MySQL打卡练习-5

文章目录

MySQL打卡练习-5

练习一：行转列

问题描述

假设有如下比赛结果

+--------------+-----------+
|    cdate     |   result  |
+--------------+-----------+
|  2021-01-01  |     胜    |
|  2021-01-01  |     负    |
|  2021-01-03  |     胜    |
|  2021-01-03  |     负    |
|  2021-01-01  |     胜    |
|  2021-01-03  |     负    |
+------------+-----------+

请使用 SQL 将比赛结果转换为如下形式：

+--------------+-----+-----|
|  比赛日期     | 胜  | 负  |
+--------------+-----------+
|  2021-01-01  |  2  |  1  |
|  2021-01-03  |  1  |  2  |
+------------+-----------+

解题答案

创建数据

CREATE TABLE compete
(
    c_date DATE,
    result CHAR
);

INSERT INTO compete
VALUES (20210101, "胜"),
       (20210101, "负"),
       (20210103, "胜"),
       (20210103, "负"),
       (20210101, "胜"),
       (20210103, "负");

解题答案

SELECT c_date                                as "比赛日期",
       COUNT(IF(result = "胜", result, NULL)) as "胜",
       COUNT(IF(result = "负", result, NULL)) as "负"
FROM compete
GROUP BY c_date;

练习二：列转行

问题描述

假设有如下比赛结果

+--------------+-----+-----|
|  比赛日期     | 胜  | 负  |
+--------------+-----------+
|  2021-01-01  |  2  |  1  |
|  2021-01-03  |  1  |  2  |
+------------+-----------+

请使用 SQL 将比赛结果转换为如下形式：

+--------------+-----------+
|    cdate     |   result  |
+--------------+-----------+
|  2021-01-01  |     胜    |
|  2021-01-01  |     负    |
|  2021-01-03  |     胜    |
|  2021-01-03  |     负    |
|  2021-01-01  |     胜    |
|  2021-01-03  |     负    |
+------------+-----------+

解题答案

创建数据

CREATE TABLE compete_score
(
    比赛日期 DATE,
    胜    INT,
    负    INT
);
INSERT INTO compete_score
VALUES ("20210101", 2, 1),
       ("20210103", 1, 2);

解题答案

暂无

练习三：连续登录

问题描述

有用户表行为记录表t_act_records表，包含两个字段：uid（用户ID），imp_date（日期）

计算2021年每个月，每个用户连续登录的最多天数
计算2021年每个月，连续2天都有登录的用户名单
计算2021年每个月，连续5天都有登录的用户数

构造表mysql如下：

DROP TABLE if EXISTS t_act_records;
CREATE TABLE t_act_records
(uid  VARCHAR(20),
imp_date DATE);

INSERT INTO t_act_records VALUES('u1001', 20210101);
INSERT INTO t_act_records VALUES('u1002', 20210101);
INSERT INTO t_act_records VALUES('u1003', 20210101);
INSERT INTO t_act_records VALUES('u1003', 20210102);
INSERT INTO t_act_records VALUES('u1004', 20210101);
INSERT INTO t_act_records VALUES('u1004', 20210102);
INSERT INTO t_act_records VALUES('u1004', 20210103);
INSERT INTO t_act_records VALUES('u1004', 20210104);
INSERT INTO t_act_records VALUES('u1004', 20210105);

解题答案

解题答案

# 解题方法:https://zhuanlan.zhihu.com/p/373033198
# step0:连续登录下imp_date - ranking 所得值相同
SELECT uid, imp_date, ROW_NUMBER() over (PARTITION BY uid ORDER BY imp_date) as "ranking"
FROM t_act_records;

# step1:做减法
SELECT *, DATE_SUB(imp_date, INTERVAL ranking DAY) as ds
FROM (
         SELECT uid, imp_date, ROW_NUMBER() over (PARTITION BY uid ORDER BY imp_date) as "ranking"
         FROM t_act_records) as test;

# step2:统计连续登录的用户天数
SELECT uid, ds, COUNT(1)
FROM (SELECT *, DATE_SUB(imp_date, INTERVAL ranking DAY) as ds
      FROM (
               SELECT uid, imp_date, ROW_NUMBER() over (PARTITION BY uid ORDER BY imp_date) as "ranking"
               FROM t_act_records) as test) as sub_ds
GROUP BY uid, ds
;
# 问题：3.1. 计算2021年每个月，每个用户连续登录的最多天数
SELECT DISTINCT uid, ds, COUNT(1)
FROM (SELECT *, DATE_SUB(imp_date, INTERVAL ranking DAY) as ds
      FROM (
               SELECT uid, imp_date, ROW_NUMBER() over (PARTITION BY uid ORDER BY imp_date) as "ranking"
               FROM t_act_records) as test) as sub_ds
GROUP BY uid, ds
;
# 问题 3.2. 计算2021年每个月，连续2天都有登录的用户名单
SELECT DISTINCT uid, ds, COUNT(1)
FROM (SELECT *, DATE_SUB(imp_date, INTERVAL ranking DAY) as ds
      FROM (
               SELECT uid, imp_date, ROW_NUMBER() over (PARTITION BY uid ORDER BY imp_date) as "ranking"
               FROM t_act_records) as test) as sub_ds
GROUP BY uid, ds
HAVING COUNT(1) >= 3
;
# 问题 3.3. 计算2021年每个月，连续5天都有登录的用户数
SELECT DISTINCT uid, ds, COUNT(1)
FROM (SELECT *, DATE_SUB(imp_date, INTERVAL ranking DAY) as ds
      FROM (
               SELECT uid, imp_date, ROW_NUMBER() over (PARTITION BY uid ORDER BY imp_date) as "ranking"
               FROM t_act_records) as test) as sub_ds
GROUP BY uid, ds
HAVING COUNT(1) >= 5
;