Mysql FOREIGN KEY error 150 - mysql

Question

CREATE TABLE IF NOT EXISTS `questions` (
`question_id` int(11) NOT NULL AUTO_INCREMENT,
`createddate` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
`updateddate` timestamp NULL DEFAULT NULL,
`active_flag` tinyint(4) NOT NULL DEFAULT '0',
PRIMARY KEY (`question_id`),
UNIQUE KEY `id_UNIQUE` (`question_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 AUTO_INCREMENT=1 ;
CREATE TABLE `alarts` (
`alart_id` BIGINT(20) unsigned NOT NULL AUTO_INCREMENT,
`alart_name` varchar(45) NOT NULL,
`interval` int(10) unsigned NOT NULL,
`alart_sent_counter` int(10) unsigned NOT NULL,
`alart_types_id` BIGINT(20) unsigned NOT NULL,
`contact_group_id` int(10) unsigned NOT NULL,
PRIMARY KEY (`alart_id`),
FOREIGN KEY (`alart_types_id`) REFERENCES alart_types(`alart_types_id`)
) ENGINE=InnoDB AUTO_INCREMENT=20 DEFAULT CHARSET=utf8;
I want to create new table with two FOREIGN KEY like this:
CREATE TABLE `alart_question_mapping` (
`alart_question_mapping_id` BIGINT(20) unsigned NOT NULL AUTO_INCREMENT,
`question_id` int(11) NOT NULL,
`alart_id` BIGINT(20) unsigned NOT NULL,
PRIMARY KEY (`alart_question_mapping_id`),
FOREIGN KEY (`question_id`) REFERENCES questions(`question_id`),
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
) ENGINE=InnoDB AUTO_INCREMENT=20 DEFAULT CHARSET=utf8;
but I am getting error:
Error Code: 1005. Can't create table 'alart_question_mapping' (errno: 150)
How can I create this table ?
Thank's.

Answer 1

Chane the statement:
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
to
FOREIGN KEY (`alart_id`) REFERENCES alarts(`alart_id`)

Answer 2

The only thing I can see is you are referencing a table in your CREATE TABLE statement that is not present in what you provided:
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
If you remove this reference the table will create. See SQL Fiddle with Demo
Edit #1, based on your update the problem is you are referencing the wrong field in your last table:
Change this:
CREATE TABLE `alart_question_mapping` (
`alart_question_mapping_id` BIGINT(20) unsigned NOT NULL AUTO_INCREMENT,
`question_id` int(11) NOT NULL,
`alart_id` BIGINT(20) unsigned NOT NULL,
PRIMARY KEY (`alart_question_mapping_id`),
FOREIGN KEY (`question_id`) REFERENCES questions(`question_id`),
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
)
To this:
CREATE TABLE `alart_question_mapping` (
`alart_question_mapping_id` BIGINT(20) unsigned NOT NULL AUTO_INCREMENT,
`question_id` int(11) NOT NULL,
`alart_id` BIGINT(20) unsigned NOT NULL,
PRIMARY KEY (`alart_question_mapping_id`),
FOREIGN KEY (`question_id`) REFERENCES questions(`question_id`),
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_types_id`)
) ENGINE=InnoDB AUTO_INCREMENT=20 DEFAULT CHARSET=utf8;
So you are changing this line:
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
to this:
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_types_id`)
If you are referencing the alart_types table then you will want to reference the alart_types_id not the alart_id
see SQL Fiddle with Demo

Answer 3

It can't find table alart_types.
From MySQL Foreign Key Constraints
If you re-create a table that was dropped, it must have a definition
that conforms to the foreign key constraints referencing it. It must
have the right column names and types, and it must have indexes on the
referenced keys, as stated earlier. If these are not satisfied, MySQL
returns error number 1005 and refers to error 150 in the error
message.
I think you mean
FOREIGN KEY (`alart_id`) REFERENCES alart(`alart_id`)
instead of
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
Hope this makes sense.

Answer 4

In this table
CREATE TABLE alart_types (
alart_types_id BIGINT(20) unsigned NOT NULL AUTO_INCREMENT,
alarts_types_name varchar(45) NOT NULL,
PRIMARY KEY (alart_types_id)
) ENGINE=InnoDB AUTO_INCREMENT=20 DEFAULT CHARSET=utf8;
it doesn't really make sense to have an autoincrement id number without also having a unique constraint on alarts_types_name. Without that unique constraint, you're almost certain to end up with a table whose data looks like this.
1 Warning
2 Critical
3 Warning
4 Warning
This constraint references a column that doesn't exist.
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
It should be
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_types_id`)