Question

我有这个查询，它做它应该做的事情。它查找表1中表2中匹配的所有记录，并指定伪列值1。然后它联合一个查询，该查询检索表1中表2中没有匹配项的所有记录，并分配伪列值0。这有效，但我确信可以更有效地完成它。

SELECT aplac.*, 1 AS selected
FROM aux_placements aplac
JOIN product_placements pplac
ON pplac.placement_id = aplac.placement_id
WHERE pplac.product_id = 1
UNION
SELECT distinct aplac.*, 0 AS selected
FROM aux_placements aplac
WHERE placement_id NOT IN 
    (SELECT aplac.placement_id
    FROM aux_placements aplac
    JOIN product_placements pplac
    ON pplac.placement_id = aplac.placement_id
    WHERE pplac.product_id = 1)

我在想这样的事可能有用：

SELECT aplac.*,
       CASE WHEN aplac.placement_id IS NULL 
       THEN 0 
       ELSE 1 
       END AS selected
FROM aux_placements aplac 
LEFT OUTER JOIN product_placements pplac
ON pplac.placement_id = aplac.placement_id
WHERE pplac.product_id = 1;

任何建议表示赞赏。提前致谢

编辑：sqlfiddle在这里可用：http://sqlfiddle.com/#!2/57013a/1

Answer 1

您希望产品ID选择成为加入条件的一部分 - 将WHERE更改为AND。
您想要检查联接是否在第二个表中找到了匹配的记录 - 将aplac.placement_id更改为pplac.placement_id。

所以下一个查询对你有用：

SELECT aplac.*,
       CASE WHEN pplac.placement_id IS NULL 
       THEN 0 
       ELSE 1 
       END AS selected
FROM aux_placements aplac 
LEFT OUTER JOIN product_placements pplac
ON pplac.placement_id = aplac.placement_id
AND pplac.product_id = 1;

需要建议formore漂亮的SQL查询

1 个答案: