MySQL Fehler 150

Question

Ich versuche ein Skript zu erstellen, um eine andere Tabelle in MySQL zu normalisieren. Im Folgenden finden Sie, was ich habe:

USE hw7; 

SET foreign_key_checks = 0; 
DROP TABLE IF EXISTS airport_codes; 
DROP TABLE IF EXISTS airport_locations; 
DROP TABLE IF EXISTS airport_codenames; 
SET foreign_key_checks = 1; 

CREATE TABLE airport_codes(
airport_code char(3) not null, 
airline_code char(2) not null, 
primary key (airport_code, airline_code) 
); 

INSERT INTO airport_codes SELECT DISTINCT airport_code, airline_code 
    FROM airport_airlines; 

CREATE TABLE airport_locations(
airport_code char(3) not null, 
city varchar(20) not null, 
state char(2) not null, 
primary key (airport_code), 
constraint ap_code_fk 
    foreign key (airport_code) 
    references airport_codes(airport_code) 
); 

INSERT INTO airport_locations SELECT DISTINCT airport_code, city, state 
    FROM airport_airlines; 

CREATE TABLE airport_codenames(
airline_code char(2) not null, 
name varchar(20) not null, 
primary key (airline_code), 
constraint al_code_fk 
    foreign key (airline_code) 
    references airport_codes(airline_code) 
); 

INSERT INTO airport_codenames SELECT DISTINCT airline_code, name 
    FROM airport_airlines; 

Dieser Code führt zu diesem Fehler:

Can't create table hw7.airport_codenames errno:150

Answer 1

Da airport_codes mehrere mögliche Zeilen pro airport_code und airline_code (als zusammengesetzter Schlüssel) hat, kann sie nicht von anderen Fremdschlüsseln referenziert werden. Verschieben Sie die FK-Beziehungen in airport_codes und zeigen Sie auf airport_locations und airport_codenames.

USE hw7; 

SET foreign_key_checks = 0; 
DROP TABLE IF EXISTS airport_codes; 
DROP TABLE IF EXISTS airport_locations; 
DROP TABLE IF EXISTS airport_codenames; 
SET foreign_key_checks = 1; 


CREATE TABLE airport_locations(
airport_code char(3) not null, 
city varchar(20) not null, 
state char(2) not null, 
primary key (airport_code) 
); 

INSERT INTO airport_locations SELECT DISTINCT airport_code, city, state 
    FROM airport_airlines; 

CREATE TABLE airport_codenames(
airline_code char(2) not null, 
name varchar(20) not null, 
primary key (airline_code) 
); 

INSERT INTO airport_codenames SELECT DISTINCT airline_code, name 
    FROM airport_airlines; 


/* airport_codes moved after the other 2 tables, and FKs defined here */ 
CREATE TABLE airport_codes(
airport_code char(3) not null, 
airline_code char(2) not null, 
primary key (airport_code, airline_code), 
/* FK relationships are defined here, rather than in the other tables, 
    since the PKs for airport_code and airline_code are defined in the 
    other tables. 
*/ 
constraint ap_code_fk 
    foreign key (airport_code) 
    references airport_locations (airport_code), 
constraint al_code_fk 
    foreign key (airline_code) 
    references airport_codenames (airline_code) 
); 

INSERT INTO airport_codes SELECT DISTINCT airport_code, airline_code 
    FROM airport_airlines; 

Answer 2

Es sollte kommen von Ihnen Tabelle Auftrag. Wie mysql doc sagen

InnoDB does not permit you to drop a table that is referenced by a FOREIGN KEY constraint, unless you do SET foreign_key_checks = 0

So die Fremdschlüsselprüfung gesetzt zu Recht oder Ihre Drop, um in ändern:

DROP TABLE IF EXISTS airport_locations; 
DROP TABLE IF EXISTS airport_codenames; 
DROP TABLE IF EXISTS airport_codes; 

Sie Fremd Drop Schlüssel vor der Referenz.