Question

在MySQL数据库中，我需要创建一个新的closure表（称为closure_new），它将两列外键集成到另一个表concept。这意味着将行添加到不在closure_new中的closure。如何设置SQL来完成此任务？

这是我第一次尝试填充closure_new：

的代码

INSERT INTO `closure_new`
SELECT o.subtypeId, d.id, d.effectiveTime
  FROM concept d
  JOIN closure o
  ON o.subtypeId = d.id;

请注意，我的第一次尝试仅针对subtypeId/subtype_effectiveTime，可能无法完全解决。 SQL还需要合并supertypeId/supertype_effectiveTime。如何编写SQL以填充closure_new表，其中包含与每个effectiveTime和每个subtypeId相关联的每个supertypeId值的记录？

这是concept表：

CREATE TABLE `concept` (
`id` BIGINT NOT NULL DEFAULT 0,
`effectiveTime` VARCHAR(8) NOT NULL DEFAULT '',
`some other fields`,
PRIMARY KEY (`id`,`effectiveTime`)
) ENGINE=InnoDB;

这是旧closure表：

CREATE TABLE `closure` (
    `id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
    `subtypeId` BIGINT(20) NOT NULL ,
    `supertypeId` BIGINT(20) NOT NULL ,
    PRIMARY KEY (`id`)
);

这是closure_new表，需要填充我上面开始编写的脚本：

CREATE TABLE `closure_new` (
`id` int(11) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
`subtypeId` BIGINT(20) NOT NULL ,
`subtype_effectiveTime` VARCHAR(8) NOT NULL DEFAULT '',
`supertypeId` BIGINT(20) NOT NULL ,
`supertype_effectiveTime` VARCHAR(8) NOT NULL DEFAULT '',
FOREIGN KEY (`supertypeId`, `supertype_effectiveTime`) references concept(`id`, `effectiveTime`),
FOREIGN KEY (`subtypeId`, `subtype_effectiveTime`) references concept(`id`, `effectiveTime`)
); ENGINE=InnoDB;

Answer 1

试试这个：

insert into closure_new 
(subtypeId, subtype_effectiveTime, supertypeId, supertype_effectiveTime) 
select cl.id, co.effectiveTime, co.id, co.effectiveTime from closure cl inner join concept co

您的数据更匹配，或者您将遇到一些外键约束问题

Answer 2

不确定我是否完全理解你所追求的是什么，但是如何：

INSERT INTO `closure_new` (subtypeId, subtype_effectiveTime, supertypeId, supertype_effectiveTime)
SELECT subCon.id, subCon.effectiveTime, superCon.id, superCOn.effectiveTimed.effectiveTime
  FROM closure o, concept subCon, concept superCon
 where subcon.Id = o.subtypeId and supercon.Id = o.supertypeId

或者可能，您可以使用该选择语句create view。

来自同一个表的两个复合键

2 个答案: