更新时如何防止重复输入键

Question

问题说明

我不会更新 3 个主键连接的最后一个主键。但问题有时是多个记录的第一个和第二个主键相同。在这种情况下，当我设置我的新值时，即使我使用子请求来避免这个问题，我也会有一个重复的条目键。

一些代码

架构

create table llx_element_contact
(
    rowid             int auto_increment
        primary key,
    datecreate        datetime           null,
    statut            smallint default 5 null,
    element_id        int                not null,
    fk_c_type_contact int                not null,
    fk_socpeople      int                not null,
    constraint idx_element_contact_idx1
        unique (element_id, fk_c_type_contact, fk_socpeople)
)

更新请求

此请求返回重复键错误

update llx_element_contact lec
set lec.fk_socpeople = 64
where 
-- Try to avoid the error by non including the values that are the same
(select count(*)
 from llx_element_contact ec
 where ec.fk_socpeople = 64
   and ec.element_id = lec.element_id
   and ec.fk_c_type_contact = lec.fk_c_type_contact) = 0

测试数据

rowid, datecreate, statut, element_id, fk_c_type_contact, fk_sockpeople
65,2015-08-31 18:59:18,4,65,160,30
66,2015-08-31 18:59:18,4,66,159,12
67,2015-08-31 18:59:18,4,67,160,12
15283,2016-03-23 11:47:15,4,6404,160,39
15284,2016-03-23 11:51:30,4,6404,160,58

Answer 1

当您尝试将相同的值分配给 3d 成员时，您应该只检查唯一约束的其他两个成员。最多只能存在一行具有相同的两个成员。

update llx_element_contact lec
set lec.fk_socpeople = 64
where 
-- Try to avoid the error by non including the values that are the same
(select count(*)
 from llx_element_contact ec
 where ec.element_id = lec.element_id
   and ec.fk_c_type_contact = lec.fk_c_type_contact) <=1

或

update llx_element_contact lec
set lec.fk_socpeople = 64
where 
-- Try to avoid the error by non including the values that are the same
 not exists (select 1
 from llx_element_contact ec
 where ec.element_id = lec.element_id
   and ec.fk_c_type_contact = lec.fk_c_type_contact
   and lec.fk_socpeople != ec.fk_socpeople)

Answer 2

你可以使用：

您可以使用left join 检查相应的行是否不存在，以防止unique 冲突：

update llx_element_contact lec left join
       (select element_id, fk_c_type_contact
        from llx_element_contact lec2
        where lec2.fk_socpeople = 64
        group by element_id, fk_c_type_contact
       ) lec2
       using (element_id, fk_c_type_contact)
    set lec.fk_socpeople = 64
    where lec2.element_id is null;

您的查询中有未解释的附加逻辑。对于您的要求，这不是必需的。

Answer 3

WITH cte AS (
SELECT rowid, 
       SUM(fk_socpeople = 64) OVER (PARTITION BY element_id, fk_c_type_contact) u_flag,
       ROW_NUMBER() OVER (PARTITION BY element_id, fk_c_type_contact ORDER BY datecreate DESC) u_rn
FROM llx_element_contact
)
update llx_element_contact lec
JOIN cte USING (rowid)
set lec.fk_socpeople = 64 
where cte.u_flag = 0
  AND cte.u_rn = 1

https://dbfiddle.uk/?rdbms=mysql_8.0&fiddle=08e20328ccc6187716084ce9d78816b0