total_in_price)/sum(total_sales_ripec)) profit_rate from ( select dtl.product_id,info.in_price*cnt total_in_price,price*cnt total_sales_ripec from tb_order_detail dtl join tb_product_info info on dtl.product_id=info.product_id join tb_order_overall oa on oa.order_id=dtl.order_id where 1=1 and oa.status=1 and info.shop_id='901' and event_time >= '2021-10-01 00:00:00')t group by product_id with rollup having profit_rate > 0.249 OR product_id IS NULL order by product_id asc) t;-- coalesce(product_id,'店铺汇总') SQL16 零食类商品中复购率top3高的商品数据DROP TABLE IF EXISTS tb_order_overall;CREATE TABLE tb_order_overall ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', uid INT NOT NULL COMMENT '用户ID', event_time datetime COMMENT '下单时间', total_amount DECIMAL NOT NULL COMMENT '订单总金额', total_cnt INT NOT NULL COMMENT '订单商品总件数', `status` TINYINT NOT NULL COMMENT '订单状态') CHARACTER SET utf8 COLLATE utf8_bin;DROP TABLE IF EXISTS tb_product_info;CREATE TABLE tb_product_info ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', product_id INT NOT NULL COMMENT '商品ID', shop_id INT NOT NULL COMMENT '店铺ID', tag VARCHAR(12) COMMENT '商品类别标签', in_price DECIMAL NOT NULL COMMENT '进货价格', quantity INT NOT NULL COMMENT '进货数量', release_time datetime COMMENT '上架时间') CHARACTER SET utf8 COLLATE utf8_bin;DROP TABLE IF EXISTS tb_order_detail;CREATE TABLE tb_order_detail ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', product_id INT NOT NULL COMMENT '商品ID', price DECIMAL NOT NULL COMMENT '商品单价', cnt INT NOT NULL COMMENT '下单数量') CHARACTER SET utf8 COLLATE utf8_bin;INSERT INTO tb_product_info(product_id, shop_id, tag, in_price, quantity, release_time) VALUES (8001, 901, '零食', 60, 1000, '2020-01-01 10:00:00'), (8002, 901, '零食', 140, 500, '2020-01-01 10:00:00'), (8003, 901, '零食', 160, 500, '2020-01-01 10:00:00');INSERT INTO tb_order_overall(order_id, uid, event_time, total_amount, total_cnt, `status`) VALUES (301001, 101, '2021-09-30 10:00:00', 140, 1, 1), (301002, 102, '2021-10-01 11:00:00', 235, 2, 1), (301011, 102, '2021-10-31 11:00:00', 250, 2, 1), (301003, 101, '2021-11-02 10:00:00', 300, 2, 1), (301013, 105, '2021-11-02 10:00:00', 300, 2, 1), (301005, 104, '2021-11-03 10:00:00', 170, 1, 1);INSERT INTO tb_order_detail(order_id, product_id, price, cnt) VALUES (301001, 8002, 150, 1), (301011, 8003, 200, 1), (301011, 8001, 80, 1), (301002, 8001, 85, 1), (301002, 8003, 180, 1), (301003, 8002, 140, 1), (301003, 8003, 180, 1), (301013, 8002, 140, 2), (301005, 8003, 180, 1);题目场景逻辑说明： 用户将购物车中多件商品一起下单时，订单总表会生成一个订单（但此时未付款， status-订单状态-订单状态为0表示待付款），在订单明细表生成该订单中每个商品的信息； 当用户支付完成时，在订单总表修改对应订单记录的status-订单状态-订单状态为1表示已付款； 若用户退货退款，在订单总表生成一条交易总金额为负值的记录（表示退款金额，订单号为退款单号，订单状态为2表示已退款）。问题：请统计零食类商品中复购率top3高的商品。注：复购率指用户在一段时间内对某商品的重复购买比例，复购率越大，则反映出消费者对品牌的忠诚度就越高，也叫回头率此处我们定义：某商品复购率 = 近90天内购买它至少两次的人数 ÷ 购买它的总人数近90天指包含最大日期（记为当天）在内的近90天。结果中复购率保留3位小数，并按复购率倒序、商品ID升序排序输出示例：示例数据的输出结果如下：product_id repurchase_rate8001 1.0008002 0.5008003 0.333解释：商品8001、8002、8003都是零食类商品，8001只被用户102购买了两次，复购率1.000；商品8002被101购买了两次，被105购买了1次，复购率0.500；商品8003被102购买两次，被101和105各购买1次，复购率为0.333。