Hi i'm trying to create a simple MySQL database. I write the following script
CREATE TABLE office (
num_office varchar(10) NOT NULL,
name varchar(45),
address varchar(45),
numder varchar(4),
city varchar(45)
);
CREATE TABLE office_phone(
n_office varchar(10) NOT NULL,
phone varchar(15) NOT NULL,
PRIMARY KEY(n_office,phone),
FOREIGN KEY (n_office) REFERENCES office(num_office)
ON DELETE RESTRICT
ON UPDATE CASCADE
);
but Error Code 1822 is appeared.
How can i fix it?
thank you in andvance
Foreign keys can reference only columns that are primary keys or at least have an index on them (thanks #Uueerdo). For example:
ALTER TABLE OFFICE ADD CONSTRAINT offnum_unq UNIQUE (NUM_OFFICE);
This forces the referenced column to have unique values and also creates an index on it. Or you could also do:
CREATE INDEX offnum_idx ON OFFICE(NUM_OFFICE);
Now your second CREATE should work.
Reference column (Column you referenced as foreign key) should have the index added to it. You can add Index to the column "num_office" in "office" table and try creating the "office_phone" table. Refer the following queries
Create Table "office" with the index added to "num_office"
CREATE TABLE office (
num_office varchar(10) NOT NULL,
name varchar(45),
address varchar(45),
numder varchar(4),
city varchar(45),
INDEX num_office_idx (num_office)
);
OR you can add index after creating the "office" table using
ALTER TABLE office Add Index num_office_indx (num_office);
This will let you to create the table "office_phone" successfully with the reference key pointing to "num_office" column of "office" table.
CREATE TABLE office_phone(
n_office varchar(10) NOT NULL,
phone varchar(15) NOT NULL,
PRIMARY KEY(n_office,phone),
FOREIGN KEY (n_office) REFERENCES office(num_office)
ON DELETE RESTRICT
ON UPDATE CASCADE
);
Refer this page for more details
Related
I am trying to create tables with a primary key and foreign key in Postgresql. Originally was created in MySQL but I do not know how to convert can someone assist.
I converted the first table but the instructor table is giving me problems
CREATE TABLE instructor_detail (
id SERIAL PRIMARY KEY,
youtube_channel VARCHAR(128) NULL,
hobby VARCHAR(45) NULL
);
DROP TABLE IF EXISTS instructor;
`CREATE TABLE instructor (
id SERIAL PRIMARY KEY,
first_name VARCHAR(45) NULL,
last_name VARCHAR(45) NULL,
email VARCHAR(45) NULL,
instructor_detail_id INT NULL,
CREATE INDEX fk_detail_idx ON instructor (instructor_detail_id), CONSTRAINT fk_detail FOREIGN KEY (instructor_detail_id) REFERENCES instructor_detail (id) ON DELETE NO ACTION ON UPDATE NO ACTION
);`
ERROR: syntax error at or near "CREATE"
LINE 7: CREATE INDEX fk_detail_idx ON instructor (instructor_detai...
The foreign key constraint will work in PostgreSQL, no change is required.
The KEY clause is a MySQL extension to create an index within CREATE TABLE, which PostgreSQL does not support.
You'll have to convert the KEY clause to a separate CREATE INDEX statement in PostgresSQL.
I am getting error when I create table with foreign key
create table _users(_id int(20) unsigned NOT NULL AUTO_INCREMENT,
_user_fullname varchar(50)not null,
_user_username varchar(160) not null,
_user_password varchar(200) not null,_user_remember_me tinyint,
_user_email varchar(30),
_user_mobile varchar(15),
_user_age varchar(10)
,primary key(_id,_user_email,_user_mobile));
_users table created successfully..there were no error..
But When I want to create employee table :
CREATE TABLE employee ( _Id INT NOT NULL AUTO_INCREMENT,
_user_mobile VARCHAR(15) not null,
_name varchar(15),
_org varchar(10),
PRIMARY KEY (_Id),
foreign key (_user_mobile) references _users(_user_mobile));
Its showing error:
ERROR 1005 (HY000): Can't create table 'DB.employee' (errno: 150)
What am I doing wrong??
Hey In this case you just need to do one thing ,
you just need to add index to the reference column of the user table and then run the create table for employee
ALTER TABLE `_users` ADD INDEX (`_user_mobile`);
After running above query just run the below query :-
CREATE TABLE `employee`(
`_Id` INT(11) NOT NULL AUTO_INCREMENT,
`_user_mobile` VARCHAR(15) NOT NULL,
`_name` VARCHAR(15),
`_org` VARCHAR(10),
PRIMARY KEY (`_Id`),
FOREIGN KEY (`_user_mobile`) REFERENCES `_users`(`_user_mobile`) );
In this way you will get rid of the error 1005 of mysql which says that you need to have index on the reference column of parent table.
150 is a foreign key error:
C:\>perror 150
MySQL error code 150: Foreign key constraint is incorrectly formed
Getting the exact error message is very tricky. You need to run this query:
show engine innodb status
... and search in the output:
------------------------
LATEST FOREIGN KEY ERROR
------------------------
160627 14:09:32 Error in foreign key constraint of table test/employee:
foreign key (_user_mobile) references _users(_user_mobile)):
Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
Note that the internal storage type of ENUM and SET changed in
tables created with >= InnoDB-4.1.12, and such columns in old tables
cannot be referenced by such columns in new tables.
See http://dev.mysql.com/doc/refman/5.5/en/innodb-foreign-key-constraints.html
for correct foreign key definition.
Once you know that, it'd be easy to add the missing index:
ALTER TABLE `_users`
ADD UNIQUE INDEX `_user_email` (`_user_email`);
But I wouldn't if I were you. It's weird to use mobile phone number as key. Instead, just simplify the primary key:
create table _users(_id int(20) unsigned NOT NULL AUTO_INCREMENT,
_user_fullname varchar(50)not null,
_user_username varchar(160) not null,
_user_password varchar(200) not null,_user_remember_me tinyint,
_user_email varchar(30),
_user_mobile varchar(15),
_user_age varchar(10)
,primary key(_id));
... and use in the linked table:
CREATE TABLE employee ( _Id INT NOT NULL AUTO_INCREMENT,
_user_id int(20) unsigned not null,
_name varchar(15),
_org varchar(10),
PRIMARY KEY (_Id),
foreign key (_user_id) references _users(_id));
The problem is in the foreign key part. If you remove that, table will be created without a problem.
If you need to use that foreign key, you need to use InnoDB as the storage engine of MySQL. InnoDB allows a foreign key constraint to reference a non-unique key as can be seen in here.
I'm trying to make a simple SQL schema, but I'm having some problem with defining foreign keys. I really don't have that much MySQL knowledge, so I thought I'd ask her for some help. I get Error Code 1215 when I try to create the foreign key roomID and 'guestEmail' in the HotelManagement.Reservation table creation.
CREATE database HotelManagement;
CREATE TABLE HotelManagement.Room (
roomID INT not null auto_increment,
roomTaken TINYINT(1),
beds INT not null,
size INT not null,
roomRank INT not null,
PRIMARY KEY(roomID));
CREATE TABLE HotelManagement.HotelTask (
taskType INT not null,
taskStatus TINYINT(1) not null,
whichRoom INT not null,
note VARCHAR(255),
PRIMARY KEY (taskType),
FOREIGN KEY (whichRoom) REFERENCES HotelManagement.Room(roomID));
CREATE TABLE HotelManagement.Guest (
firstName varchar(25) not null,
lastName varchar(25) not null,
userPassword varchar(25) not null,
email varchar(25) not null,
reservation INT,
PRIMARY KEY (userPassword, email));
CREATE TABLE HotelManagement.Reservation (
reservationID INT not null,
id_room INT not null,
guestEmail varchar(25) not null,
fromDate DATE not null,
toDate DATE not null,
PRIMARY KEY (reservationID),
FOREIGN KEY (guestEmail)
REFERENCES HotelManagement.Guest(email),
FOREIGN KEY (id_room)
REFERENCES HotelManagement.Room(roomID)
);
ALTER TABLE HotelManagement.Guest
ADD CONSTRAINT res_constr FOREIGN KEY (reservation)
REFERENCES HotelManagement.Reservation(reservationID);
Updated the .sql
In the hoteltask table you have already defined a foreign key named roomid. Foreign key names also have to be unique, so just give a different name to the 2nd foreign key or omit the name completely:
If the CONSTRAINT symbol clause is given, the symbol value, if used,
must be unique in the database. A duplicate symbol will result in an
error similar to: ERROR 1022 (2300): Can't write; duplicate key in
table '#sql- 464_1'. If the clause is not given, or a symbol is not
included following the CONSTRAINT keyword, a name for the constraint
is created automatically.
