MySQL JOIN两个表最接近的匹配值

Question

我有以下情况：

tableA
+-------+-------+
| id    | Value |
+-------+-------+
| 1     | 1000  |
| 2     | 20    |
| 3     | 62    |
| 4     | 0     |
+-------+-------+

tableB
+-------+--------+
| Value | Lookup |
+-------+--------+
|    10 | a      |
|    20 | b      |
|    30 | b      |
|    40 | g      |
|    50 | h      |
|    60 | f      |
|    70 | a      |
|    80 | a      |
|    90 | v      |
|   100 | b      |
+-------+--------+

我需要返回表 B 中与表 A 中值字段最匹配的查找。例如。

+-------+-------+--------+
| id    | Value | Lookup |
+-------+-------+--------+
| 1     | 1000  | b      |
| 2     | 20    | b      |
| 3     | 62    | f      |
| 4     | 0     | a      |
+-------+-------+--------+

我该怎么做呢？

Answer 1

这是一个使用连接的选项：

SELECT
    a.Id, a.Value, b.Lookup
FROM tableA a
CROSS JOIN tableB b
INNER JOIN
(
    SELECT a.Id, MIN(ABS(a.Value - b.Value)) AS min_abs_value
    FROM tableA a
    CROSS JOIN tableB b
    GROUP BY a.Id
) t
    ON a.Id = t.Id AND
       ABS(a.Value - b.Value) = t.min_abs_value;

演示

虽然此查询确实连接到子查询，但子查询不相关。

Answer 2

一种方法是使用相关子查询：

SELECT a.Id, a.Value, 
       (SELECT b.Lookup
       FROM TableB AS b
       ORDER BY ABS(a.Value - b.Value) LIMIT 1)
FROM TableA AS a

演示在这里