mysql - INSERT...SELECT...WHERE...ON DUPLICATE...mysql 查询

Question

请帮忙。我正在尝试用我们产品的当前 Assets 数量更新这张表。如果该产品已存在于表中，则应更新该产品的最新计数。但是，使用下面的查询，mysql 正在返回我

"ERROR 1111 (HY000): Invalid use of group function".

我无法确定我的错误是什么，或者在“重复键”函数中使用计数是否真的有效:

INSERT INTO report_count_assets
    SELECT products.product_id, 
    count(product_assets.asset_id),
    count(case when assets.asset_type_id=1 THEN 1 END), 
    count(case when assets.asset_type_id=2 THEN 1 END), 
    count(case when assets.asset_type_id=3 THEN 1 END), 
    count(case when assets.asset_type_id=11 THEN 1 END) 
    FROM products 
    LEFT JOIN product_assets USING (product_id) 
    LEFT JOIN assets USING (asset_id)
    WHERE products.brand_id=671

ON DUPLICATE KEY UPDATE
    asset_count = count(product_assets.asset_id),
    asset_type_image = count(case when assets.asset_type_id=1 THEN 1 END), 
    asset_type_video = count(case when assets.asset_type_id=2 THEN 1 END), 
    asset_type_sound = count(case when assets.asset_type_id=3 THEN 1 END), 
    asset_type_install = count(case when assets.asset_type_id=11 THEN 1 END);

Answer 1

我认为您不能在 ON DUPLICATE 中使用聚合函数. MySQL 看到你的 SQL 有点像这样:

insert into report_count_assets
expr
on duplicate key update
...

ON DUPLICATE 不知道 expr 中发生了什么，它只知道它有一个重复的行要处理。不知道 expr 内部发生了什么，就没有 count 操作的上下文。此外，您应该在更新中使用 values:

You can use the VALUES(col_name) function in the UPDATE clause to refer to column values from the INSERT portion of the INSERT ... ON DUPLICATE KEY UPDATE statement.

并且 values(count(x)) 不是有效语法。但是 values(column_name) 是有效的，所以这应该有效:

INSERT INTO report_count_assets
(product_id, asset_count, asset_type_image, asset_type_video, asset_type_sound, asset_type_install)
    SELECT products.product_id, 
    count(product_assets.asset_id),
    count(case when assets.asset_type_id=1 THEN 1 END), 
    count(case when assets.asset_type_id=2 THEN 1 END), 
    count(case when assets.asset_type_id=3 THEN 1 END), 
    count(case when assets.asset_type_id=11 THEN 1 END) 
    FROM products 
    LEFT JOIN product_assets USING (product_id) 
    LEFT JOIN assets USING (asset_id)
    WHERE products.brand_id=671
ON DUPLICATE KEY UPDATE
    asset_count = values(asset_count),
    asset_type_image = values(asset_type_image), 
    asset_type_video = values(asset_type_video), 
    asset_type_sound = values(asset_type_sound), 
    asset_type_install = values(asset_type_install);

我不得不猜测 report_count_assets 中 product_id 列的名称。

如果这不起作用(显然它不起作用)，那么您可以通过预先计算 SELECT 来以困难的方式完成它。创建一个临时表:

create temporary table t (
    product_id int,
    product_count int,
    assets1 int,
    assets2 int,
    assets3 int,
    assets11 int
)

填充它:

INSERT INTO t (product_id, product_count, assets1, assets2, assets3, assets11)
SELECT products.product_id, 
count(product_assets.asset_id),
count(case when assets.asset_type_id=1 THEN 1 END), 
count(case when assets.asset_type_id=2 THEN 1 END), 
count(case when assets.asset_type_id=3 THEN 1 END), 
count(case when assets.asset_type_id=11 THEN 1 END) 
FROM products 
LEFT JOIN product_assets USING (product_id) 
LEFT JOIN assets USING (asset_id)
WHERE products.brand_id=671

然后使用那个临时表来做你真正想做的插入:

insert into report_count_assets
    select product_id, product_count, assets1, assets2, assets3, assets11
    from t
on duplicate key update
    asset_count = values(product_count),
    asset_type_image = values(assets1),
    asset_type_video = values(assets2),
    asset_type_sound = values(assets3),
    asset_type_install = values(assets11)