Question

我使用以下查询：

SELECT shop_entrys.id, shop_images.path FROM shop_entrys,shop_images 
WHERE shop_entrys.id = shop_images.s_id AND 
shop_images.pos = 0 AND 
shop_entrys.category_id = 1 
ORDER BY shop_entrys.pos ASC

但是， shop_entrys 中的一行可能存在，并且没有链接到 shop_images 中的行。因此， ... WHERE shop_entrys.id = shop_images.s_id ... 将无法满足。在这种情况下，我仍然想要返回一个结果。例如：

shop_entrys.id shop_images.path 1 "/img1.jpg" ... ... 42 "not found"

如何更改上述查询仍然返回结果？

Answer 1

使用LEFT OUTER JOIN和COALESCE为第2列提供默认值

SELECT shop_entrys.id, COALESCE(shop_images.path, 'NOT FOUND' )
FROM shop_entrys
LEFT OUTER JOIN shop_images 
ON shop_entrys.id = shop_images.s_id AND shop_images.pos = 0 
WHERE shop_entrys.category_id = 1 
ORDER BY shop_entrys.pos ASC

Answer 2

使用LEFT JOIN和COALESCE()：

SELECT shop_entrys.id,
        COALESCE(shop_images.path,'NOT FOUND') 
FROM shop_entrys
LEFT JOIN shop_images 
 ON(shop_entrys.id = shop_images.s_id AND 
    shop_images.pos = 0)
WHERE shop_entrys.category_id = 1 
ORDER BY shop_entrys.pos

请避免使用隐式连接语法（以逗号分隔），因为它已被弃用且混乱，并且很多时候都会导致错误。仅使用正确的连接语法！

即使条目不存在，也会返回值

2 个答案: