UPDATE SET Increment INNER JOIN，我的选择有效但不更新答案

【问题标题】：UPDATE SET Increment INNER JOIN, my select works but not updateUPDATE SET Increment INNER JOIN，我的选择有效但不更新
【发布时间】：2021-12-26 07:35:45
【问题描述】：

我有以下 SQL 命令。有两张表，

名为 rbhl_linkednodes 的表如下所示

Id, Node1, Node2
1 , 6 , 7

名为 rbhl_nodelist 的表有一堆数据，包括 id 和 r，我需要更改 r 值

Id, R
6, 15
7, 15

选择命令有效并显示了上述结果。当我尝试用更新换掉它时，它不起作用。

UPDATE rbhl_nodelist
SET nl.r = nl.r - 3
FROM rbhl_nodelist nl
INNER JOIN rbhl_linkednodes ln
ON ln.node1 = nl.id or ln.node2 = nl.id
WHERE ln.id = 1;

SELECT nl.r
FROM rbhl_nodelist nl
INNER JOIN rbhl_linkednodes ln
ON ln.node1 = nl.id or ln.node2 = nl.id
WHERE ln.id = 1;

目标是使用一个命令一次将两个值减 3。

【问题讨论】：

标签： php sql sql-update inner-join

【解决方案1】：

试试Not Exists

update rbhl_nodelist
set r = r - 3
from rbhl_nodelist t1
where exists (
select 1 from rbhl_linkednodes t2
where t2.ID = 1
and t2.node1 = t1.id
or t2.node2 = t1.id
)

【讨论】：

【解决方案2】：

你几乎拥有它。

创建测试用例：

CREATE TABLE rbhl_linkednodes (
    id int auto_increment primary key,
    node1 varchar(255),
    node2 varchar(255)
);

CREATE TABLE rbhl_nodelist (
    id int,
    r int
);

insert into rbhl_linkednodes set node1 = 6, node2 = 7;
insert into rbhl_linkednodes set node1 = 16, node2 = 17;
insert into rbhl_linkednodes set node1 = 26, node2 = 27;

insert into rbhl_nodelist set id = 6, r = 15;
insert into rbhl_nodelist set id = 7, r = 15;
insert into rbhl_nodelist set id = 16, r = 15;
insert into rbhl_nodelist set id = 17, r = 15;
insert into rbhl_nodelist set id = 26, r = 15;
insert into rbhl_nodelist set id = 27, r = 15;


select * from rbhl_linkednodes;
+----+-------+-------+
| id | node1 | node2 |
+----+-------+-------+
|  1 | 6     | 7     |
|  2 | 16    | 17    |
|  3 | 26    | 27    |
+----+-------+-------+
select * from rbhl_nodelist;
+------+------+
| id   | r    |
+------+------+
|    6 |   15 |
|    7 |   15 |
|   16 |   15 |
|   17 |   15 |
|   26 |   15 |
|   27 |   15 |
+------+------+

更新查询：

UPDATE rbhl_nodelist nl
INNER JOIN rbhl_linkednodes ln
ON ln.node1 = nl.id or ln.node2 = nl.id
SET nl.r = nl.r - 3
WHERE ln.id = 1;

结果

select * from rbhl_linkednodes;
+----+-------+-------+
| id | node1 | node2 |
+----+-------+-------+
|  1 | 6     | 7     |
|  2 | 16    | 17    |
|  3 | 26    | 27    |
+----+-------+-------+
select * from rbhl_nodelist;
+------+------+
| id   | r    |
+------+------+
|    6 |   12 | <--- OK
|    7 |   12 | <--- OK
|   16 |   15 |
|   17 |   15 |
|   26 |   15 |
|   27 |   15 |
+------+------+

所以顺序是：
更新
加入
设置在哪里

【讨论】：

谢谢！！！我要疯了，想弄清楚。上帝保佑 stackoverflow 和给予他们帮助的人们！