mySQL'無效列'錯誤

當'產品'表（master_categories_id = 500）滿足條件時，我試圖在表'products_to_categories'中插入一個關係我的猜測是因爲我的SELECT語句應該是一個JOIN？mySQL'無效列'錯誤

table: products_to_categories 
columns: products_id, categories_id 

table: products 
columns: products_id, master_categories_id 

INSERT INTO products_to_categories (products_id, categories_id) 
SELECT products.products_id, 500 FROM products 
WHERE products.master_categories_id=500 HAVING count(categories_id) < 1;

錯誤返回：categories_id is an invalid column

來源

2012-04-13 Byron Claiborne

幾個方面：

SELECT products.products_id, 500, products.categories_id FROM products 
WHERE products.master_categories_id=500 HAVING count(categories_id) < 1;

或者

SELECT products.products_id, 500, count(products.categories_id) as cat_count FROM products 
WHERE products.master_categories_id=500 HAVING cat_count < 1;

來源

2012-04-13 23:49:27

是的我試過了，categories_id不在'產品'表中。這就是爲什麼我猜測我需要用JOIN重寫。或者你可以簡單地在SELECT中說products_to_categories.categories_id？ – 2012-04-13 23:54:29

我只想用一個連接 – 2012-04-13 23:57:29

這是我最好的猜測，原諒我，通常我只是用PHP來處理這種事情，試圖用更高級的查詢來獲得更多的舒適，以避免腳本編寫。 INSERT INTO products_to_categories（products_id，categories_id） SELECT products.products_id，products.master_categories_id，products_to_categories.categories_id INNER JOIN產品ON products.products_id = products_to_categories.products_id FROM產品WHERE products.master_categories_id = 500，其具有計數（categories_id）< 1; – 2012-04-14 00:10:54

的categories_id列需要在一個表中提供您的SELECT。在這種情況下，只有products在那裏，所以唯一有效的列是或master_categories_id。

來源

2012-04-13 23:49:45 Tenner

mySQL'無效列'錯誤

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