Error 1005 Hy000 Errno 150 Mysql
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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 and policies of this site About Us Learn error 1005 hy000 can t create table errno 150 more about Stack Overflow the company Business Learn more about hiring developers or mysql error code 1005 errno 150 posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community mysql error 1005 errno 150 Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up MySQL Foreign Key Error 1005 errno 150 up vote 31 error 1005 (hy000) can't create table error 150 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 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
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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` FOREIGN KEY (`CivicoImmobile` ) REFERENCES `dbimmobili`.`Immobile` (`CivicoImmobile` ) ON DELETE CASCADE ON UPDATE CASCADE, CONSTRAINT `InternoImmobile` FOREIGN KEY (`InternoImmobile` ) REFERENCES `dbimmobili`.`Immobile` (`InternoImmobile` ) ON DELETE CASCADE ON UPDATE CASCADE ) ENGINE = InnoDB Showing the Engine Status: Error in foreign key constraint of table dbimmobili/valutazionimercato: Cannot find an index
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Errno 121 Mysql
more about Stack Overflow the company Business Learn more about hiring developers or error no 150 in mysql posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community mysql can't create table (errno: -1) Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up error 1005 (hy000) can't create table (errno 150) in http://stackoverflow.com/questions/4063141/mysql-foreign-key-error-1005-errno-150 mysql up vote -1 down vote favorite 1 As in screenshot, both the fields: cust_id in customer table and Store_ID in ShelfLocation table have the same signature, still I am not able to reference cust_id from Store_ID. Any help? mysql sql foreign-keys share|improve this question edited Nov 24 '12 at 22:29 marc_s 452k938641029 asked Nov 24 '12 at 21:41 Vivek Sethi 731210 Try giving your foreign http://stackoverflow.com/questions/13545895/error-1005-hy000-cant-create-table-errno-150-in-mysql key a unique name –Laurence Nov 24 '12 at 21:49 Does the table ShelfLocation already contain rows when you try to add the key? If it does and any of row has a Store_ID which does not match customer.cust_id, you'll get errors (though I cannot remember for certain if that causes err150) –Michael Berkowski Nov 24 '12 at 22:34 @MichaelBerkowski: No, table ShelfLocation is empty, still I am not able to refer. –Vivek Sethi Nov 24 '12 at 22:42 @Laurence: what unique name?? as in how?? –Vivek Sethi Nov 24 '12 at 22:44 @VivekSethi Alter Table ShelfLocation Add Constraint FK_GiveLozABeer Foreign Key (Store_ID) References Customer (cust_id); –Laurence Nov 24 '12 at 23:11 | show 1 more comment 2 Answers 2 active oldest votes up vote 0 down vote accepted This solved it for me: Add before first line: SET FOREIGN_KEY_CHECKS = 0; Add after last line: SET FOREIGN_KEY_CHECKS = 1; share|improve this answer answered Dec 2 '13 at 17:40 Ronny Sherer 2,9721116 add a comment| up vote 0 down vote The solution is checking to make sure Primary_Key and Foreign_Key exactly match their data types. If one is signed and the other is uns
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 this site About Us Learn more about http://dba.stackexchange.com/questions/80435/mysql-create-table-shows-error-1005-errno-150 Stack Overflow the company Business Learn more about hiring developers or posting ads with us Database Administrators Questions Tags Users Badges Unanswered Ask Question _ Database Administrators Stack Exchange is a question and answer site for https://mariadb.org/mariadb-innodb-foreign-key-constraint-errors/ database professionals who wish to improve their database skills and learn from others in the community. Join them; it only takes a minute: Sign up Here's how it works: Anybody can ask a question Anybody can error 1005 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 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) t create table ); 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 107k15138274 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 611214 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 details. share|improve this answer answered Oct 17 '14 at 7:30 Musakkhir Sayyed 240314 add a comment| up vote 0 down vote Here is the quick fix: Just add a UNIQUE KEY on tagName in the cardTags table SAMPLE SCRIPT DROP DATABASE IF
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