I am creating 2 tables in my database:
DROP TABLE IF EXISTS `med_pharmacy`;
CREATE TABLE IF NOT EXISTS `med_pharmacy` (
`med_pharmacy_id` int(11) NOT NULL AUTO_INCREMENT,
`med_id` int(11) NOT NULL,
`med_barcode` varchar(45) DEFAULT NULL,
`med_received` date DEFAULT NULL,
`med_expiry` date DEFAULT NULL,
`med_tablet` int(11) DEFAULT NULL,
`med_pill` int(11) DEFAULT NULL,
`clinic_id` varchar(45) DEFAULT NULL,
PRIMARY KEY (`med_pharmacy_id`),
KEY `fk_med_pharmacy_medication1_idx` (`med_id`)
) ENGINE=InnoDB AUTO_INCREMENT=1261 DEFAULT CHARSET=utf8mb4;
AND:
DROP TABLE IF EXISTS `medication`;
CREATE TABLE `medication` (
`med_id` int(11) NOT NULL,
`med_name` varchar(75) NOT NULL,
`med_date_added` date DEFAULT NULL,
`clinic_id` varchar(45) DEFAULT NULL,
`med_type` varchar(15) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
And when I run the queries in wamp I got this error:
SQL query:
ALTER TABLE `med_pharmacy`
ADD CONSTRAINT `fk_med_pharmacy_medication1`
FOREIGN KEY (`med_id`)
REFERENCES
`medication` (`med_id`) ON DELETE CASCADE ON UPDATE CASCADE MySQL
said: Documentation
#1822 - Failed to add the foreign key constaint. Missing index for constraint 'fk_med_pharmacy_medication1' in the referenced table
'medication'
The tables already exists but I changed one field.
The column referenced in a foreign key must be indexed. You need to add an index on medication.med_id. In fact, this should probably be the primary key of the table.
ALTER TABLE medication ADD PRIMARY KEY (med_id);
if you are giving
ADD CONSTRAINT `fk_med_pharmacy_medication1`
FOREIGN KEY (`med_id`)
A FOREIGN KEY constraint does not have to be linked only to a PRIMARY KEY constraint in another table; it can also be defined to reference the columns of a UNIQUE constraint in another table.
so med_id should have primary key in medication or reference the columns of a UNIQUE constraint
Related
I am creating 2 tables in my database:
DROP TABLE IF EXISTS `med_pharmacy`;
CREATE TABLE IF NOT EXISTS `med_pharmacy` (
`med_pharmacy_id` int(11) NOT NULL AUTO_INCREMENT,
`med_id` int(11) NOT NULL,
`med_barcode` varchar(45) DEFAULT NULL,
`med_received` date DEFAULT NULL,
`med_expiry` date DEFAULT NULL,
`med_tablet` int(11) DEFAULT NULL,
`med_pill` int(11) DEFAULT NULL,
`clinic_id` varchar(45) DEFAULT NULL,
PRIMARY KEY (`med_pharmacy_id`),
KEY `fk_med_pharmacy_medication1_idx` (`med_id`)
) ENGINE=InnoDB AUTO_INCREMENT=1261 DEFAULT CHARSET=utf8mb4;
AND:
DROP TABLE IF EXISTS `medication`;
CREATE TABLE `medication` (
`med_id` int(11) NOT NULL,
`med_name` varchar(75) NOT NULL,
`med_date_added` date DEFAULT NULL,
`clinic_id` varchar(45) DEFAULT NULL,
`med_type` varchar(15) DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8mb4;
And when I run the queries in wamp I got this error:
SQL query:
ALTER TABLE `med_pharmacy`
ADD CONSTRAINT `fk_med_pharmacy_medication1`
FOREIGN KEY (`med_id`)
REFERENCES
`medication` (`med_id`) ON DELETE CASCADE ON UPDATE CASCADE MySQL
said: Documentation
#1822 - Failed to add the foreign key constaint. Missing index for constraint 'fk_med_pharmacy_medication1' in the referenced table
'medication'
The tables already exists but I changed one field.
The column referenced in a foreign key must be indexed. You need to add an index on medication.med_id. In fact, this should probably be the primary key of the table.
ALTER TABLE medication ADD PRIMARY KEY (med_id);
if you are giving
ADD CONSTRAINT `fk_med_pharmacy_medication1`
FOREIGN KEY (`med_id`)
A FOREIGN KEY constraint does not have to be linked only to a PRIMARY KEY constraint in another table; it can also be defined to reference the columns of a UNIQUE constraint in another table.
so med_id should have primary key in medication or reference the columns of a UNIQUE constraint
CREATE TABLE `class` (
`class_id` int(11) NOT NULL AUTO_INCREMENT,
`section_name` varchar(50) NOT NULL,
`class_alias` varchar(200) NOT NULL,
`grading_scheme` int(11) NOT NULL DEFAULT '0',
`year` year(4) NOT NULL,
`grade_calc_method_id` varchar(20) DEFAULT NULL,
PRIMARY KEY (`class_id`)
) ENGINE=InnoDB AUTO_INCREMENT=48819 DEFAULT CHARSET=latin1;
CREATE TABLE `teachers` (
`teacher_id` int(11) NOT NULL AUTO_INCREMENT,
`user_id` int(11) NOT NULL,
`teacher_subject` varchar(20) NOT NULL DEFAULT 'None',
PRIMARY KEY (`teacher_id`),
KEY `user_id` (`user_id`,`school_id`)
) ENGINE=InnoDB AUTO_INCREMENT=48606 DEFAULT CHARSET=latin1;
CREATE TABLE `teacher_classes` (
`teacher_class_id` int(11) NOT NULL AUTO_INCREMENT,
`teacher_id` int(11) NOT NULL,
`class_id` int(11) NOT NULL,
PRIMARY KEY (`teacher_class_id`),
UNIQUE KEY `teacher_id_class_id` (`teacher_id`,`class_id`),
KEY `teacher_id` (`teacher_id`,`class_id`)
) ENGINE=InnoDB AUTO_INCREMENT=46707 DEFAULT CHARSET=latin1;
Trying to insure data consistency between the tables by using foreign key so that the DBMS can check for errors.I have another junction table teacher_classes
Here is my query to add foreign keys constraint
ALTER TABLE teacher_classes
ADD CONSTRAINT `tc_fk_class_id` FOREIGN KEY (`class_id`)
REFERENCES class (`class_id`) ON UPDATE NO ACTION ON DELETE NO ACTION,
ADD CONSTRAINT `tc_fk_teacher_id` FOREIGN KEY (`teacher_id`)
REFERENCES teachers (`teacher_id`) ON UPDATE NO ACTION ON DELETE NO ACTION;
've seen the other posts on this topic, but no luck, getting following error.
Cannot add or update a child row: a foreign key constraint fails
(DB_NAME.#sql-403_12, CONSTRAINT
tc_fk_teacher_id FOREIGN KEY (teacher_id) REFERENCES teachers
(teacher_id) ON DELETE NO ACTION ON UPDATE NO ACTION)
Too late to Answer. I just had the same problem the solution is easy.
You're getting this error because you're trying to or UPDATE a row to teacher_classes doesn't match the id in table teachers.
A simple solution is disable foreign key checks before performing any operation on the table.
SET FOREIGN_KEY_CHECKS = 0;
After you are done with the table enable it again.
SET FOREIGN_KEY_CHECKS = 1;
Or you can remove not null constraint and insert a NULL value in it.
That's most probably the column definition doesn't match properly. For table teachers the PK column definition is as below.
`teacher_id` int(11) NOT NULL AUTO_INCREMENT
Make sure you have the same definition in your child table teacher_classes
Does anybody know why I have the error
Error Code: 1215. Cannot add foreign key constraint
When I try
ALTER TABLE hermanos ADD CONSTRAINT fk_hno_provincia FOREIGN KEY (provincia) REFERENCES p_provincias (id)
On these tables:
CREATE TABLE IF NOT EXISTS `hermanos` (
`codigo` int(11) NOT NULL AUTO_INCREMENT,
`nombre` varchar(255) NOT NULL,
`apellidos` varchar(255) NOT NULL,
`direccion` varchar(255) NOT NULL,
`codigoPostal` int(11) NOT NULL,
`provincia` int(11) NOT NULL,
`numeroHermano` int(11) NOT NULL,
`dni` varchar(9) NOT NULL,
`tipoCuota` int(11) NOT NULL,
`sexo` int(11) NOT NULL,
PRIMARY KEY (`codigo`),
KEY `sexo` (`sexo`),
KEY `pk_hno_cuota` (`tipoCuota`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;
CREATE TABLE IF NOT EXISTS `p_provincias` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`nombre` varchar(125) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=53 ;
Thanks
The parent and child tables must use the same storage engine.
I noticed that hermanos uses InnoDB and p_provincias uses MyISAM.
For more info, see here
Foreign keys definitions are subject to the following conditions:
Foreign key relationships involve a parent table that holds the
central data values, and a child table with identical values pointing
back to its parent. The FOREIGN KEY clause is specified in the child
table. The parent and child tables must use the same storage engine.
They must not be TEMPORARY tables.
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.
Chane the statement:
FOREIGN KEY (`alart_id`) REFERENCES alart_types(`alart_id`)
to
FOREIGN KEY (`alart_id`) REFERENCES alarts(`alart_id`)
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
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.
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`)
I have two tables I have created and I'm adding the foreign key constraint after the fact.
The two tables are defined as such:
CREATE TABLE `user` (
`user_id` int(11) NOT NULL auto_increment,
`user_ad_id` varchar(500) default NULL,
`user_name` varchar(100) NOT NULL,
`login_id` varchar(100) default NULL,
`email` varchar(256) NOT NULL,
`personal_config` int(10) NOT NULL,
PRIMARY KEY (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
and
CREATE TABLE IF NOT EXISTS personal_config (
config_id INT(10) NOT NULL AUTO_INCREMENT,
last_updated TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
configuration TEXT(25600) NOT NULL,
PRIMARY KEY (config_id)
)ENGINE=InnoDB DEFAULT CHARSET=utf8;
ALTER TABLE personal_config ADD CONSTRAINT personal_config_fk_user FOREIGN KEY
(config_id) REFERENCES user(personal_config);
And I keep getting the same error but can't figure it out. I've searched all the related threads to this.
Your FK config_id can't be an autoincrement field, that doesn't make much sense right? That field reflects a value in the foreign table, it cannot be set arbitrarily in the local table.
I think this is what you want:
ALTER TABLE user ADD CONSTRAINT personal_config_fk_user FOREIGN KEY (personal_config) REFERENCES personal_config(config_id);
Your ALTER TABLE statement is backward. Since personal_config.config_id is an auto_increment primary key, the foreign key should be defined in the users table against personal_config, not in personal_config against the users table.
ALTER TABLE users ADD CONSTRAINT user_fk_personal_config
FOREIGN KEY (personal_config)
REFERENCES personal_config(config_id);
if you set your user table field personal_config is primary key then it is possible to execute
CREATE TABLE IF NOT EXISTS personal_config (
config_id INT(10) NOT NULL AUTO_INCREMENT,
last_updated TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
configuration TEXT(25600) NOT NULL,
PRIMARY KEY (config_id), FOREIGN KEY
(config_id) REFERENCES user(personal_config)
)ENGINE=InnoDB DEFAULT CHARSET=utf8;