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.
Related
Im doing an assignment in school which, that includes creating a local database for a booking system. I created the tables like this :
SET foreign_key_checks = 0;
DROP TABLE IF EXISTS Users;
CREATE TABLE Users(
username VARCHAR(20) NOT NULL UNIQUE,
name VARCHAR(20) NOT NULL,
phoneNbr VARCHAR(20) NOT NULL,
address VARCHAR(255),
PRIMARY KEY (username)
);
DROP TABLE IF EXISTS Theatres;
CREATE TABLE Theatres(
theatreName VARCHAR(20) NOT NULL UNIQUE,
seats INT NOT NULL,
PRIMARY KEY (theatreName)
);
DROP TABLE IF EXISTS Movies;
CREATE TABLE Movies(
movieTitle VARCHAR(255) NOT NULL,
PRIMARY KEY (movieTitle)
);
DROP TABLE IF EXISTS Bookings;
CREATE TABLE Bookings(
bookingNbr INT NOT NULL AUTO_INCREMENT,
userName VARCHAR(20) NOT NULL,
day date NOT NULL,
movieTitle VARCHAR(255) NOT NULL,
PRIMARY KEY (bookingNbr),
FOREIGN KEY (userName) REFERENCES Users(userName),
FOREIGN KEY (movieTitle) REFERENCES Movies(movieTitle),
FOREIGN KEY (movieTitle, day) REFERENCES Performances(movieTitle, day),
CONSTRAINT oneUserOneTicket UNIQUE (movieTitle, userName)
);
DROP TABLE IF EXISTS Performances;
CREATE TABLE Performances(
movieTitle VARCHAR(255) NOT NULL,
bookedSeats INT,
theatreName VARCHAR(20) NOT NULL,
day DATE NOT NULL,
PRIMARY KEY(movieTitle, day),
FOREIGN KEY (movieTitle) REFERENCES Movies(movieTitle),
FOREIGN KEY (theatreName) REFERENCES Theatres(theatreName),
CONSTRAINT oneMovieOneDay UNIQUE (movieTitle, day)
);
SET foreign_key_checks = 1;
After creating theese tables, the query SELECT * FROM Performances;
gives me theese outputs in mySQL workbench and vsc respectivly
https://gyazo.com/4dd670122582d7ed5340e3b5a03eafbe
https://gyazo.com/366687f14944134610839cbb7a790081
Is there any way i can make the date format in the output in vsc look like the one in mySQL workbench?
Ive tried different convert statements and nls_format something something although i couldnt get it to work. Im very new to SQL as a whole.
Hi I'm not very familiar with MySQL as I have only started using it today and I keep getting this syntax error and am not really sure what the problem is. I have attached a screenshot of the code and also pasted it below with the error in bold.
I'm sorry if this is a silly error that is easily fixed I'm just not sure how to fix it and would be very appreciative of any help.
CREATE TABLE copy (
`code` INT NOT NULL,
isbn CHAR(17) NOT NULL,
duration TINYINT NOT NULL,
CONSTRAINT pkcopy PRIMARY KEY (isbn, `code`),
CONSTRAINT fkcopy FOREIGN KEY (isbn) REFERENCES book (isbn));
CREATE TABLE student (
`no` INT NOT NULL,
`name` VARCHAR(30) NOT NULL,
school CHAR(3) NOT NULL,
embargo BIT NOT NULL,
CONSTRAINT pkstudent PRIMARY KEY (`no`));
CREATE TABLE loan (
`code` INT NOT NULL,
`no` INT NOT NULL,
taken DATE NOT NULL,
due DATE NOT NULL,
`return` DATE NULL,
CONSTRAINT pkloan PRIMARY KEY (taken, `code`, `no`),
CONSTRAINT fkloan FOREIGN KEY (`code`, `no`) REFERENCES copy, student **(**`code`, `no`));
Create the tables first, then use the ALTER TABLE statement to add the foreign keys one by one. You won't be able to call two different tables on the foreign key, so you'll have to use an ID that maps to both. Here is an example to add the foreign keys after the table has been created:
Add a new table named vendors and change the products table to include the vendor id field:
USE dbdemo;
CREATE TABLE vendors(
vdr_id int not null auto_increment primary key,
vdr_name varchar(255)
)ENGINE=InnoDB;
ALTER TABLE products
ADD COLUMN vdr_id int not null AFTER cat_id;
To add a foreign key to the products table, you use the following statement:
ALTER TABLE products
ADD FOREIGN KEY fk_vendor(vdr_id)
REFERENCES vendors(vdr_id)
ON DELETE NO ACTION
ON UPDATE CASCADE;
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 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 stuck with this error no 150 problem in mysql and I know there have been questions
which discuss this problem but I still can't find where I am wrong. Here is the database I am trying to create:
create table business (
ident varchar(40) NOT NULL,
name varchar(50) NOT NULL,
rating INT UNSIGNED NOT NULL,
PRIMARY KEY(ident)
) ENGINE=InnoDB;
create table deals (
business_id varchar(40) NOT NULL,
deals_id varchar(20) NOT NULL,
deals_title varchar(50) NOT NULL,
PRIMARY KEY (business_id, deals_id),
FOREIGN KEY (business_id) REFERENCES business(ident) ON DELETE CASCADE
) ENGINE=InnoDB;
create table d_options (
business_id varchar(40) NOT NULL,
dealid varchar(20) NOT NULL,
option_title varchar(40) NOT NULL,
PRIMARY KEY(business_id, dealid, option_title),
FOREIGN KEY(business_id) REFERENCES business(ident) ON DELETE CASCADE,
FOREIGN KEY(dealid) REFERENCES deals(deals_id)
) ENGINE=InnoDB;
I get error: ERROR 1005 (HY000): Can't create table 'test.d_options' (errno: 150)
I know for foreign key constraints to be satisfied there should be a index in the parent table as per mysql documentation, but I think that there is by default indexing
on primary key.
The result of innodb status is:
120530 0:47:48 Error in foreign key constraint of table test/d_options:
FOREIGN KEY(dealid) REFERENCES deals(deals_id)
) ENGINE=InnoDB:
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/5.1/en/innodb-foreign-key-constraints.html
for correct foreign key definition.
Any help is appriciated.
You have a compound primary key on (business_id, deal_id) and they are indexed as a pair, but to satisfy the FK, you need another index on deal_id alone:
create table deals (
business_id varchar(40) NOT NULL,
deals_id varchar(20) NOT NULL,
deals_title varchar(50) NOT NULL,
PRIMARY KEY (business_id, deals_id),
FOREIGN KEY (business_id) REFERENCES business(ident) ON DELETE CASCADE,
/* Add an index on deals_id, separate from the compound PK */
INDEX idx_deals_id (deals_id)
) ENGINE=InnoDB;