文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

表: Visits

+---------------+---------+ | Column Name | Type | +---------------+---------+ | user_id | int | | visit_date | date | +---------------+---------+ (user_id, visit_date) 是該表的主鍵 該表的每行表示 user_id 在 visit_date 訪問了銀行

表: Transactions

+------------------+---------+ | Column Name | Type | +------------------+---------+ | user_id | int | | transaction_date | date | | amount | int | +------------------+---------+ 該表沒有主鍵，所以可能有重復行 該表的每一行表示 user_id 在 transaction_date 完成了一筆 amount 數(shù)額的交易 可以保證用戶 (user) 在 transaction_date 訪問了銀行 (也就是說 Visits 表包含 (user_id, transaction_date) 行)

銀行想要得到銀行客戶在一次訪問時的交易次數(shù)和相應的在一次訪問時該交易次數(shù)的客戶數(shù)量的圖表

寫一條 SQL 查詢多少客戶訪問了銀行但沒有進行任何交易，多少客戶訪問了銀行進行了一次交易等等

結果包含兩列：

• transactions_count： 客戶在一次訪問中的交易次數(shù)
• visits_count： 在 transactions_count 交易次數(shù)下相應的一次訪問時的客戶數(shù)量
  transactions_count 的值從 0 到所有用戶一次訪問中的 max(transactions_count)

按 transactions_count 排序

下面是查詢結果格式的例子：

Visits 表: +---------+------------+ | user_id | visit_date | +---------+------------+ | 1 | 2020-01-01 | | 2 | 2020-01-02 | | 12 | 2020-01-01 | | 19 | 2020-01-03 | | 1 | 2020-01-02 | | 2 | 2020-01-03 | | 1 | 2020-01-04 | | 7 | 2020-01-11 | | 9 | 2020-01-25 | | 8 | 2020-01-28 | +---------+------------+ Transactions 表: +---------+------------------+--------+ | user_id | transaction_date | amount | +---------+------------------+--------+ | 1 | 2020-01-02 | 120 | | 2 | 2020-01-03 | 22 | | 7 | 2020-01-11 | 232 | | 1 | 2020-01-04 | 7 | | 9 | 2020-01-25 | 33 | | 9 | 2020-01-25 | 66 | | 8 | 2020-01-28 | 1 | | 9 | 2020-01-25 | 99 | +---------+------------------+--------+ 結果表: +--------------------+--------------+ | transactions_count | visits_count | +--------------------+--------------+ | 0 | 4 | | 1 | 5 | | 2 | 0 | | 3 | 1 | +--------------------+--------------+ * 對于 transactions_count = 0, * visits 中 (1, "2020-01-01"), (2, "2020-01-02"), * (12, "2020-01-01") 和 (19, "2020-01-03") * 沒有進行交易，所以 visits_count = 4 。 * 對于 transactions_count = 1, * visits 中 (2, "2020-01-03"), (7, "2020-01-11"), * (8, "2020-01-28"), (1, "2020-01-02") * 和 (1, "2020-01-04") 進行了一次交易， * 所以 visits_count = 5 。 * 對于 transactions_count = 2, * 沒有客戶訪問銀行進行了兩次交易， * 所以 visits_count = 0 。 * 對于 transactions_count = 3, * visits 中 (9, "2020-01-25") 進行了三次交易， * 所以 visits_count = 1 。 * 對于 transactions_count >= 4, * 沒有客戶訪問銀行進行了超過3次交易， * 所以我們停止在 transactions_count = 3 。

如下是這個例子的圖表：

來源：力扣（LeetCode）
鏈接：https://leetcode-cn.com/problems/number-of-transactions-per-visit
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2. 解題

• 先生成第一列，多了沒關系，一會篩選，注意加個0

select 0 transactions_count union all select row_number() over(order by transaction_date) transactions_count from Transactions

• with as 創(chuàng)建 臨時表 t

select distinct v.user_id, visit_date, transaction_date,count(*) over(partition by user_id, visit_date, visit_date=transaction_date) times from Visits v left join Transactions tr on v.user_id=tr.user_id and v.visit_date=tr.transaction_date {"headers": ["user_id", "visit_date", "transaction_date", "times"], "values": [ [1, "2020-01-01", null, 1], [1, "2020-01-02", "2020-01-02", 1], [1, "2020-01-04", "2020-01-04", 1], [2, "2020-01-02", null, 1], [2, "2020-01-03", "2020-01-03", 1], [7, "2020-01-11", "2020-01-11", 1], [8, "2020-01-28", "2020-01-28", 1], [9, "2020-01-25", "2020-01-25", 3], [12, "2020-01-01", null, 1], [19, "2020-01-03", null, 1]]}

• 統(tǒng)計數(shù)量

select 0 transactions_count, count(*) visits_count from t where transaction_date is null union all select times transactions_count, count(*) visits_count from t where transaction_date is not null group by times {"headers": ["transactions_count", "visits_count"], "values": [[0, 4], [1, 5], [3, 1]]}

• 左連接，篩選數(shù)據(jù)

# Write your MySQL query statement below with t as (select distinct v.user_id, visit_date, transaction_date,count(*) over(partition by user_id, visit_date, visit_date=transaction_date) timesfrom Visits v left join Transactions tron v.user_id=tr.user_id and v.visit_date=tr.transaction_date )select t1.transactions_count, ifnull(t2.visits_count,0) visits_count from (select 0 transactions_countunion allselect row_number() over(order by transaction_date) transactions_countfrom Transactions ) t1 left join (select 0 transactions_count, count(*) visits_countfrom twhere transaction_date is nullunion allselect times transactions_count, count(*) visits_countfrom twhere transaction_date is not nullgroup by times ) t2 on t1.transactions_count = t2.transactions_count where t1.transactions_count <= (select max(transactions_count) from (select 0 transactions_count, count(*) visits_countfrom twhere transaction_date is nullunion allselect times transactions_count, count(*) visits_countfrom twhere transaction_date is not nullgroup by times) t3)

評論區(qū)簡潔解答

# Write your MySQL query statement below with temp1 as (select transactions_count, count(user_id) visits_countfrom (select v.user_id, count(t.user_id) transactions_countfrom Visits v left join Transactions ton v.user_id = t.user_id and visit_date = transaction_dategroup by v.user_id, v.visit_date) agroup by transactions_count )select temp2.transactions_count, ifnull(temp1.visits_count,0) visits_count from (select 0 transactions_countunionselect row_number() over (order by transaction_date) transactions_countfrom Transactions ) temp2 left join temp1 on temp2.transactions_count = temp1.transactions_count where temp2.transactions_count <= (select max(transactions_count)from temp1)

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