SQLselect product_id,round(avg(cnt_uid),3) ratio from ( select uid,product_id, if(count(distinct order_id)>=2,1,0) cnt_uid from ( select dtl.order_id,oa.uid,dtl.product_id from tb_order_detail dtl join tb_product_info info on dtl.product_id=info.product_id join tb_order_overall oa on oa.order_id=dtl.order_id where 1=1 and oa.status in(0,1) and info.tag='零食' and event_time >= date_sub((select max(event_time) from tb_order_overall),interval 89 day) ) t group by product_id,uid ) t group by product_id order by ratio desc,product_id limit 3; SQL17 10月的新户客单价和获客成本数据DROP TABLE IF EXISTS tb_order_overall;CREATE TABLE tb_order_overall ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', uid INT NOT NULL COMMENT '用户ID', event_time datetime COMMENT '下单时间', total_amount DECIMAL NOT NULL COMMENT '订单总金额', total_cnt INT NOT NULL COMMENT '订单商品总件数', `status` TINYINT NOT NULL COMMENT '订单状态') CHARACTER SET utf8 COLLATE utf8_bin;DROP TABLE IF EXISTS tb_product_info;CREATE TABLE tb_product_info ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', product_id INT NOT NULL COMMENT '商品ID', shop_id INT NOT NULL COMMENT '店铺ID', tag VARCHAR(12) COMMENT '商品类别标签', in_price DECIMAL NOT NULL COMMENT '进货价格', quantity INT NOT NULL COMMENT '进货数量', release_time datetime COMMENT '上架时间') CHARACTER SET utf8 COLLATE utf8_bin;DROP TABLE IF EXISTS tb_order_detail;CREATE TABLE tb_order_detail ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', product_id INT NOT NULL COMMENT '商品ID', price DECIMAL NOT NULL COMMENT '商品单价', cnt INT NOT NULL COMMENT '下单数量') CHARACTER SET utf8 COLLATE utf8_bin;INSERT INTO tb_product_info(product_id, shop_id, tag, in_price, quantity, release_time) VALUES (8001, 901, '日用', 60, 1000, '2020-01-01 10:00:00'), (8002, 901, '零食', 140, 500, '2020-01-01 10:00:00'), (8003, 901, '零食', 160, 500, '2020-01-01 10:00:00'), (8004, 902, '零食', 130, 500, '2020-01-01 10:00:00');INSERT INTO tb_order_overall(order_id, uid, event_time, total_amount, total_cnt, `status`) VALUES (301002, 102, '2021-10-01 11:00:00', 235, 2, 1), (301003, 101, '2021-10-02 10:00:00', 300, 2, 1), (301005, 104, '2021-10-03 10:00:00', 160, 1, 1);INSERT INTO tb_order_detail(order_id, product_id, price, cnt) VALUES (301002, 8001, 85, 1), (301002, 8003, 180, 1), (301003, 8004, 140, 1), (301003, 8003, 180, 1), (301005, 8003, 180, 1);题目问题：请计算2021年10月商城里所有新用户的首单平均交易金额（客单价）和平均获客成本（保留一位小数）。注：订单的优惠金额 = 订单明细里的{该订单各商品单价×数量之和} - 订单总表里的{订单总金额} 。输出示例：示例数据的输出结果如下avg_amount avg_cost231.7 23.3解释：2021年10月有3个新用户，102的首单为301002，订单金额为235，商品总金额为85+180=265，优惠金额为30；101的首单为301003，订单金额为300，商品总金额为140+180=320，优惠金额为20；104的首单为301005，订单金额为160，商品总金额为180，优惠金额为20；平均首单客单价为(235+300+160)/3=231.7，平均获客成本为(30+20+20)/3=23.3SQL-- 所有新用户及首单 select uid,order_id,total_amount from tb_order_overall oa1 where event_time =(select min(event_time) from tb_order_overall oa2 where oa1.uid=oa2.uid)and date_format(event_time,'%Y-%m')='2021-10'; -- 订单总金额select order_id,sum(price)from tb_order_detailgroup by order_id;select round(sum(total_amount_coupon)/count(*),1) avg_amount,round(sum(total_amount-total_amount_coupon)/count(*),1) avg_cost from ( select uid,order_id,total_amount total_amount_coupon from tb_order_overall oa1 where event_time =(select min(event_time) from tb_order_overall oa2 where oa1.uid=oa2.uid) and date_format(event_time,'%Y-%m')='2021-10' )t1join ( select order_id,sum(price) total_amount from tb_order_detail group by order_id) t2 on t1.order_id=t2.order_id; SQL18 店铺901国庆期间的7日动销率和滞销率 完全不会，看了题解要用不等式连接数据DROP TABLE IF EXISTS tb_order_overall;CREATE TABLE tb_order_overall ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', uid INT NOT NULL COMMENT '用户ID', event_time datetime COMMENT '下单时间', total_amount DECIMAL NOT NULL COMMENT '订单总金额', total_cnt INT NOT NULL COMMENT '订单商品总件数', `status` TINYINT NOT NULL COMMENT '订单状态') CHARACTER SET utf8 COLLATE utf8_bin;DROP TABLE IF EXISTS tb_product_info;CREATE TABLE tb_product_info ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', product_id INT NOT NULL COMMENT '商品ID', shop_id INT NOT NULL COMMENT '店铺ID', tag VARCHAR(12) COMMENT '商品类别标签', in_price DECIMAL NOT NULL COMMENT '进货价格', quantity INT NOT NULL COMMENT '进货数量', release_time datetime COMMENT '上架时间') CHARACTER SET utf8 COLLATE utf8_bin;DROP TABLE IF EXISTS tb_order_detail;CREATE TABLE tb_order_detail ( id INT PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID', order_id INT NOT NULL COMMENT '订单号', product_id INT NOT NULL COMMENT '商品ID', price DECIMAL NOT NULL COMMENT '商品单价', cnt INT NOT NULL COMMENT '下单数量') CHARACTER SET utf8 COLLATE utf8_bin;INSERT INTO tb_product_info(product_id, shop_id, tag, in_price, quantity, release_time) VALUES (8001, 901, '日用', 60, 1000, '2020-01-01 10:00:00'), (8002, 901, '零食', 140, 500, '2020-01-01 10:00:00'), (8003, 901, '零食', 160, 500, '2020-01-01 10:00:00');INSERT INTO tb_order_overall(order_id, uid, event_time, total_amount, total_cnt, `status`) VALUES (301004, 102, '2021-09-30 10:00:00', 170, 1, 1), (301005, 104, '2021-10-01 10:00:00', 160, 1, 1), (301003, 101, '2021-10-02 10:00:00', 300, 2, 1), (301002, 102, '2021-10-03 11:00:00', 235, 2, 1);INSERT INTO tb_order_detail(order_id, product_id, price, cnt) VALUES (301004, 8002, 180, 1), (301005, 8002, 170, 1), (301002, 8001, 85, 1), (301002, 8003, 180, 1), (301003, 8002, 150, 1), (301003, 8003, 180, 1);题目问题：请计算店铺901在2021年国庆头3天的7日动销率和滞销率，结果保留3位小数，按日期升序排序。注： 动销率定义为店铺中一段时间内有销量的商品占当前已上架总商品数的比例（有销量的商品/已上架总商品数)。 滞销率定义为店铺中一段时间内没有销量的商品占当前已上架总商品数的比例。（没有销量的商品/已上架总商品数)。 只要当天任一店铺有任何商品的销量就输出该天的结果，即使店铺901当天的动销率为0。 输出示例：示例数据的输出结果如下：dt sale_rate unsale_rate2021-10-01 0.333 0.6672021-10-02 0.667 0.3332021-10-03 1.000 0.000解释：10月1日的近7日（9月25日---10月1日）店铺901有销量的商品有8002，截止当天在售商品数为3，动销率为0.333，滞销率为0.667；10月2日的近7日（9月26日---10月2日）店铺901有销量的商品有8002、8003，截止当天在售商品数为3，动销率为0.667，滞销率为0.333；10月3日的近7日（9月27日---10月3日）店铺901有销量的商品有8002、8003、8001，截止当天店铺901在售商品数为3，动销率为1.000，滞销率为0.000；SQL1SELECT dt, ROUND(cnt / total_cnt, 3) AS sale_rate, ROUND(1 - cnt / total_cnt, 3) AS unsale_rateFROM( SELECT DISTINCT DATE(event_time) AS dt, ( SELECT COUNT(DISTINCT (IF(shop_id != 901, null, product_id))) FROM tb_order_overall JOIN tb_order_detail USING (order_id) JOIN tb_product_info USING (product_id) WHERE TIMESTAMPDIFF(DAY, event_time, to1.event_time) BETWEEN 0 AND 6 -- 当是10.1号时，统计的是订单表1号及之前已售出的产品id -- 当是10.3号时，统计的是订单表3号及之前已售出的产品id ) AS cnt, ( SELECT COUNT(DISTINCT product_id) FROM tb_product_info WHERE shop_id = 901 -- 数据刚好都是3，所以结果正确 ) AS total_cnt FROM tb_order_overall to1 WHERE DATE(event_time) BETWEEN '2021-10-01' AND '2021-10-03') AS t0ORDER BY dt;SQL2/*问题：请计算店铺901 WHERE shop_id=901在2021年国庆头3天的7日动销率和滞销率，结果保留3位小数，按日期升序排序。 