mysql - LEFT JOIN 删除行而不过滤原始表

Question

下面的查询让我感到困惑。如果我执行 WHERE (inventory_to_pos.POSID IS NULL OR inventory_to_pos.POSID = ?) 并绑定(bind)，它不会显示 products 表中的所有行inventory_to_pos 中可能存在也可能不存在的 POSID。当 LEFT JOIN-ing 表时，我是否应该从原始表中获取 所有 行，当我只对原始表进行过滤并使用 IS NULL 在 LEFT JOIN'ed 表的任何条件下？

SELECT products.ID,products.NAME,products.VOLUME,productcombinations.PRODUCTID,productcombinations.PART,inventory_to_pos.FULLCOUNT
FROM products
LEFT JOIN productcombinations ON products.ID = productcombinations.PARTOF
LEFT JOIN inventory_to_pos ON products.ID = inventory_to_pos.PRODUCT 
WHERE products.INVENTORY = 1
AND products.AVAILABLE = 1
AND products.ID > 0
AND (inventory_to_pos.POSID IS NULL OR inventory_to_pos.POSID = ?);

如果给定产品和 POSID 不存在 inventory_to_pos.PRODUCT 和 inventory_to_pos.POSID，则没有行。为什么？

Answer 1

将所有相关的 invetory_to_pos 子句移动到 LEFT JOIN 中，即:

SELECT
    products.ID,
    products. NAME,
    products.VOLUME,
    productcombinations.PRODUCTID,
    productcombinations.PART,
    inventory_to_pos.FULLCOUNT
FROM
    products
LEFT JOIN productcombinations ON products.ID = productcombinations.PARTOF
LEFT JOIN inventory_to_pos ON products.ID = inventory_to_pos.PRODUCT AND (
    inventory_to_pos.POSID IS NULL
    OR inventory_to_pos.POSID = ?
)
WHERE
    products.INVENTORY = 1
AND products.AVAILABLE = 1
AND products.ID > 0