I was trying to feed the following commands to MySQL CLI with
, which seems to be good, however when I added one more table, it complained there was an error in syntax.
Here are the commands
CREATE TABLE IF NOT EXISTS User(
uid INT,
name VARCHAR(64) UNIQUE,
birthday date,
PRIMARY KEY(uid)
) ENGINE = InnoDB ;
CREATE TABLE IF NOT EXISTS UserEmail(
uid INT,
email VARCHAR(64),
PRIMARY KEY(uid, email),
FOREIGN KEY(uid) REFERENCES User(uid)
);
However if I wanted to add one more table, it said there's an syntax error near the ')' at the line where PRIMARY KEY(uid) lies.
CREATE TABLE IF NOT EXISTS User(
uid INT,
name VARCHAR(64) UNIQUE,
birthday date,
PRIMARY KEY(uid)
) ENGINE = InnoDB ;
CREATE TABLE IF NOT EXISTS UserEmail(
uid INT,
email VARCHAR(64),
PRIMARY KEY(uid, email),
FOREIGN KEY(uid) REFERENCES User(uid)
);
CREATE TABLE friendship(
invite_uid INT,
accept_uid INT,
start_date DATE,
PRIMARY KEY(invite_uid, accept_uid),
FOREIGN KEY(invite_uid) REFERENCES User(uid),
FOREIGN KEY(accept_uid) REFERENCES User(uid),
);
Not sure where went wrong since the error was not complained in the newly added command.
--UPDATE--
Well that was fixed, but there's one more problem.
Adding the following table throws
cannot create table "estore.contains" (errorno: 150)
CREATE TABLE contains(
uid INT,
wid INT,
pid INT,
PRIMARY KEY(uid, wid, pid),
FOREIGN KEY(uid) REFERENCES User(uid),
FOREIGN KEY(wid) REFERENCES Wishlist(wid),
FOREIGN KEY(pid) REFERENCES Product(pid)
);
--UPDATE 2--
Full Tables that I want to add
CREATE TABLE IF NOT EXISTS User(
uid INT,
name VARCHAR(64) UNIQUE,
birthday date,
PRIMARY KEY(uid)
) ENGINE = InnoDB ;
CREATE TABLE IF NOT EXISTS UserEmail(
uid INT,
email VARCHAR(64),
PRIMARY KEY(uid, email),
FOREIGN KEY(uid) REFERENCES User(uid)
);
CREATE TABLE friendship(
invite_uid INT,
accept_uid INT,
start_date DATE,
PRIMARY KEY(invite_uid, accept_uid),
FOREIGN KEY(invite_uid) REFERENCES User(uid),
FOREIGN KEY(accept_uid) REFERENCES User(uid)
);
CREATE TABLE Seller(
sid INT,
name VARCHAR(64),
PRIMARY KEY(sid)
);
CREATE TABLE Product(
pid INT,
sid INT,
name VARCHAR(64),
description TEXT,
price DOUBLE,
PRIMARY KEY(pid),
FOREIGN KEY(sid) REFERENCES Seller(sid)
);
CREATE TABLE buy(
uid INT,
pid INT,
time DATE,
PRIMARY KEY(uid, pid),
FOREIGN KEY(uid) REFERENCES User(uid),
FOREIGN KEY(pid) REFERENCES Product(pid)
);
CREATE TABLE Wishlist(
uid INT,
wid INT,
start_time DATE,
end_time DATE,
PRIMARY KEY(uid,wid),
FOREIGN KEY(uid) REFERENCES User(uid)
);
CREATE TABLE conntains(
uid INT,
wid INT,
pid INT,
PRIMARY KEY(uid, wid, pid),
FOREIGN KEY(uid) REFERENCES User(uid),
FOREIGN KEY(wid) REFERENCES Wishlist(wid),
FOREIGN KEY(pid) REFERENCES Product(pid)
)ENGINE = InnoDB;
Anyone could help? Thx
Check the last line of your code
FOREIGN KEY(accept_uid) REFERENCES User(uid),
Remove the comma at the end.
remove , a end of line
FOREIGN KEY(accept_uid) REFERENCES User(uid),
--------------------------------------------^
There is an extra comma on the third table.
CREATE TABLE IF NOT EXISTS User(
uid INT,
name VARCHAR(64) UNIQUE,
birthday date,
PRIMARY KEY(uid)
) ENGINE = InnoDB ;
CREATE TABLE IF NOT EXISTS UserEmail(
uid INT,
email VARCHAR(64),
PRIMARY KEY(uid, email),
FOREIGN KEY(uid) REFERENCES User(uid)
);
CREATE TABLE friendship(
invite_uid INT,
accept_uid INT,
start_date DATE,
PRIMARY KEY(invite_uid, accept_uid),
FOREIGN KEY(invite_uid) REFERENCES User(uid),
FOREIGN KEY(accept_uid) REFERENCES User(uid), <---- Extra comma
);
FOREIGN KEY(accept_uid) REFERENCES User(uid),<-- Comma should be removed
Try this:
CREATE TABLE friendship(
invite_uid INT,
accept_uid INT,
start_date DATE,
PRIMARY KEY(invite_uid, accept_uid),
FOREIGN KEY(invite_uid) REFERENCES User(uid),
FOREIGN KEY(accept_uid) REFERENCES User(uid)
);
#Daniel, you cannot use the name CONTAINS for a table since its an SQL keyword. Please create the table with some other name.
Fortunately I figured the second problem out, the foreign key was not made right, I need to use
FOREIGN KEY(uid, wid) REFERENCES Wishlist(uid, wid),
to refer the the primary key combination in Wishlist
Related
I am having a hard time understanding why my code is giving me cannot add foreign key constraint error. I just started coding in MySQL so I am sorry for anything dumb.
drop table if exists COURSE_INSTRUCTORJG;
drop table if exists CLASSJG;
drop table if exists COURSEJG;
drop table if exists INSTRUCTORJG;
create table INSTRUCTORJG(
InstructorID int,
DepartHead int,
primary key(InstructorID)
);
create table COURSEJG(
CourseNo char(10),
ComputerNeeded smallint,
primary key(CourseNo)
);
create table CLASSJG(
Building CHAR(20),
RoomNum smallint,
Maxcap int,
primary key(Building, RoomNum)
);
create table COURSE_INSTRUCTORJG (
COURSEJG_CourseNo CHAR(10),
INSTRUCTORJG_InstructorID INT,
Semester CHAR(20),
Section INT,
CLASSJG_Building char(20),
CLASSJG_RoomNum smallint,
primary key(COURSEJG_CourseNo, INSTRUCTORJG_InstructorID, Semester, Section),
foreign key (COURSEJG_CourseNo) references COURSEJG(CourseNo),
foreign key (INSTRUCTORJG_InstructorID) references INSTRUCTORJG(InstructorID),
foreign key (CLASSJG_Building) references CLASSJG(Building),
foreign key (CLASSJG_RoomNum) references CLASSJG(RoomNum)
);
My error is coming from the line: foreign key (CLASSJG_RoomNum) references CLASSJG(RoomNum).
You can reference primary key, keys or unique comnstrqaint, but in table CLASSJG you have a double primary key, so reference both columns or define another key
create table CLASSJG(
Building CHAR(20),
RoomNum smallint,
Maxcap int,
primary key(Building, RoomNum),
KEY(RoomNum)
);
see sample fiddle
CREATE TABLE employee (
emp_id INT,
first_name VARCHAR(40),
last_name VARCHAR(40),
birth_day DATE,
sex VARCHAR(1),
salary INT,
super_id INT,
branch_id INT,
PRIMARY KEY (emp_id)
);
CREATE TABLE branch(
branch_id INT,
branch_name VARCHAR(40),
mgr_id INT,
mgr_start_date DATE,
FOREIGN KEY(mgr_id) REFERENCES employee(emp_id) ON DELETE SET NULL
);
ALTER TABLE employee
ADD FOREIGN KEY(branch_id)
REFERENCES branch(branch_id)
ON DELETE SET NULL;
I'm not sure what I'm doing wrong. When I run the lasts query, I keep getting the
"failed to add the foreign key constraint"
error saying there's a missing index for constraint 'employee_ibfk_2' in the reference table 'branch'
Columns referenced in a foreign key must be a key.
Presumably you forgot to declare branch_id as primary key?
Try:
...
CREATE TABLE branch(
branch_id INT,
branch_name VARCHAR(40),
mgr_id INT,
mgr_start_date DATE,
PRIMARY KEY (branch_id),
FOREIGN KEY(mgr_id) REFERENCES employee(emp_id) ON DELETE SET NULL
);
...
The branch_id column in your branch table is not indexed. a requirement for foreign key constraints.
Add an index like this:
ALTER TABLE `branch`
ADD INDEX `branch_id` (`branch_id` ASC) VISIBLE;
;
I try to create a table for the company database.
As I tried to add some foreign keys in there, something goes wrong.
CREATE TABLE employee (
emp_id INT PRIMARY KEY,
first_name VARCHAR(40),
last_name VARCHAR(40),
birth_day DATE,
sex VARCHAR(1),
salary INT,
super_id INT,
branch_id INT unique
);
CREATE TABLE branch (
branch_id INT PRIMARY KEY,
branch_name VARCHAR(40),
mgr_id INT,
mgr_start_date DATE,
FOREIGN KEY(mgr_id) REFERENCES employee(emp_id) ON DELETE SET NULL
);
ALTER TABLE employee
ADD FOREIGN KEY(branch_id)
REFERENCES branch(branch_id)
ON DELETE SET NULL;
ALTER TABLE employee
ADD FOREIGN KEY(super_id)
REFERENCES employee(emp_id)
ON DELETE SET NULL
The "alter table employee" part to add foreign key (branch) failed with the following statement.
ER_FK_NO_INDEX_PARENT: Failed to add the foreign key constraint. Missing index for constraint 'employee_ibfk_10' in the referenced table 'branch'
Could this be some problem with my setting?
I am trying to create 5 tables (should not be too hard), but I am having issues assigning the foreign keys. PhpMyAdmin gives me this error:
Can't create table 'databasexx.gave' (errno: 150)
Basically all the tables with no foreign keys are created.
DROP TABLE IF EXISTS oppbygging;
CREATE TABLE oppbygging (
gnr INT,
dnr INT,
ant INT,
CONSTRAINT dnr_grn_pk PRIMARY KEY (gnr, dnr)
) ENGINE=InnoDB;
DROP TABLE IF EXISTS onske;
CREATE TABLE onske (
onr INT,
pnr INT,
gnr INT,
prioriet INT,
ferdig INT,
CONSTRAINT pnr_gnr_pk PRIMARY KEY (pnr, gnr)
) ENGINE=InnoDB;
DROP TABLE IF EXISTS person;
CREATE TABLE person (
pnr INT,
fornavn VARCHAR(64),
etternavn VARCHAR(64),
fdato DATE,
CONSTRAINT pnr_pk PRIMARY KEY (pnr),
CONSTRAINT person_pnr_fk FOREIGN KEY (pnr) REFERENCES onske(pnr)
) ENGINE=InnoDB;
DROP TABLE IF EXISTS gave;
CREATE TABLE gave
(
gnr int,
navn varchar (255) UNIQUE,
prod_tid int NOT NULL,
CONSTRAINT gnr_pk PRIMARY KEY (gnr),
CONSTRAINT gave_gnr_fk FOREIGN KEY (gnr) REFERENCES oppbygging(gnr),
CONSTRAINT gave_gnr_fk FOREIGN KEY (gnr) REFERENCES onske(gnr)
) ENGINE=InnoDB;
DROP TABLE IF EXISTS del;
CREATE TABLE del (
dnr INT,
navn VARCHAR(64),
lager_ant INT NOT NULL,
CONSTRAINT dnr_pk PRIMARY KEY (dnr),
CONSTRAINT del_dnr_fk FOREIGN KEY (dnr) REFERENCES oppbygging(dnr)
) ENGINE=InnoDB;
I am sure that I am making some sort of obvious mistake, but I just cannot figure it out. Any help would be very much appreciated.
Here is the relationship view, ignore the Eiendom table:
TO CREATE ANOTHER TABLE WITH A FOREIGN KEY, THE FOREIGN KEY
SPECIFIED MUST BE THE PRIMARY KEY OF AT LEAST ONE TABLE
THAT HAS ALREADY BEEN CREATED,E.G.
CREATE TABLE 1 WITH THE PRIMARY KEY:
CREATE TABLE PERSONS(
PERSON_ID INT(10),
FNAME VARCHAR(20)NOT NULL,
LNAME VARCHAR(20),
DOB DATETIME,
AMOUNTPAID DECIMAL(15,2),
PRIMARY KEY(PERSON_ID)
);
THEN CREATE TABLE 2 WITH THE FOREIGN KEY:
CREATE TABLE COMMENTS(
COMMENTID INT AUTO_INCREMENT,
COMMENTS VARCHAR(500),
PERSON_ID INT(10),
PRIMARY KEY(COMMENTID),
FOREIGN KEY (PERSON_ID) REFERENCES PERSONS(PERSON_ID)
);
Sql wont let me create these table because of a syntax error can someone help me.
drop table users;
drop table intrest;
drop table friendships;
create table users(
id INT,
Fname char(15),
Lname char(15),
email char(20),
street char(15),
state char(2),
zip INT,
age INT,
gender char (2),
phone INT,
User_password char(15),
primary key (id),
foreign key (intrest_id) references intrest(id)
);
create table Intrests(
id INT,
description char(30),
Primary key (id),
foreign key (users_id) references users(id)
);
create table User_intrest(
foreign key (users_id) references users(id),
foreign key (intrest_id) references intrest(id)
);
create table friendships(
User_1_id INT,
User_2_id INT,
description Char(20),
foreign key (users_id) references users(id)
);
create table Intrests( id INT, description char(30),
Primary key (id),
foreign key (users_id) references users(id) );
create table User_intrest( foreign key (users_id) references users(id),
foreign key (intrest_id) references intrest(id) );
For interests table, where is the column users_id defined?
Column definitions for user_interest table?
You have a cyclical FK relationship:
Users has Interests(id) as a FK, and Interests has user(id) as a FK.
You don't need EITHER of these since you have a user_intrest table to link them!
You also have several typos with table names, like intrests and intrest.
Also, you should make both your User fields in Friendships be FK to Users, I'm not sure why you want a third, unrelated Users_Id field.
The first (of many) syntax errors you have is in
create table users(
...
primary key (id),
foreign key (intrest_id) references intrest(id) <--- there is no table intrest
);
I would suggest:
But STOP! Don't just copy-paste. Try yourself to see why MySQL gives you that error message: Key column 'intrest_id' doesn't exist in table, isn't the error you get?
Try to first fix that error. What do you have to do? Add an intrest_id field in table users, not just declare it as FOREIGN KEY. (It's not a useful field to have in the end, but do it anyway).
Then rerun your query and try to fix next error. One by one. If you realy get stuck somewhere, well, you know a site to ask questions :)
So, try to fix errors one by one, until you have a script that is running without any errors at all. You'll have learned much more than simply copying and pasting any answer.
CREATE TABLE Users(
id INT,
Fname char(15),
Lname char(15),
email char(50), <-- 20 is too small for emails
street char(15),
state char(2),
zip INT,
age INT,
gender char (2),
phone char(20), <-- phone should be char, not INT
user_password char(15),
PRIMARY KEY (id)
); <-- You don't really need a foreign key to Interests,
<-- that's why you have the User_interest "middle" table
CREATE TABLE Interests(
id INT,
description char(30),
PRIMARY KEY (id)
); <-- You don't really need a foreign key to Users,
<-- for the same reasons.
CREATE TABLE User_interest(
user_id INT, <-- These foreign keys are
interest_id INT, <-- good, but you need to
<-- declare them as fields, too
PRIMARY KEY (user_id, interest_id), <-- It's good if all tables have PK
FOREIGN KEY (user_id) REFERENCES Users(id),
FOREIGN KEY (interest_id) REFERENCES Interests(id)
);
CREATE TABLE Friendships(
user_1_id INT,
user_2_id INT,
description char(20),
PRIMARY KEY (user_1_id, user_2_id),
FOREIGN KEY (user_1_id) REFERENCES Users(id),
FOREIGN KEY (user_2_id) REFERENCES Users(id)
);