Question

我在MYSQL中有一个表：

CREATE TABLE test.tem(a INT,b INT);

以下数据：

INSERT INTO test.tem VALUES(1,2),(1,1),(1,NULL),(2,3);

现在数据应该是：

+------+------+
| a    | b    |
+------+------+
|    1 |    2 |
|    1 |    1 |
|    1 | NULL |
|    2 |    3 |
+------+------+

我想按列a。将列b更新到min（b）组。

所以SQL应该是：

UPDATE test.tem o
SET o.b = (SELECT
             MIN(b)
           FROM test.tem i
           WHERE i.a = o.a)

但是MYSQL 无法在FROM子句中指定更新目标表

所以我认为下面的SQL可以很好地解决我的问题：

UPDATE test.tem t1
  JOIN test.tem t2
    ON t1.a = t2.a
SET t1.b = t2.b
WHERE t1.b IS NULL
     OR t1.b > t2.b;

但结果是：

+------+------+
| a    | b    |
+------+------+
|    1 |    1 |
|    1 |    1 |
|    1 |    2 |
|    2 |    3 |
+------+------+

实际上我需要的结果是：

+------+------+
| a    | b    |
+------+------+
|    1 |    1 |
|    1 |    1 |
|    1 |    1 |
|    2 |    3 |
+------+------+

问题1：为什么MYSQL使用SQL计算出错误的结果？具有高效率的正确SQL应该是什么？问题2：如果我只想用b值更新b（仅更新第三条记录），SQL应该是什么？

关于问题2，我试图使用下面不正确的SQL：

UPDATE test.tem t1
  JOIN test.tem t2
    ON t1.a = t2.a
    AND t1.b IS NULL
SET t1.b = t2.b
WHERE t1.b IS NULL
     OR t1.b > t2.b;

Answer 1

您没有唯一列来标识行。因此，您的JOIN可能会按照您的想法更新更多行。

你可能想要这样的东西：

UPDATE tem AS t1 JOIN (SELECT a, MIN(b) AS m FROM tem GROUP BY a) AS t2
USING (a)
SET t1.b = t2.m;

请参阅http://sqlfiddle.com/#!2/c6a04/1

如果您只想更新 b 列中NULL的行，这只是WHERE子句的问题：

CREATE TABLE tem(a INT,b INT);    
INSERT INTO tem VALUES(1,2),(1,1),(1,NULL),(2,3);

UPDATE tem AS t1 JOIN (SELECT a, MIN(b) AS m FROM tem GROUP BY a) AS t2
USING (a)
SET t1.b = t2.m
WHERE t1.b IS NULL;

请参阅http://sqlfiddle.com/#!2/31ffb/1

Answer 2

Write it as a JOIN instead：

UPDATE tem
JOIN ( SELECT a, MIN(b) AS min_b FROM tem GROUP BY a ) AS mins USING (a)
SET tem.b = mins.min_b ;

Answer 3

您可以使用临时表来执行此操作：

create temporary table tem2 (a INT, b INT);

insert into tem2 
  select a, min(b) from tem group by a;

update tem
  inner join tem2 on tem.a = tem2.a
  set tem.b = tem2.b;

drop table tem2;

我认为这应该有效。 'drop table'并不是绝对必要的，因为tem2在连接关闭时会被丢弃，尽管它的形式很好。

mysql如何更新自己的表工作

3 个答案: