Suppose I have a table items
with columns id (PRIMARY), name(VARCHAR), section_id (BIGINT), updated_at (DATETIME)
,
and a table sections
with id (PRIMARY)
.
Naturally, items.section_id
is a foreign key that refers to sections.id
.
Suppose there is an index on items
of the columns (section_id, name)
. I believe that if you tried to drop this index, you would get an error that it is needed in a foreign key constraint
. I can accept this.
Now, I want to create a new index, like create index ix_section_id_id_updated_at on items (section_id, id, updated_at)
. MySQL lets me do this, but if I go to drop this table, I get that same error: it fails, because it is needed in a foreign key constraint
.
Why should this be? It already has one index that can be used for this foreign key check. Further, the error does NOT go away with set FOREIGN_KEY_CHECKS=0;
. Is there a way to force MySQL to not associate the new index with the foreign key, so that it is quick to drop? This is necessary because I will be running the migration on a production server with temporary downtime, and need to be able to quickly revert the migration in case of anything going wrong afterwards.
CodePudding user response:
I can reproduce your issue if I don't create an index on section_id and allow mysql to do so on the creation of a foreign key(as described in the manual). Adding a new index drops the auto generated key and if you then drop the new index an error is generated because of the requirement to have a key , and mysql does not auto generate one on a drop.. . If you manually generate a key on section_id this problem does not happen..and the newly created compound index drops successfully.
drop table if exists items;
drop table if exists sections;
create table items(id int PRIMARY key, name varchar(3), section_id BIGINT, updated_at DATETIME);
create table sections(id bigint primary key);
alter table items
add foreign key fk1(section_id) references sections(id);
show create table items;
CREATE TABLE `items` ( `id` int(11) NOT NULL,
`name` varchar(3) DEFAULT NULL,
`section_id` bigint(20) DEFAULT NULL,
`updated_at` datetime DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `fk1` (`section_id`),
CONSTRAINT `fk1` FOREIGN KEY (`section_id`) REFERENCES `sections` (`id`)) ENGINE=InnoDB DEFAULT CHARSET=latin1;
alter table items
add key key1(section_id, name);
show create table items;
CREATE TABLE `items` (
`id` int(11) NOT NULL,
`name` varchar(3) DEFAULT NULL,
`section_id` bigint(20) DEFAULT NULL,
`updated_at` datetime DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `key1` (`section_id`,`name`),
CONSTRAINT `fk1` FOREIGN KEY (`section_id`) REFERENCES `sections` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
and with manually generated key
drop table if exists items;
drop table if exists sections;
create table items(id int PRIMARY key, name varchar(3), section_id BIGINT, updated_at DATETIME);
create table sections(id bigint primary key);
alter table items
add key sid(section_id);
alter table items
add foreign key fk1(section_id) references sections(id);
show create table items;
CREATE TABLE `items` (
`id` int(11) NOT NULL,
`name` varchar(3) DEFAULT NULL,
`section_id` bigint(20) DEFAULT NULL,
`updated_at` datetime DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `sid` (`section_id`),
CONSTRAINT `fk1` FOREIGN KEY (`section_id`) REFERENCES `sections` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
alter table items
add key key1(section_id, name);
show create table items;
CREATE TABLE `items` (
`id` int(11) NOT NULL,
`name` varchar(3) DEFAULT NULL,
`section_id` bigint(20) DEFAULT NULL,
`updated_at` datetime DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `sid` (`section_id`),
KEY `key1` (`section_id`,`name`),
CONSTRAINT `fk1` FOREIGN KEY (`section_id`) REFERENCES `sections` (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;