Question

我有这个问题，应该向人们退还过去一周他们购买的东西的一半：

UPDATE main_data SET md.amount_current = md.amount_current + (bought.total / 2) 
FROM main_data AS md 
INNER JOIN (
    SELECT DISTINCT sb.user_id, SUM(sb.spend) AS total 
    FROM shopitems_bought AS sb 
    LEFT JOIN shopitems AS si 
    ON sb.shopitem_id = si.id 
    WHERE sb.date_bought <= '2012-03-09' 
    AND sb.date_bought > DATE_ADD('2012-03-09', INTERVAL -7 DAY) 
    AND si.valid = 1 
    GROUP BY sb.user_id
) AS bought ON bought.user_id = md.user_id 
WHERE md.valid = 1

SELECT部分自行执行并返回正确的数据（应该退还的ID以及他们在那一周花费的金额）。但是整个查询会抛出一个错误，说我在第2行附近的SQL语法中有错误（我引用：'FROM main_data AS md INNER JOIN（SELECT DISTINCT sb.forum_id，SUM（sb.s'）。

我看不出我做错了什么。

Answer 1

MySql uses a different syntax for join with update statements比你上面使用的要多。尝试将您的查询更改为：

UPDATE main_data md 
INNER JOIN (
    SELECT DISTINCT sb.user_id, SUM(sb.spend) AS total 
    FROM shopitems_bought AS sb 
    LEFT JOIN shopitems AS si 
    ON sb.shopitem_id = si.id 
    WHERE sb.date_bought <= '2012-03-09' 
    AND sb.date_bought > DATE_ADD('2012-03-09', INTERVAL -7 DAY) 
    AND si.valid = 1 
    GROUP BY sb.user_id
) bought ON bought.user_id = md.user_id 
SET amount_current = md.amount_current - (bought.total / 2) 
WHERE md.valid = 1

注意，我改变了

SET amount_current = md.amount_current + (bought.total / 2)

要减去而不是添加：

SET amount_current = md.amount_current - (bought.total / 2)

Answer 2

INNER JOIN中没有user_Id

UPDATE main_data SET md.amount_current = md.amount_current + (bought.total / 2) 
FROM main_data AS md 
INNER JOIN (
    SELECT sb.user_id, DISTINCT sb.forum_id, SUM(sb.spend) AS total 
    FROM shopitems_bought AS sb 
    LEFT JOIN shopitems AS si 
    ON sb.shopitem_id = si.id 
    WHERE sb.date_bought <= '2012-03-09' 
    AND sb.date_bought > DATE_ADD('2012-03-09', INTERVAL -7 DAY) 
    AND si.valid = 1 
    GROUP BY sb.user_id
) AS bought ON bought.user_id = md.user_id 
WHERE md.valid = 1

SQL查询错误，需要一些帮助

2 个答案: