Mysql Error 1005 Errno 150
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here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings mysql error 1005 can't create table and policies of this site About Us Learn more about Stack mysql error 1005 errno 121 Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs error 1005 iphone Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; error 1005 archeage it only takes a minute: Sign up MySQL Foreign Key Error 1005 errno 150 up vote 31 down vote favorite 8 I'm doing a small DataBase with MySQL Workbench. I have a main table, called "Immobili", which has a Primary Key composed by four columns: (Comune, Via, Civico, Immobile). Now, I also have three other tables, wich have the same
Error 1005 Access Denied
primary key (Comune, Via, Civico, Immobile), but these fields are also referenced to the table Immobili. First question: Can I make a Primary Key that is also a Foreign Key? Second Question: When I try to export the changes it says: Executing SQL script in server # ERROR: Error 1005: Can't create table 'dbimmobili.condoni' (errno: 150) CREATE TABLE IF NOT EXISTS `dbimmobili`.`Condoni` ( `ComuneImmobile` VARCHAR(50) NOT NULL , `ViaImmobile` VARCHAR(50) NOT NULL , `CivicoImmobile` VARCHAR(5) NOT NULL , `InternoImmobile` VARCHAR(3) NOT NULL , `ProtocolloNumero` VARCHAR(15) NULL , `DataRichiestaSanatoria` DATE NULL , `DataSanatoria` DATE NULL , `SullePartiEsclusive` TINYINT(1) NULL , `SullePartiComuni` TINYINT(1) NULL , `OblazioneInEuro` DOUBLE NULL , `TecnicoOblazione` VARCHAR(45) NULL , `TelefonoTecnico` VARCHAR(15) NULL , INDEX `ComuneImmobile` (`ComuneImmobile` ASC) , INDEX `ViaImmobile` (`ViaImmobile` ASC) , INDEX `CivicoImmobile` (`CivicoImmobile` ASC) , INDEX `InternoImmobile` (`InternoImmobile` ASC) , PRIMARY KEY (`ComuneImmobile`, `ViaImmobile`, `CivicoImmobile`, `InternoImmobile`) , CONSTRAINT `ComuneImmobile` FOREIGN KEY (`ComuneImmobile` ) REFERENCES `dbimmobili`.`Immobile` (`ComuneImmobile` ) ON DELETE CASCADE ON UPDATE CASCADE, CONSTRAINT `ViaImmobile` FOREIGN KEY (`ViaImmobile` ) REFERENCES `dbimmobili`.`Immobile` (`ViaImmobile` ) ON DELETE CASCADE ON UPDATE CASCADE, CONSTRAINT `CivicoImmobile` FOR
Communication Skills Training Interpersonal Skills Training Blog About Blog Troy Fawkes / Archives / Solved: MySQL ERROR 1005: Can't create table (errno: 150) (Foreign Key) Nov 27 Solved: MySQL ERROR
Error 1005 Iphone 6
1005: Can't create table (errno: 150) (Foreign Key) November 27, 2011 Troy can't create table (errno 150) mysql Fawkes 8 Comments Archives Share on Facebook Share 0 Share on TwitterTweet 0 Share on Google Plus Share 0 error code 1005 iphone Share on Pinterest Share 0 Share on LinkedIn Share 0 This is another stupid error. It has to do with trying to successfully set foreign keys in MySQL. ERROR 1005: Can't create http://stackoverflow.com/questions/4063141/mysql-foreign-key-error-1005-errno-150 table (errno: 150) Great, that's fantastic. Here's an example of where this error will occur. CREATE TABLE main(
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id)
);
CREATE TABLE other(
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
main_id INT NOT NULL,
PRIMARY KEY(id),
FOREIGN KEY(main_id) REFERENCES main(id)
); So I'm trying to make the table "other" reference the table "main" through the foreign key https://www.troyfawkes.com/solved-mysql-error-1005-cant-create-table-errno-150/ "main_id" and, if you try it, it'll throw an Error 150. Want the solution? The foreign key "main_id" has to have the exact same type as the primary key that it references. In the example, "main_id" in the table "other" has the type INT NOT NULL while "id" in the table "main" has the type "INT UNSIGNED NOT NULL" and also AUTO_INCREMENT, but that isn't something we have to worry about. To make things incredibly clear, here's the working example. CREATE TABLE main( id INT UNSIGNED NOT NULL AUTO_INCREMENT, PRIMARY KEY(id) ); CREATE TABLE other( id INT UNSIGNED NOT NULL AUTO_INCREMENT, main_id INT UNSIGNED NOT NULL, PRIMARY KEY(id), FOREIGN KEY(main_id) REFERENCES main(id) ); To solve ‘MySQL ERROR 1005: Can't create table (errno: 150)‘ you likely just have to ensure that your foreign key has the exact same type as the primary key. Hope it helps. Share on Facebook Share 0 Share on TwitterTweet 0 Share on Google Plus Share 0 Share on Pinterest Share 0 Share on LinkedIn Share 0 Facebook Twitter Tumblr Pinterest Google+ LinkedIn E-Mail About The Author My name is Troy Boileau but
log in tour help Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of http://dba.stackexchange.com/questions/80435/mysql-create-table-shows-error-1005-errno-150 this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Database Administrators Questions Tags Users Badges Unanswered Ask Question _ https://mariadb.org/mariadb-innodb-foreign-key-constraint-errors/ Database Administrators Stack Exchange is a question and answer site for database professionals who wish to improve their database skills and learn from others in the community. Join them; it only takes error 1005 a minute: Sign up Here's how it works: Anybody can ask a question Anybody can answer The best answers are voted up and rise to the top Mysql Create table shows ERROR 1005 errno 150 up vote 0 down vote favorite I am trying to make a junction table, but I getting a ERROR 1005 errno 150 CREATE TABLE 61furiousFistPokemon( cardNumber int NOT NULL PRIMARY mysql error 1005 KEY, pokemonName VARCHAR(12), type VARCHAR(10), stage VARCHAR(10), evolvesFrom VARCHAR(12), HP INT, retreatCost INT, weakness VARCHAR(10), weaknessAmount VARCHAR(5), resistance VARCHAR(5), resistanceAmount VARCHAR(5) ); CREATE TABLE cardTags ( tagId int NOT NULL AUTO_INCREMENT PRIMARY KEY, tagName VARCHAR(60) NOT NULL ); CREATE TABLE 61furiousFistPokemonTags ( tagId int NOT NULL AUTO_INCREMENT PRIMARY KEY, pokemonCardNumber int, tagName VARCHAR(60), FOREIGN KEY (pokemonCardNumber) REFERENCES 61furiousFistPokemon(cardNumber), FOREIGN KEY (tagName) REFERENCES cardTags(tagName) ); Any help would be greatly appreciated. mysql innodb foreign-key table share|improve this question edited Oct 17 '14 at 15:23 RolandoMySQLDBA 108k15139276 asked Oct 17 '14 at 3:43 user50479 add a comment| 3 Answers 3 active oldest votes up vote 1 down vote tagName column in table cardTags is not defined as primary key and you are declaring foreign key FOREIGN KEY (tagName) REFERENCES cardTags(tagName). share|improve this answer answered Oct 17 '14 at 8:13 Nawaz Sohail 611314 Although you did not provide a solution, +1 for revealing root cause. –RolandoMySQLDBA Oct 17 '14 at 15:19 add a comment| up vote 0 down vote Very often it happens, when the foreign key and the reference key don't have same type or same length. Click here for more
for Developers MariaDB Contributor Agreement MariaDB Contributor Agreement FAQs Community Ambassadors Events Past Events and Conferences Sponsor Sponsors List of Donors Blog About MariaDB Sponsors Governance Logos and Badges MariaDB Trademark Usage Statistics Service Providers Maintenance Policy Security Policy Download Learn Get Involved Social Media Getting Started for Developers MariaDB Contributor Agreement MariaDB Contributor Agreement FAQs Community Ambassadors Events Past Events and Conferences Sponsor Sponsors List of Donors Blog HomeGeneralMariaDB: InnoDB foreign key constraint errors MariaDB: InnoDB foreign key constraint errors 2015-08-07 4 Comments Written by Jan Lindstrom Introduction A foreign key is a field (or collection of fields) in one table that uniquely identifies a row of another table. The table containing the foreign key is called the child table, and the table containing the candidate key is called the referenced or parent table. The purpose of the foreign key is to identify a particular row of the referenced table. Therefore, it is required that the foreign key is equal to the candidate key in some row of the primary table, or else have no value (the NULL value). This is called a referential integrity constraint between the two tables. Because violations of these constraints can be the source of many database problems, most database management systems provide mechanisms to ensure that every non-null foreign key corresponds to a row of the referenced table. Consider following simple example: create table parent ( id int not null primary key, name char(80) ) engine=innodb; create table child ( id int not null, name char(80), parent_id int, foreign key(parent_id) references parent(id) ) engine=innodb; As far as I know, the following storage engines for MariaDB and/or MySQL support foreign keys: InnoDB (both innodb_plugin and XtraDB) PBXT (https://mariadb.com/kb/en/mariadb/about-pbxt/) SolidDB for MySQL (http://sourceforge.net/projects/soliddb/) ScaleDB (https://mariadb.com/kb/en/mariadb/scaledb/ and http://scaledb.com/pdfs/TechnicalOverview.pdf) MySQL Cluster NDB 7.3 or later (https://dev.mysql.com/doc/refman/5.6/en/mysql-cluster-ndb-innodb-engines.html) MariaDB foreign key syntax is documented at https://mariadb.com/kb/en/mariadb/foreign-keys/ (and MySQL at http://dev.mysql.com/doc/refman/5.5/en/innodb-foreign-key-constraints.html). While most of the syntax is parsed and checked when the CREATE TABLE or ALTER TABLE clause is parsed, there are still several error cases that can happen inside InnoDB. Yes, InnoDB has its own internal foreign key constraint parser (in dict0dict.c function dict_create_foreign_con