使用MySQL计算不同日期交易积分余额及差值
MySQL计算每日积分总余额及日差值
需求说明
按交易日期汇总当日积分总余额,并计算当日总余额与前一日总余额的差值。
样本数据
假设数据表名为transaction_points,字段包括:
id:交易编号(INT)transaction_date:交易日期(VARCHAR,格式为DD-MM-YYYY)points:交易积分(INT,正为增加,负为扣除)
样本数据如下:
| id | transaction_date | points |
|---|---|---|
| 1 | 09-08-2023 | 10 |
| 2 | 09-08-2023 | -5 |
| 3 | 09-08-2023 | 8 |
| 4 | 09-08-2023 | -3 |
| 5 | 09-08-2023 | 15 |
| 6 | 09-08-2023 | -8 |
| 7 | 09-08-2023 | 5 |
| 8 | 09-08-2023 | -2 |
| 9 | 09-08-2023 | 12 |
| 10 | 09-08-2023 | -6 |
| 11 | 10-08-2023 | 11 |
| 12 | 10-08-2023 | -4 |
| 13 | 10-08-2023 | 7 |
| 14 | 10-08-2023 | -3 |
| 15 | 10-08-2023 | 13 |
| 16 | 10-08-2023 | -7 |
| 17 | 10-08-2023 | 6 |
| 18 | 10-08-2023 | -2 |
| 19 | 10-08-2023 | 10 |
| 20 | 10-08-2023 | -5 |
| 21 | 11-08-2023 | 9 |
| 22 | 11-08-2023 | -4 |
| 23 | 11-08-2023 | 8 |
| 24 | 11-08-2023 | -3 |
| 25 | 11-08-2023 | 14 |
| 26 | 11-08-2023 | -7 |
解决方案SQL
SELECT transaction_date, daily_total AS 总积分, daily_total - LAG(daily_total) OVER (ORDER BY STR_TO_DATE(transaction_date, '%d-%m-%Y')) AS 差值 FROM ( -- 先按日期汇总每日总积分 SELECT transaction_date, SUM(points) AS daily_total FROM transaction_points GROUP BY transaction_date ) AS daily_summary ORDER BY STR_TO_DATE(transaction_date, '%d-%m-%Y');
代码解释
- 内层子查询
daily_summary:通过GROUP BY transaction_date分组,用SUM(points)计算当日积分总余额,得到每日的总积分数据。 - 外层查询:
- 使用
LAG(daily_total) OVER (ORDER BY STR_TO_DATE(transaction_date, '%d-%m-%Y'))获取前一日的总积分值,ORDER BY子句确保日期按时间顺序排序(原始日期为字符串格式,需用STR_TO_DATE转换为日期类型保证排序逻辑正确)。 - 用当日总积分减去前一日总积分得到差值;第一日无前置数据,差值为
NULL。
- 使用
执行结果
| transaction_date | 总积分 | 差值 |
|---|---|---|
| 09-08-2023 | 26 | NULL |
| 10-08-2023 | 26 | 0 |
| 11-08-2023 | 17 | -9 |
内容的提问来源于stack exchange,提问作者geroge
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