Question

我有一张名为＆＃39; language＆＃39;如下图所示

personid |     lang
-------------------
1       |    english
1       |    french
1       |    italian
2       |    italian
3       |    french
3       |    italian
4       |    japanese

我希望用人格1知道的任何语言选择所有人格。（即任何具有英语，法语或italaian值的人物）

我使用了以下查询并获得了解决方案。但似乎有点慢。（我认为这是由于＆＃39; IN＆＃39;条款）。是否有任何其他查询选项可以加快执行速度。

SELECT distinct personid FROM language WHERE personid!=1 AND lang IN (SELECT lang FROM language WHERE personid=1)

Answer 1

您可以自行加入language表：

SELECT DISTINCT l2.personid AS personid FROM
language l1 INNER JOIN language l2
ON l1.lang = l2.lang
WHERE l1.personid = 1;

<强>输出：

+----------+
| personid |
+----------+
|    2     |
|    3     |
+----------+

Answer 2

您可以使用EXISTS：

SELECT distinct personid 
FROM language AS l1
WHERE personid != 1 AND EXISTS (SELECT 1 
                                FROM language AS l2
                                WHERE personid = 1 AND l1.lang = l2.lang)

＆＃39; IN＆＃39; mysql中的子句使执行速度变慢，还有其他选择吗？

2 个答案: