如何在子查询中使用MySQL WHERE IN语句？

表如何在子查询中使用MySQL WHERE IN语句？

mysql> SELECT * FROM swaps_products; 

+----+---------+-----------+------------+-----------+---------------------+ 
| id | swap_id | holder_id | swap_to_id | product_id| added    | 
+----+---------+-----------+------------+-----------+---------------------+ 
| 1 |  9 |   3 |   0 |  32 | 2017-01-06 10:43:52 | 
| 2 |  11 |  13 |   0 |   3 | 2017-01-06 11:03:45 | 
| 3 |  11 |  13 |   0 |   4 | 2017-01-06 11:03:45 | 
| 4 |  11 |   3 |   0 |  32 | 2017-01-06 11:03:45 | 
| 5 |  11 |   3 |   0 |  31 | 2017-01-06 11:03:45 | 
| 6 |  11 |   3 |   0 |  30 | 2017-01-06 11:03:45 | 
| 7 |  12 |   3 |   0 |  32 | 2017-01-06 14:16:13 | 
| 8 |  12 |   3 |   0 |  31 | 2017-01-06 14:16:13 | 
| 9 |  12 |   2 |   0 |   2 | 2017-01-06 14:16:13 | 
| 10 |  12 |   2 |   0 |   1 | 2017-01-06 14:16:13 | 
| 11 |  13 |  13 |   3 |   3 | 2017-01-12 14:31:44 | 
| 12 |  13 |  13 |   3 |   4 | 2017-01-12 14:31:44 | 
| 13 |  13 |   3 |   13 |  32 | 2017-01-12 14:31:44 | 
| 14 |  13 |   3 |   13 |  31 | 2017-01-12 14:31:44 |

要swap_id = 13有涉及四款产品 - 3,4,31,32。我需要一个查询，它会给我所有其他交换（swap_id的列表），其中有3,4,31,32个产品中的任何一个交换了。

我有这个

SELECT sp.swap_id as swap_id, sp.holder_id as holder_id 
    -> FROM swaps_products as sp 
    -> JOIN swaps as s 
    -> ON s.id=sp.swap_id 
    -> WHERE (sp.swap_id != 13) AND 
    -> (s.rejected IS NULL) AND 
    -> ((s.swapped IS NULL) OR (s.swapped2 IS NULL)) AND 
    -> (sp.product_id IN (3,4,31,32));

连接是没有问题的重要。这个查询的工作原理，但问题是，我只知道swap_id。所以，我想这一点：

(sp.product_id IN (SELECT product_id FROM swaps_products WHERE swap_id = 13));

，但得到的错误：

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'JOIN swaps as s 
ON s.id=sp.swap_id 
WHERE (sp.swap_id != 13) AND 
(s.rejected IS N' at line 2

我认为问题是，子查询不返回数组，它是从声明的预期。

来源

2017-01-12 Kārlis Janisels

更新了错误消息的问题 –

IN语句要求结果是要检查的值的单列向量。

直接回答您的查询：

select * from swaps_products where product_id in (SELECT product_id FROM swaps_products where swap_id = 13) and swap_id != 13

此外，基于上述错误信息，您所遇到的第2行指向的语法错误;我怀疑你在where...in查询中遗漏了FROM swaps_products as sp部分。

来源

2017-01-12 16:06:02

我遗漏了FROM部分。 –

发生在我们身边:)快乐的编码！ –

SELECT DISTINCT y.* 
      FROM swaps_products x 
      JOIN swaps_products y 
      ON y.product_id = x.product_id 
      AND y.swap_id <> 13 
      WHERE x.swap_id = 13;

来源

2017-01-12 16:42:52 Strawberry

如何在子查询中使用MySQL WHERE IN语句？

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