禁止在MySQL的更新子句中使用相同的目标表
P粉982009874
P粉982009874 2024-04-02 10:49:18
0
1
367

我有一个名为“employee”的表。表创建代码如下:

create table employee(name varchar(50),ph_no varchar(10),e_id varchar(5),pay_scale varchar(5),year varchar(4));

表格内容如下:

insert into employee(name,ph_no,pay_scale,year) values('AMIT','123456','PL-10','2019');
insert into employee(name,ph_no,pay_scale,year) values('AMIT','123456','PL-10','2020');
insert into employee(name,ph_no,pay_scale,year) values('AMIT','123456','PL-11','2021');
insert into employee(name,ph_no,pay_scale,year) values('AMIT','123456','PL-11','2022');

+------+--------+------+-----------+------+
| name | ph_no  | e_id | pay_scale | year |
+------+--------+------+-----------+------+
| AMIT | 123456 | NULL | PL-10     | 2019 |
| AMIT | 123456 | NULL | PL-10     | 2020 |
| AMIT | 123456 | NULL | PL-11     | 2021 |
| AMIT | 123456 | NULL | PL-11     | 2022 |
+------+--------+------+-----------+------+

现在我想更新'e_id',首先它会检查表中是否有相同的e_id,如果不在表中那么它只会更新给定e_id的行,否则不会要去更新了。 因此,我的升级查询如下:

update employee 
set e_id='0132' 
where concat_ws(',',name,ph_no,pay_scale)=concat_ws(',','AMIT','123456','PL-10') 
  and not exists (select e_id 
                  from employee 
                  group by e_id 
                  having count(*)>=1);

但它给出了以下错误:

错误 1093 (HY000):您无法在 FROM 子句中指定要更新的目标表“employee” 我尝试过以下查询:

update employee set e_id='0132' where
concat_ws(',',name,ph_no,pay_scale)=concat_ws(',','AMIT','123456','PL-10') and 
e_id not in 
    (select e_id from 
    (select e_id from employee group by e_id having count(*)>=1) as t);

但这也无法更新表格并显示以下结果:

Query OK, 0 rows affected (0.01 sec)
匹配的行:0 更改:0 警告:0

还尝试了以下代码:

update employee set 
employee.e_id='0132' where 
employee.e_id not in (select * from
    (select f.e_id from 
    employee f inner join employee b on 
    b.name=f.name and b.ph_no=f.ph_no and b.pay_scale=f.pay_scale) as tmp) 
and employee.name='AMIT' and employee.ph_no='123456' and employee.pay_scale='PL-10';

但这也无法更新表格并给出以下结果: 查询正常,0 行受影响(0.00 秒) 匹配的行:0 更改:0 警告:0 请帮忙。预先感谢您。

P粉982009874
P粉982009874

全部回复(1)
P粉604669414

NULL 的播放方式与某些人期望的 NOT IN 不同:https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=24c176ff4d4e2c52309aaca14cc121c5 因此,只需将 WHERE e_id IS NOT NULL 放在子中询问。另外,HAVING COUNT(*) >= 1 可以删除,因为它总是返回 1 或更多的值...

update
  employee
set
  e_id='0132'
where
     name      = 'AMIT'
 and ph_no     = '123456'
 and pay_scale = 'PL-10'
 and e_id      not in (select e_id from 
                        (select distinct e_id
                           from employee
                          where e_id IS NOT NULL
                        )
                        as t
                       );

https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=2a0b036a7d1db9138e3ab29af3d346f8 一个>

热门教程
更多>
最新下载
更多>
网站特效
网站源码
网站素材
前端模板