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.
Related
Got an odd problem I cant solve after browsing dozens of forum posts, and my local SQL Books.
I've got two tables, and want to add a foreign key to one of them. The foreign key and primary key share the same datatype and charset and yet I cannot add the Foreign Key at all.
addon_account
name
type
comments
id
int(11)
Primary Key
name
varchar(60)
Primary Key
label
varchar(255)
shared
int(11)
addon_account_data
name
type
comments
id
int(11)
Primary Key
account_name
varchar(60)
Primary Key
money
double
owner
varchar()
The query I ran:
ALTER TABLE `addon_account_data` ADD FOREIGN KEY (`account_name`) REFERENCES `addon_account`(`name`) ON DELETE RESTRICT ON UPDATE RESTRICT;
Can't get it to work. Tosses out the same issue the entire time.
You are creating a foreign key on addon_account_data(account_name) that references addon_account(name). You have a composite primary the referred table : addon_account(id, name).
This is not allowed in MySQL, as explained in the documentation:
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.
Possible solutions:
add an additional column in the referring table: addon_account_data(account_id, account_name) and create a composite primary key to the corresponding columns in addon_account
create an index on addon_account(name) (probably the simplest solution)
change the order of the columns in the primary key of the referred table, like: addon_account(name, id) (you might want to first consider the impacts this may have in terms of performance)
I am not exactly a MySQL guy, but:
I believe the problem is that you are referencing only part of the primary key:
Your table addon_account has a composite key PK(id, name).
So, to put your relationship to work, you will need to add 'account_id' as part of the foreign key as well:
ALTER TABLE addon_account_data ADD FOREIGN KEY (account_id, account_name) REFERENCES addon_account(id, name)
This thread deals with something similar.
I hope this helps.
EDITED
I have installed a MySQL server instance on my local machine... (MySQL 8).
I have run the script below, and it worked (giving warnings about integer display being a deprecated feature, so I would recommend ommitting it):
CREATE TABLE addon_account(
id INT(11) NOT NULL,
`name` VARCHAR(60) NOT NULL,
label VARCHAR(255),
shared INT(11),
CONSTRAINT pk_addon_account PRIMARY KEY(id, `name`));
CREATE TABLE addon_account_data (
id INT(11) NOT NULL,
account_name VARCHAR(60) NOT NULL,
account_id INT(11),
money DOUBLE,
`owner` VARCHAR(255),
CONSTRAINT pk_addon_account_data PRIMARY KEY(id, account_name),
CONSTRAINT fk_addon_account_account_data FOREIGN KEY(account_id, account_name)
REFERENCES addon_account(id, `name`));
Could you try it and see if this works for you?
I am not that familiar with MySQL.
make sure that the 2 tables have the same collation
like
COLLATE='utf8_general_ci'
I'm having trouble adding a foreign key field that references another table.
First I created the users table as so:
CREATE TABLE users (
user_id INT(11) NOT NULL AUTO_INCREMENT PRIMARY KEY,
userName VARCHAR(256) NOT NULL,
userEmail VARCHAR (256) NOT NULL,
userPwd VARCHAR(256) NOT NULL,
);
then I'd like the quizzes table to have a foreign key that references the user_id from the first table
CREATE TABLE quizzes (
quizId INT(11) NOT NULL AUTO_INCREMENT PRIMARY KEY,
quizName VARCHAR(128) NOT NULL,
quizMax SMALLINT(6) NOT NULL,
FOREIGN KEY (user_id) REFERENCES users (user_id)
);
This is throwing the error: 'Key column 'user_id' doesn't exist in table.
Other answers advised to check that DB is InnoDB, which I did, and it is.
Can't understand why it's telling me that user_id doesn't exist, when it clearly does exist in the users table.
Firstly check if table user is created successfully, due to the additional ',' on last column!
Secondly, the column you reffered in FOREIGN KEY(user_id) is not defined in table quizzes, you need to add this column in quizzes table.
First: You do not need the last comma - , in the first CREATE statement. -
Second: you have to create the columns before you can use them in a foreign key constraint and user_id does not exist in the second table at the moment of constraint creation.
Take a look at the example below. The last create succeeds when user_id column is added before the constraint is created:
I am unable to create these tables in MySQL. Everything looks complete to me but I keep getting errors.
Here is my SQL:
CREATE TABLE Movies(
title char PRIMARY KEY NOT NULL,
Years date NOT NULL,
length decimal not null,
genre char NOT NULL,
academy_award char not null,
FOREIGN KEY(the_name) REFERENCES Studio,
FOREIGN KEY(directorName) REFERENCES Director);
CREATE TABLE StarsIn(
FOREIGN KEY(title) REFERENCES Movies,
FOREIGN KEY(Years) REFERENCES Movies,
FOREIGN KEY(the_name) REFERENCES MovieStar);
CREATE TABLE Director(
the_name char PRIMARY KEY NOT NULL,
birthdate date NOT NULL,
nationality char NOT NULL);
CREATE TABLE MovieStar(
the_name char PRIMARY KEY NOT NULL,
birthdate date NOT NULL,
address char NOT NULL,
gender char NOT NULL);
CREATE TABLE Studio(
the_name char PRIMARY KEY NOT NULL,
address char NOT NULL);
This isn't working because your tables are not being created in the proper order.
The parent table must always be created first, because otherwise you will be trying to reference a column that does not exist.
For example, look at this part here:
CREATE TABLE Movies(title char PRIMARY KEY NOT NULL, Years date NOT NULL,
length decimal not null, genre char NOT NULL, academy_award char not null,
FOREIGN KEY(the_name) REFERENCES Studio, FOREIGN KEY(directorName) REFERENCES Director);
You can't reference tables studio or director because they don't even exist.
The proper order could be:
Studio
Movie Star
Director
Movies
StarsIn
In addition, you have a lot of other problems going on here.
You do not define any columns in the Stars IN table, you only declare foreign keys:
CREATE TABLE StarsIn(
FOREIGN KEY(title) REFERENCES Movies,
FOREIGN KEY(Years) REFERENCES Movies,
FOREIGN KEY(the_name) REFERENCES MovieStar);
Not only are you not defining any columns, you aren't referencing anything. This will also throw an error. Your foreign key title must reference some column in the movies table. And in addition to that, you can't create a foreign key reference for Years. Mostly because you cannot create a foreign key to a column that is not primary key in another table, and also why should you need two foreign keys to the same table?
Those problems exist in other create table statements as well, so given everything I've said please go back and look at each of your create table statements.
Here is a working SQL Fiddle you can use for reference though.
EDIT
I would like to also add that I do not recommend using the char datatype. Please look into using VARCHAR().
The problem you are having is that you are referencing columns in tables that do not exist.
Although you create these tables later, the DBMS engine does not know they will exist.
The solution - create the tables without the FK's that do not exist, and add these later using the "ALTER TABLE" command.
It seems like you know the syntax, so I'll let you inform me if you cannot find it.
your syntax should look like (All I did is re-order, and change char to be archer...):
CREATE TABLE Studio(the_name varchar(50) PRIMARY KEY NOT NULL, address varchar(50) NOT NULL);
CREATE TABLE Director(the_name varchar(50) PRIMARY KEY NOT NULL, birthdate date NOT NULL, nationality varchar(50) NOT NULL);
CREATE TABLE Movies(title varchar(50) PRIMARY KEY NOT NULL, Years date NOT NULL,
length decimal not null, genre varchar(50) NOT NULL, academy_award varchar(50) not null,
the_name REFERENCES Studio(the_name), directorname REFERENCES Director(the_name));
CREATE TABLE MovieStar(the_name char PRIMARY KEY NOT NULL, birthdate date NOT NULL, address char NOT NULL, gender char NOT NULL);
CREATE TABLE StarsIn(title REFERENCES Movies(title),movie REFERENCES Movies(title), moviestar REFERENCES MovieStar(the_name));
A little note, depends on the version - you'd might have to use the FOREIGN KEY(col_name) instead of col_name. I like more the syntax without it, but some versions force you to use FOREIGN KEY(title) instead of title in the last SQL for example.
The syntax you used - Foreign key NAME References TABLE (Column) -- you forgot the column-- Allows you to name the FK. If you don't use the "Foreign Key" the system will randomly assign a name. This is usually not important, unless you want the DB design to be "clean" and to be able to reference it by name.
P.S. - I did not run it, I trust you to inform if there's any other issue - I did not check the syntax, just fixed the error you reported- references that do not exist...
P.S. P.S. Always check syntax... http://dev.mysql.com/doc/refman/5.1/en/create-table.html
i got these two succesfull queries:
create table Donors (
donor_id int not null auto_increment primary key,
gender varchar(1) not null,
date_of_birth date not null,
first_name varchar(20) not null,
middle_name varchar(20),
last_name varchar(30) not null,
home_phone tinyint(10),
work_phone tinyint(10),
cell_mobile_phone tinyint(10),
medical_condition text,
other_details text );
and
create table Donors_Medical_Condition (
donor_id int not null,
condition_code int not null,
seriousness text,
primary key(donor_id, condition_code),
foreign key(donor_id) references Donors(donor_id) );
but when i try this one:
create table Medical_Conditions (
condition_code int not null,
condition_name varchar(50) not null,
condition_description text,
other_details text,
primary key(condition_code),
foreign key(condition_code) references Donors_Medical_Condition(condition_code) );
i get "Error Code: 1215, cannot add foreign key constraint"
i dont know what am i doing wrong.
In MySql, a foreign key reference needs to reference to an index (including primary key), where the first part of the index matches the foreign key field. If you create an an index on condition_code or change the primary key st that condition_code is first you should be able to create the index.
To define a foreign key, the referenced parent field must have an index defined on it.
As per documentation on foreign key constraints:
REFERENCES tbl_name (index_col_name,...)
Define an INDEX on condition_code in parent table Donors_Medical_Condition and it should be working.
create table Donors_Medical_Condition (
donor_id int not null,
condition_code int not null,
seriousness text,
KEY ( condition_code ), -- <---- this is newly added index key
primary key(donor_id, condition_code),
foreign key(donor_id) references Donors(donor_id) );
But it seems you defined your tables order and references wrongly.
You should have defined foreign key in Donors_Medical_Condition table but not in Donors_Medical_Conditions table. The latter seems to be a parent.
Modify your script accordingly.
They should be written as:
-- create parent table first ( general practice )
create table Medical_Conditions (
condition_code int not null,
condition_name varchar(50) not null,
condition_description text,
other_details text,
primary key(condition_code)
);
-- child table of Medical_Conditions
create table Donors_Medical_Condition (
donor_id int not null,
condition_code int not null,
seriousness text,
primary key(donor_id, condition_code),
foreign key(donor_id) references Donors(donor_id),
foreign key(condition_code)
references Donors_Medical_Condition(condition_code)
);
Refer to:
MySQL Using FOREIGN KEY Constraints
[CONSTRAINT [symbol]] FOREIGN KEY
[index_name] (index_col_name, ...)
REFERENCES tbl_name (index_col_name,...)
[ON DELETE reference_option]
[ON UPDATE reference_option]
reference_option:
RESTRICT | CASCADE | SET NULL | NO ACTION
A workaround for those who need a quick how-to:
FYI: My issue was NOT caused by the inconsistency of the columns’ data types/sizes, collation or InnoDB storage engine.
How to:
Download a MySQL workbench and use it’s GUI to add foreign key. That’s it!
Why:
The error DOES have something to do with indexes. I learned this from the DML script automatically generated by the MySQL workbench. Which also helped me to rule out all those inconsistency possibilities.It applies to one of the conditions to which the foreign key definition subject. That is: “MySQL requires indexes on foreign keys and referenced keys so that foreign key checks can be fast and not require a table scan.” Here is the official statement: http://dev.mysql.com/doc/refman/5.7/en/create-table-foreign-keys.html
I did not get the idea of adding an index ON the foreign key column(in the child table), only paid attention to the referenced TO column(in the parent table).
Here is the auto-generated script(PHONE.PERSON_ID did not have index originally):
ALTER TABLE `netctoss`.`phone`
ADD INDEX `personfk_idx` (`PERSON_ID` ASC);
ALTER TABLE `netctoss`.`phone`
ADD CONSTRAINT `personfk`
FOREIGN KEY (`PERSON_ID`)
REFERENCES `netctoss`.`person` (`ID`)
ON DELETE NO ACTION
ON UPDATE NO ACTION;
I think you've got your tables a bit backwards. I'm assuming that Donors_Medical_Condtion links donors and medical conditions, so you want a foreign key for donors and conditions on that table.
UPDATED
Ok, you're also creating your tables in the wrong order. Here's the entire script:
create table Donors (
donor_id int not null auto_increment primary key,
gender varchar(1) not null,
date_of_birth date not null,
first_name varchar(20) not null,
middle_name varchar(20),
last_name varchar(30) not null,
home_phone tinyint(10),
work_phone tinyint(10),
cell_mobile_phone tinyint(10),
medical_condition text,
other_details text );
create table Medical_Conditions (
condition_code int not null,
condition_name varchar(50) not null,
condition_description text,
other_details text,
primary key(condition_code) );
create table Donors_Medical_Condition (
donor_id int not null,
condition_code int not null,
seriousness text,
primary key(donor_id, condition_code),
foreign key(donor_id) references Donors(donor_id),
foreign key(condition_code) references Medical_Conditions(condition_code) );
I got the same issue and as per given answers, I verified all datatype and reference but every time I recreate my tables I get this error. After spending couple of hours I came to know below command which gave me inside of error-
SHOW ENGINE INNODB STATUS;
LATEST FOREIGN KEY ERROR
------------------------
2015-05-16 00:55:24 12af3b000 Error in foreign key constraint of table letmecall/lmc_service_result_ext:
there is no index in referenced table which would contain
the columns as the first columns, or the data types in the
referenced table do not match the ones in table. Constraint:
,
CONSTRAINT "fk_SERVICE_RESULT_EXT_LMC_SERVICE_RESULT1" FOREIGN KEY ("FK_SERVICE_RESULT") REFERENCES "LMC_SERVICE_RESULT" ("SERVICE_RESULT") ON DELETE NO ACTION ON UPDATE NO ACTION
I removed all relation using mysql workbench but still I see same error. After spending few more minutes, I execute below statement to see all constraint available in DB-
select * from information_schema.table_constraints where
constraint_schema = 'XXXXX'
I was wondering that I have removed all relationship using mysql workbench but still that constraint was there. And the reason was that because this constraint was already created in db.
Since it was my test DB So I dropped DB and when I recreate all table along with this table then it worked. So solution was that this constraint must be deleted from DB before creating new tables.
Check that both fields are the same size and if the referenced field is unsigned then the referencing field should also be unsigned.
I am not sure about this , but do I need to create foreign key explicitly in the SQL command?
This guy did this:
CREATE TABLE languages (
lang_id TINYINT UNSIGNED NOT NULL AUTO_INCREMENT,
lang VARCHAR(60) NOT NULL,
lang_eng VARCHAR(20) NOT NULL,
PRIMARY KEY (lang_id),
UNIQUE (lang)
);
CREATE TABLE threads (
thread_id INT UNSIGNED NOT NULL AUTO_INCREMENT,
lang_id TINYINT(3) UNSIGNED NOT NULL,
user_id INT UNSIGNED NOT NULL,
subject VARCHAR(150) NOT NULL,
PRIMARY KEY (thread_id),
INDEX (lang_id),
INDEX (user_id)
);
In this case, does it mean that INDEX(lang_id) becomes FOREIGN KEY automatically? I know INDEX makes search go faster, but I don't understand the part about foreign key
I would really appreciate any answer
No. An index is just that... an index on a field. A foreign key tells MySQL that "this particular field MUST have a matching record in that table over there".
MySQL's internal design requires that all fields used as foreign keys be indexed, but modern versions will automatically create that index for you.
The converse is not true, whoever. Adding an index to a field does not turn it into a foreign key - a foreign key definition must also include what the foreign table/field is, and a simple index declaration has none of that information.
For your sample table, you'd need to have
...
INDEX (lang_id),
FOREIGN KEY (lang_id) REFERENCES languages (lang_id),
...
to produce a foreign key.
A foreign key means that the value(s) must exist in the referenced column(s). It is not automatic - you need to write it explicitly.
FOREIGN KEY lang_id REFERENCES languages (lang_id)
No a foreign key has to be explicitly declared
CREATE TABLE threads (
thread_id INT UNSIGNED NOT NULL AUTO_INCREMENT,
lang_id TINYINT(3) UNSIGNED NOT NULL FOREIGN KEY FK_1 REFERENCES languages(lang_id),
user_id INT UNSIGNED NOT NULL,
subject VARCHAR(150) NOT NULL,
PRIMARY KEY (thread_id),
INDEX (lang_id),
INDEX (user_id)
);
What you now have are two tables with primary keys and indexes on those primary key values.
You could stop here if you want but you won't have declared referential integrity enforcing that relationship between laguage and threads tables.
To do that you would explicitly create a foreign key relationship as explained here - http://dev.mysql.com/doc/refman/5.5/en/innodb-foreign-key-constraints.html
The differences are as follows :-
The primary key identifies a record uniquely in a table with multiple rows.
An index is a generic term, where by you can create more than one index for a table, in this case the database creates indexes based on the columns that you specified, so that when you query the appropriate index will kick in and give you results faster.
A foreign key on the other hand says that this column in table b, is the primary column in table A, so that whenever you enter rows into table B the databse will check that the specified column/data exists in table A otherwise it will throw an error.