Question

我有两张桌子：

purchase_mis(id, user_id, total_purchased, date)
daily_purchase(id, user_id, product_id, paid_amount, purchase_date)

我有一个每晚运行的CRON文件，它会计算"daily_purchase"表中的每日购买量并运行insert into "purchase_mis"。

例如：

SELECT
    COUNT(*) AS purchase_count,
    purchase_date
FROM daily_purchase
GROUP BY user_id;

这将为每个用户返回purchase_count，然后将其插入"purchase_mis"表。

INSERT INTO
    purchase_mis(user_id, total_purchased, date)
VALUES
    ('2', 'purchase_count', 'purchase_date');

但在插入之前，需要检查某个日期“user_id = 2”的purchase_date的购买信息是否已经插入，因此不应再次插入。

我想要类似下面的查询：

INSERT INTO
    purchase_mis(user_id, total_purchased, date)
VALUES
    ('2', 'purchase_count', 'purchase_date')
WHERE date NOT EXISTS (SELECT date FROM purchase_mis WHERE user_id = '2');

Answer 1

在日期列上创建一个键，然后使用INSERT IGNORE或INSERT ... ON DUPLICATE KEY UPDATE语法。有关详情，请参阅此处："INSERT IGNORE" vs "INSERT ... ON DUPLICATE KEY UPDATE"

此外，您可以使用触发器或过程而不是cron作业;可能会让生活变得更轻松。

Answer 2

insert into purchase_mis
   (user_id, total_purchased, date)
select * 
from (
   select 2 as id, 
          100 as user_id, 
          str_to_date('2012-12-04', '%Y-%m-%d') as purchase_date
) t 
where not exists (SELECT 1 
                  FROM purchase_mis pm
                  WHERE pm.user_id = t.id 
                  and pm.date = t.purchase_date);

MySQL数据库相关

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