UPDATE
The email field in the guest table is the rightmost column of the primary key, this way the pk cannot be used to independently look up email in that table. Either change the order or fields in the pk, or have a separate index on email field in the guest table. Quote from the same link as above:
MySQL requires indexes on foreign keys and referenced keys so that
foreign key checks can be fast and not require a table scan. In the
referencing table, there must be an index where the foreign key
columns are listed as the first columns in the same order. Such an
index is created on the referencing table automatically if it does not
exist. This index might be silently dropped later, if you create
another index that can be used to enforce the foreign key constraint.
index_name, if given, is used as described previously.
Pls read through the entire documentation I linked before proceeding with creating the fks!
Side note 2 (1st is in the comments below): you should probably have a unique numeric guest id because that is lot more efficient than using email. Even if you decide to stick with email as id, I would restrict the pk in the guest table to email only. With the current pk I can register with the same email multiple times if I use different password.
I'm not sure what I'm doing wrong here, but I get an error:
Error Code: 1005. Can't create table 'erm.section' (emo:150)
Here is the code. The 'course' table is created successfully. I tried modifying the name of the course_number attritube in the 'section' table but that didn't work.
USE erm;
CREATE TABLE course
(
course_name VARCHAR(30) NOT NULL,
course_number VARCHAR(20) NOT NULL,
credit_hours INT NOT NULL,
department VARCHAR(10),
CONSTRAINT course_pk PRIMARY KEY (course_name)
);
CREATE TABLE section
(
section_identifier INT NOT NULL,
course_number VARCHAR(20),
semester VARCHAR(10) NOT NULL,
school_year VARCHAR(4) NOT NULL,
instructor VARCHAR(25),
CONSTRAINT section_pk PRIMARY KEY (section_identifier),
CONSTRAINT section_fk FOREIGN KEY (course_number)
REFERENCES course (course_number)
ON DELETE SET NULL
ON UPDATE CASCADE
);
course_number is not a primary key in the table course.
the foreign key in the table section must reference a primary key form another table.
You have to create an index on the column referenced by the foreign key:
alter table course add index (course_number);
(It doesn't have to be a primary key index.) From the MySQL documentation:
InnoDB permits a foreign key to reference any index column or group of
columns. However, in the referenced table, there must be an index
where the referenced columns are listed as the first columns in the
same order.
A FOREIGN KEY in one table points to a PRIMARY KEY in another table.
AFAIK, dropping the CONSTRAINT, then rename, then add the CONSTRAINT back is the only way. Backup first!
to drop it use this
ALTER TABLE section
DROP FOREIGN KEY course_number
and to add it again use this
ALTER TABLE section
ADD FOREIGN KEY (course_number)
REFERENCES course (course_number)
I want to create a table relationship with MYSQL PHPMYADMIN.
I have this Create table:
CREATE TABLE students(code_students int(8)not null AUTO_INCREMENT,
name_students varchar(25),
age_students int(3),
degree_program varchar(25),
code_advisor int(8)not null,
primary key(code_students, code_advisor)
);
and i want to make a create table named advise relationship between code_students, code_advisor.
Ok this is my tryout.
CREATE TABLE advise (
code_students int(8),
code_advisor int(8),
primary key(code_students, code_advisor),
foreign key(code_students)references students(code_students),
foreign key(code_advisor)references students(code_advisor)
);
mySQL says :
A FOREIGN KEY constraint that references a non-UNIQUE key is not standard SQL. It is an InnoDB extension to standard SQL
Try adding the UNIQUE keyword to your first table:
CREATE TABLE students(
code_students int(8)not null unique AUTO_INCREMENT,
name_students varchar(25),
age_students int(3),
degree_program varchar(25),
code_advisor int(8)not null unique,
primary key(code_students, code_advisor)
);
Check out this sqlFiddle and see how it works, then try removing the UNIQUE keywords and see that you get the same error that you mentioned. ( Hit the Build Schema button )
http://sqlfiddle.com/#!2/46b69