Mysql Workbench Error 1005 Errno 150
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Mysql Error 1005 Errno 150
Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like mysql error 1005 errno 121 you, helping each other. Join them; it only takes a minute: Sign up Error Code: 1005. Can't create table '…' (errno: 150) up vote 74 down vote favorite 30 I searched for a solution to this problem
Error Code 1005 Iphone
on internet and checked the SO questions but no solution worked for my case. I want to create a foreign key from table sira_no to metal_kod. ALTER TABLE sira_no ADD CONSTRAINT METAL_KODU FOREIGN KEY(METAL_KODU) REFERENCES metal_kod(METAL_KODU) ON DELETE SET NULL ON UPDATE SET NULL ; This script returns: Error Code: 1005. Can't create table 'ebs.#sql-f48_1a3' (errno: 150) I tried adding index to the referenced table: CREATE INDEX METAL_KODU_INDEX ON metal_kod (METAL_KODU); I checked METAL_KODU on error code 10005 both tables (charset and collation). But couldn't find a solution to this problem. Does anyone have any idea? Thanks in advance. EDIT: Here is the metal_kod table: METAL_KODU varchar(4) NO PRI DURUM bit(1) NO METAL_ISMI varchar(30) NO AYAR_YOGUNLUK smallint(6) YES 100 mysql mysql-error-1005 share|improve this question edited Jan 27 '14 at 7:57 Melon 786617 asked Jan 26 '12 at 13:14 lamostreta 85122046 Can you show the schema for the metal_kod table ... which field in that table should the Foreign Key reference ? –ManseUK Jan 26 '12 at 13:28 add a comment| 9 Answers 9 active oldest votes up vote 199 down vote accepted Error Code: 1005 -- there is a wrong primary key reference in your code usually it's due to a reference FK field not exist. might be you have typo mistake,or check case it should be same, or there's a field-type mismatch. FK-linked fields must match definitions exactly. Some Known causes may be : The two key fields type and/or size doesn’t match exactly. For example, if one is INT(10) the key field needs to be INT(10) as well and not INT(11) or TINYINT. You may want to confirm the field size using SHOW CREATE TABLE because Query Browser will sometimes visually show just INTEGER for both INT(10) and INT(11). You should also check that one is not
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about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss can't create table (errno 150) Join the Stack Overflow Community Stack Overflow is a community of 6.2 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up MySQL error code http://stackoverflow.com/questions/9018584/error-code-1005-cant-create-table-errno-150 1005 / errno: 150 up vote 1 down vote favorite I'm using a XAMPP 5.6.19-0 as a MySQL Database server and managing via MySQL Workbench (Mac OS 10.11.3). I'm trying to create some tables, with foreign keys, and im getting this error: Error Code: 1005. Can't create table imobiliaria24h.proprietario (errno: 150 "Foreign key constraint is incorrectly formed") This is the code I'm trying to use: http://stackoverflow.com/questions/36453597/mysql-error-code-1005-errno-150 CREATE TABLE IF NOT EXISTS PROPRIETARIO ( NOME VARCHAR(30) NOT NULL, CPF INT NOT NULL, TELEFONE INT NOT NULL, CORRETOR_CPF INT NOT NULL, PRIMARY KEY (CPF), FOREIGN KEY (CORRETOR_CPF) REFERENCES CORRETOR(CPF) )ENGINE=InnoDB; CREATE TABLE IF NOT EXISTS IMOVEL( ENDERECO VARCHAR(100) NOT NULL, TIPO ENUM('ALUGUEL', 'VENDA'), INQUILINO_CPF INT, PROPRIETARIO_CPF INT, VALOR_ALUGUEL_PRECOVENDA INT NOT NULL, NUM_QUARTOS INT NOT NULL, METRAGEM INT NOT NULL, NUM_BANHEIROS INT NOT NULL, VALOR_CONDOMINIO INT NOT NULL, IDADE INT NOT NULL, HORA_REGISTRO TIME NOT NULL, PRIMARY KEY(ENDERECO), FOREIGN KEY (INQUILINO_CPF) REFERENCES INQUILINO(CPF), FOREIGN KEY (PROPRIETARIO_CPF) REFERENCES PROPRIETARIO(CPF) )ENGINE=InnoDB; CREATE TABLE IF NOT EXISTS INQUILINO( NOME VARCHAR(30) NOT NULL, CPF INT NOT NULL, TELEFONE INT NOT NULL, FIADOR BOOLEAN NOT NULL, INADIMPLENCIA INT NOT NULL, CORRETOR_CPF INT NOT NULL, PRIMARY KEY (CPF), FOREIGN KEY (CORRETOR_CPF) REFERENCES CORRETOR(CPF) )ENGINE=InnoDB; CREATE TABLE IF NOT EXISTS CORRETOR( NOME VARCHAR(30) NOT NULL, CPF INT NOT NULL, TELEFONE INT NOT NULL, ANOS_CASA SMALLINT NOT NULL, TRABALHA_FDS BOOLEAN NOT NULL, TRABALHA_NOITE BOOLEAN NOT NULL, PRIMARY KEY (CPF) )ENGINE=InnoDB; The interesting thing is that, if I create the table separately, only the ones that references each other, for example, I created first PROPRIETARIO and CORRETOR, second INQUILINO, a
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 1005: Can't create table (errno: 150) (Foreign Key) November 27, https://www.troyfawkes.com/solved-mysql-error-1005-cant-create-table-errno-150/ 2011 Troy Fawkes 8 Comments Archives Share on Facebook Share 0 Share on TwitterTweet http://dba.stackexchange.com/questions/80435/mysql-create-table-shows-error-1005-errno-150 0 Share on Google Plus Share 0 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 table (errno: 150) Great, that's fantastic. Here's an example of where this error will occur. CREATE TABLE main(
id INT error 1005 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 "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 error 1005 errno 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 I go by Troy Fawkes. I'm a digital marketer working in Toronto. On top of that, my passions include social skills, networking and dating; subjects that I've been semi-professionally involved with for almost five years. Top LinksGet The Conversation Handbook How To Improve Conversation Skills How to Speak Clearly & Confidently Communication Skills Training Categories Archives Development Romance SEO Social Sk
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 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 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 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) ); 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 108k15141276 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 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 i