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Home/ Questions/Q 6666253
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T02:48:33+00:00 2026-05-26T02:48:33+00:00

here are my two tables CREATE TABLE IF NOT EXISTS `carslibrary` ( `CarID` int(10)

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here are my two tables

CREATE TABLE IF NOT EXISTS `carslibrary` (   
  `CarID` int(10) unsigned NOT NULL AUTO_INCREMENT,   
  `CarName` varchar(255) NOT NULL,  
  `colorslibrary_ID` int(11) unsigned NOT NULL,   
  PRIMARY KEY (`CarID`),
  KEY `colorslibrary_ID` (`colorslibrary_ID`) 
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ;

CREATE TABLE IF NOT EXISTS `colorslibrary` (   
  `ColorID` int(11) unsigned NOT NULL AUTO_INCREMENT,   
  `ColorName` varchar(255) NOT NULL,
  PRIMARY KEY (`ColorID`) 
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ;

I get an error on the following query:

ALTER TABLE  `carslibrary` ADD FOREIGN KEY (  `colorslibrary_ID` )
REFERENCES  `cars2`.`colorslibrary` (`ColorID` );

MySQL says:

#1452 – Cannot add or update a child row: a foreign key constraint
fails (`cars2`.<result 2 when explaining filename
‘#sql-cf8_41a’>, CONSTRAINT `#sql-cf8_41a_ibfk_1` FOREIGN KEY
(`colorslibrary_ID`) REFERENCES `colorslibrary` (`ColorID`))

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T02:48:34+00:00Added an answer on May 26, 2026 at 2:48 am

    Your tables aren’t empty, therefore a constraint fails (reference not found) when you create it.

    Use SET FOREIGN_KEY_CHECKS = 0; and re-run your alter table.

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