1149. Article Views II
# 0.33
# 注意count（）中應該是文章article的id，不能是count（*）會存在重複行。
# 題目中問的也是 more than one article
select distinct viewer_id id
from views
group by viewer_id,view_date
having count(distinct article_id)>1
order by id
1158. Market Analysis I
# 自己的方法：0.8667
select user_id buyer_id, join_date, ifnull(num,0) orders_in_2019
from
users left join
(select buyer_id, count(*) num
from orders
where order_date between date('2019-01-01') and date('2019-12-31')
group by buyer_id) t
on users.user_id=t.buyer_id
1159. Market Analysis II
# 注意一個問題：兩次都是left join, 儘管第二個表的第三個表之間是 inneer join
# 0.6344
select a.user_id seller_id, case when a.favorite_brand=b.item_brand then 'yes'
else 'no' end 2nd_item_fav_brand
from
users a
left join
(select order_date, item_id, seller_id, case when @id=seller_id then @cnt:=@cnt+1
when @id:=seller_id then @cnt:=1 end cnt
from orders,(select @id:=null, @cnt:=null) t1
order by seller_id,order_date) t3
on t3.cnt=2 and a.user_id=t3.seller_id
left join items b
on t3.item_id=b.item_id
order by seller_id
1164. Product Price at a Given Date
我對這個題產生了誤解，其中一個條件是：如果該產品在2019-8-16之前沒有最新價格，默認爲10
# 常規方法 0.3402
# 通過select distinct product_id from Products 獲取完整產品表的思路真的很好
select a.product_id,ifnull(b.new_price,10) as price
from
(select distinct product_id from Products) a
left join
(select product_id, new_price from Products
where (product_id,change_date) in (select product_id,max(change_date) as change_date from Products
where change_date<='2019-08-16' group by product_id)) b
on a.product_id=b.product_id
# 方法2 0.74 思路很好
SELECT product_id, new_price AS price
FROM Products
WHERE (product_id, change_date) IN (
SELECT product_id, MAX(change_date)
FROM Products
WHERE change_date <= '2019-08-16'
GROUP BY product_id)
UNION all
SELECT DISTINCT product_id, 10 AS price
FROM Products
WHERE product_id NOT IN (SELECT product_id FROM Products WHERE change_date <= '2019-08-16')
# 方法3 0.9125 寫法確實漂亮
# SUM(CASE WHEN b.change_date IS NOT NULL THEN new_price ELSE 0 END) 很好
SELECT a.product_id, CASE WHEN MAX(b.change_date) IS NULL THEN 10
ELSE SUM(CASE WHEN b.change_date IS NOT NULL THEN new_price ELSE 0 END) END AS price
FROM Products a
LEFT JOIN
(SELECT product_id, MAX(change_date) AS change_date
FROM Products a
WHERE a.change_date <= '2019-08-16'
GROUP BY product_id) b
ON a.product_id = b.product_id AND a.change_date = b.change_date
GROUP BY a.product_id
1173. Immediate Food Delivery I
# 一道簡單題，但是自己的方法不好。0.33
select round(number1/count(delivery_id)*100,2) immediate_percentage
from delivery,
(select count(*) number1
from delivery
where order_date= customer_pref_delivery_date) t
# 注意sum函數的使用方式，和上個題有相同思路 0.87
SELECT round(SUM(CASE WHEN a.order_date = a.customer_pref_delivery_date THEN 1 ELSE 0 END)
* 100 / COUNT(*), 2) AS immediate_percentage
FROM Delivery a
# 方法3：也可以這麼寫，但是感覺不太好
SELECT ROUND(sum(order_date=customer_pref_delivery_date)/count(*)*100,2) AS immediate_percentage
FROM Delivery
1174. Immediate Food Delivery II
# 自己的方法 0.85
select round(SUM(CASE WHEN a.order_date = a.customer_pref_delivery_date THEN 1 ELSE 0 END)
* 100 / COUNT(*), 2) immediate_percentage
from delivery a
join
(select customer_id,min(order_date) order_date
from delivery
group by customer_id) b
on a.customer_id=b.customer_id and a.order_date=b.order_date
1179. 重新格式化部門表
# 方法1
SELECT DISTINCT id AS "id",
SUM(IF (month = "Jan", revenue, null)) AS "Jan_Revenue",
SUM(IF (month = "Feb", revenue, null)) AS "Feb_Revenue",
SUM(IF (month = "Mar", revenue, null)) AS "Mar_Revenue",
SUM(IF (month = "Apr", revenue, null)) AS "Apr_Revenue",
SUM(IF (month = "May", revenue, null)) AS "May_Revenue",
SUM(IF (month = "Jun", revenue, null)) AS "Jun_Revenue",
SUM(IF (month = "Jul", revenue, null)) AS "Jul_Revenue",
SUM(IF (month = "Aug", revenue, null)) AS "Aug_Revenue",
SUM(IF (month = "Sep", revenue, null)) AS "Sep_Revenue",
SUM(IF (month = "Oct", revenue, null)) AS "Oct_Revenue",
SUM(IF (month = "Nov", revenue, null)) AS "Nov_Revenue",
SUM(IF (month = "Dec", revenue, null)) AS "Dec_Revenue"
FROM Department
GROUP BY id
# 方法2
SELECT a.id,
MAX((CASE WHEN a.month = 'Jan' THEN a.revenue ELSE NULL END)) as 'Jan_Revenue',
MAX((CASE WHEN a.month = 'Feb' THEN a.revenue ELSE NULL END)) as 'Feb_Revenue',
MAX((CASE WHEN a.month = 'Mar' THEN a.revenue ELSE NULL END)) as 'Mar_Revenue',
MAX((CASE WHEN a.month = 'Apr' THEN a.revenue ELSE NULL END)) as 'Apr_Revenue',
MAX((CASE WHEN a.month = 'May' THEN a.revenue ELSE NULL END)) as 'May_Revenue',
MAX((CASE WHEN a.month = 'Jun' THEN a.revenue ELSE NULL END)) as 'Jun_Revenue',
MAX((CASE WHEN a.month = 'Jul' THEN a.revenue ELSE NULL END)) as 'Jul_Revenue',
MAX((CASE WHEN a.month = 'Aug' THEN a.revenue ELSE NULL END)) as 'Aug_Revenue',
MAX((CASE WHEN a.month = 'Sep' THEN a.revenue ELSE NULL END)) as 'Sep_Revenue',
MAX((CASE WHEN a.month = 'Oct' THEN a.revenue ELSE NULL END)) as 'Oct_Revenue',
MAX((CASE WHEN a.month = 'Nov' THEN a.revenue ELSE NULL END)) as 'Nov_Revenue',
MAX((CASE WHEN a.month = 'Dec' THEN a.revenue ELSE NULL END)) as 'Dec_Revenue'
FROM Department a GROUP BY a.id;
1193. 每月交易 I
這個題考察了date_format的使用方式
# 注意data_format函數的使用方式 0.4612
SELECT DATE_FORMAT(trans_date,'%Y-%m') AS month,
country,
Count(*) as trans_count,
Sum(IF(state='approved',1,0)) AS approved_count,
Sum(amount) AS trans_total_amount,
Sum(IF(state='approved',amount,0)) AS approved_total_amount
FROM Transactions
GROUP BY DATE_FORMAT(trans_date,'%Y-%m'), country
ORDER BY month,country;