ORDER BY DATE(event_time) ASC注：动销率定义为店铺中一段时间内有销量的商品占当前已上架总商品数的比例（有销量的商品/已上架总商品数)。 COUNT(DISTINCT product_id)/BIT_COUNT(DISTINCT produc_id)滞销率定义为店铺中一段时间内没有销量的商品占当前已上架总商品数的比例。（没有销量的商品/已上架总商品数)。只要当天任一店铺有任何商品的销量就输出该天的结果，即使店铺901当天的动销率为0。需要获取每天出售了哪些商品，继而退出前七天的商品种类数 可以用where exists 来表示。*/WITH tmp AS (SELECT DATE(too1.event_time) AS sale_day, tod1.product_id ,tpi1.shop_id FROM tb_order_detail AS tod1 LEFT JOIN tb_order_overall AS too1 ON tod1.order_id = too1.order_id LEFT JOIN tb_product_info AS tpi1 ON tod1.product_id = tpi1.product_id) -- 这就是一个简单的链接查询，把三个表链接起来了SELECT t1.sale_day , ROUND( COUNT(DISTINCT CASE WHEN t2.shop_id=901 THEN t2.product_id ELSE NULL END) -- 为了实现“只要当天任一店铺有任何商品的销量就输出该天的结果”这一条件就不能在where中将其他店铺的记录删去，可以放在CASE 表达式中，不是901则不计。 /(SELECT COUNT(DISTINCT product_id) FROM tb_product_info WHERE shop_id=901 AND DATE(release_time)<=t1.sale_day),3) AS sale_rate -- 一个子查询查询901店铺当天一共上架了多少商品 , ROUND(1-COUNT(DISTINCT CASE WHEN t2.shop_id=901 THEN t2.product_id ELSE NULL END) /(SELECT COUNT(DISTINCT product_id) FROM tb_product_info WHERE shop_id=901 AND DATE(release_time)<=t1.sale_day),3) AS unsale_rate -- 动销率和滞消率相加正好等于1 FROM tmp AS t1 LEFT JOIN tmp AS t2 ON t2.sale_day BETWEEN DATE_SUB(t1.sale_day, INTERVAL 6 DAY) AND t1.sale_day -- 用自连接的方法获得前7天的产品销售情况。 WHERE t1.sale_day BETWEEN '2021-10-01' AND '2021-10-03' GROUP BY t1.sale_day ORDER BY t1.sale_day ASCSQL3-- 国庆前3天with t1 as(select date(event_time) dtfrom tb_order_overall where date(event_time) between '2021-10-01' and '2021-10-03'group by dt),-- 901 25 26 27 28 29 30 1 2 3 这几天的有效订单t2 as(select date(oa.event_time) dt,info.shop_id, dtl.product_id from tb_order_detail dtl join tb_product_info info on dtl.product_id=info.product_id join tb_order_overall oa on oa.order_id=dtl.order_id where info.shop_id='901' and oa.status='1' and date(oa.event_time) between '2021-09-25' and '2021-10-03' order by dt asc)select dt, round(cnt_sale/cnt_all,3) sale_rate,1-round(cnt_sale/cnt_all,3) unsale_rate from( select t1.dt, count(distinct t2.product_id) cnt_sale,-- count(distinct -- case -- when t2.shop_id='901' then t2.product_id else null -- end )cnt_sale, (select count(info.product_id) from tb_product_info info where info.shop_id='901' and date(info.release_time) <= t1.dt) cnt_all from t1 left join t2 on datediff(t1.dt,t2.dt) between 0 and 6 group by t1.dt) t order by dt asc| 2021-10-01 || 2021-10-02 || 2021-10-03 | 2021-09-30 | 901 | 8002 || 2021-10-01 | 901 | 8002 || 2021-10-02 | 901 | 8002 || 2021-10-02 | 901 | 8003 || 2021-10-03 | 901 | 8001 || 2021-10-03 | 901 | 8003 || 2021-10-01 | 2021-09-30 | 901 | 8002 || 2021-10-01 | 2021-10-01 | 901 | 8002 |-- 10.1及之前| 2021-10-02 | 2021-09-30 | 901 | 8002 || 2021-10-02 | 2021-10-01 | 901 | 8002 || 2021-10-02 | 2021-10-02 | 901 | 8003 || 2021-10-02 | 2021-10-02 | 901 | 8002 |-- 10.2及之前| 2021-10-03 | 2021-09-30 | 901 | 8002 || 2021-10-03 | 2021-10-01 | 901 | 8002 || 2021-10-03 | 2021-10-02 | 901 | 8003 || 2021-10-03 | 2021-10-02 | 901 | 8002 || 2021-10-03 | 2021-10-03 | 901 | 8003 || 2021-10-03 | 2021-10-03 | 901 | 8001 |-- 10.3及之前 收藏（0）