Question

我正在构建一个连接多个表的查询，一个表prodRecipe将包含某些行的数据，但不是全部，但tmp_inv1将包含所有内容的计数信息。问题是，tmp_inv1.count取决于项目是否在prodRecipe表中。

即。如果存在prodRecipe.product，则tmp_inv1.count基于此，否则它基于product.prodID。

在JOIN中使用OR似乎可以根据任何给出有效结果的方式进行填充，但是我想知道这是否会导致我不知道的不必要或灾难性的结果？

- 我一时兴起尝试使用OR，因为我还没有找到更好的解决方案。但最重要的是它对PHP很友好，所以我不能使用多个或嵌套的选择查询.--

SELECT invTrack AS track, varID, varSKU, revelSku, revelInvID, p.prodID AS prodID, pr.ingredient AS ing, pr.main_product AS mainProduct, pr.product AS product, i.invID AS invC, i.count,  pr.qty AS Xply
FROM shopifyProd sp
LEFT JOIN product p ON sp.revelSku = p.sku AND p.location = 1
LEFT JOIN prodRecipe pr ON p.prodID = pr.main_product
LEFT JOIN tmp_inv1 i ON pr.product = i.prodID OR p.prodID = i.prodID
WHERE 1

Answer 1

要避免多次匹配，您需要加入两次（每个条件一次）并选择优先级匹配。

SELECT invTrack AS track, varID, varSKU, revelSku, revelInvID, p.prodID AS prodID, pr.ingredient AS ing, pr.main_product AS mainProduct, pr.product AS product
, CASE WHEN p_i.prodID IS NOT NULL THEN p_i.invID ELSE pr_i.invID END AS invC
, CASE WHEN p_i.prodID IS NOT NULL THEN p_i.count ELSE pr_i.count END AS `count`
, pr.qty AS Xply
FROM shopifyProd sp
LEFT JOIN product p ON sp.revelSku = p.sku AND p.location = 1
LEFT JOIN prodRecipe pr ON p.prodID = pr.main_product
LEFT JOIN tmp_inv1 p_i ON p.product = p_i.prodID
LEFT JOIN tmp_inv1 pr_i ON pr.product = pr_i.prodID
WHERE 1

可以在LEFT JOIN table1 ON条件1或条件2中使用OR导致不必要的问题吗？

1 个答案: