mysql - 如何将 LEFT JOIN 限制为一个结果并从结果集中返回最小值？

Question

假设我有这样的查询：

SELECT t1.name, t2.likes
FROM table_1 t1
LEFT JOIN table_2 t2 ON t1.name = t2.name

如何修改左连接以使其执行以下操作：

SELECT likes FROM table_2 WHERE name = [name from table 1] ORDER BY likes ASC

我尝试编写子查询，但它不起作用，因为（我假设）您不能在子查询内部使用子查询外部的变量（例如名称）（名称不是常量，作为查询的结果集将不止一个）。

那有可能吗？

编辑：表格将是这样的：

表格1：

+------+--------+
| id   |  name  |
+------+--------+
|  0   |  cat   |
|  1   |  dog   |
+------+--------+

表2：

+------+--------+---------+
| id   |  name  |  likes  |
+------+--------+---------+
|  0   |  cat   |   23    |
|  1   |  cat   |   2     |
|  2   |  cat   |   53    |
|  3   |  dog   |   25    |
|  4   |  dog   |   12    |
+------+--------+---------+

所以，我想要的是：

+--------+---------+
|  name  |  likes  |
+--------+---------+
|  cat   |   2     |
|  dog   |   12    |
+--------+---------+

Answer 1

SELECT 
    t1.name, t2.likes
FROM table_1 t1
LEFT JOIN (
    SELECT
    name,
    MIN(likes) as likes 
    FROM table_2 
    GROUP BY name
) as t2 ON t1.name = t2.name

您也可以使用子查询来做到这一点，但我认为这会更慢：

SELECT 
    t1.name,
    (SELECT MIN(likes) FROM table_2 WHERE table_2.name = t1.name)
FROM table_1 t1

Answer 2

likes每个名字的最低要求：

SELECT   name, MIN(likes) AS likes
FROM     table_2
GROUP BY name

如果您需要有关 的更多信息name，请连接到table_1表，将上述查询合并到子选择中：

SELECT a.name, b.likes
FROM   table_1 a
LEFT JOIN
(
    SELECT   name, MIN(likes) AS likes
    FROM     table_2
    GROUP BY name
) b ON a.name = b.name