#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax

Question

I'm working a project with Yiiframwork and I have this table in my data base project

 CREATE TABLE IF NOT EXISTS `tbl_annonce` (
`id` mediumint(8) unsigned NOT NULL AUTO_INCREMENT,
`idEntreprise` tinyint(3) unsigned NOT NULL,
`titre` varchar(100) NOT NULL,
`detailleDiscription` varchar(5000) NOT NULL,
`categorie` varchar(100) DEFAULT NULL,
`typePoste` varchar(100) NOT NULL,
`salaireMin` varchar(80) NOT NULL,
`salaireMax` varchar(80) NOT NULL,
`niveauEtude` varchar(80) NOT NULL,
`niveauExperience` varchar(80) NOT NULL,
`langue` varchar(50) DEFAULT NULL,
`poste` varchar(50) NOT NULL,
`pays` varchar(50) NOT NULL,
`ville` varchar(50) NOT NULL,
`adresse` varchar(80) NOT NULL,
`datePublication` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`etat` varchar(50) NOT NULL,
`photo` varchar(255)  NULL,
`nometr` text NOT NULL,
PRIMARY KEY (`id`),
CONSTRAINT 'fk_idEntrepriseAnn' FOREIGN KEY (idEntreprise) REFERENCES tbl_entreprise(id)ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;

I get the following error message:

1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''fk_idEntrepriseAnn' FOREIGN KEY (idEntreprise) REFERENCES tbl_entreprise(id) ' at line 22

Here is some info on what I'm working with

•Server type: MySQL •Server version: 5.5.32 - MySQL Community Server

•phpMyAdmin: 4.0.4.1, latest stable version: 4.1.7

can any one help me plz !!

Answer 1

I think the problem is with wrong single quote in your query.Instead of putting 'fk_idEntrepriseAnn' please try

`fk_idEntrepriseAnn`

Answer 2

try to use "``" on 'fk_idEntrepriseAnn' like

CREATE TABLE IF NOT EXISTS `tbl_annonce` (
`id` mediumint(8) unsigned NOT NULL AUTO_INCREMENT,
`idEntreprise` tinyint(3) unsigned NOT NULL,
`titre` varchar(100) NOT NULL,
`detailleDiscription` varchar(5000) NOT NULL,
`categorie` varchar(100) DEFAULT NULL,
`typePoste` varchar(100) NOT NULL,
`salaireMin` varchar(80) NOT NULL,
`salaireMax` varchar(80) NOT NULL,
`niveauEtude` varchar(80) NOT NULL,
`niveauExperience` varchar(80) NOT NULL,
`langue` varchar(50) DEFAULT NULL,
`poste` varchar(50) NOT NULL,
`pays` varchar(50) NOT NULL,
`ville` varchar(50) NOT NULL,
`adresse` varchar(80) NOT NULL,
`datePublication` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`etat` varchar(50) NOT NULL,
`photo` varchar(255)  NULL,
`nometr` text NOT NULL,
PRIMARY KEY (`id`),
CONSTRAINT `fk_idEntrepriseAnn` FOREIGN KEY (idEntreprise) REFERENCES tbl_entreprise(id)ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;

Answer 3

I think you are new in Stackoverflow ! next time try to correct it yourself ! because Stackoverflow is for much more difficult question ! I think this is the right answer try this :

 CREATE TABLE IF NOT EXISTS `tbl_annonce` (
`id` mediumint(8) unsigned NOT NULL AUTO_INCREMENT,
`idEntreprise` tinyint(3) unsigned NOT NULL,
`titre` varchar(100) NOT NULL,
`detailleDiscription` varchar(5000) NOT NULL,
`categorie` varchar(100) DEFAULT NULL,
`typePoste` varchar(100) NOT NULL,
`salaireMin` varchar(80) NOT NULL,
`salaireMax` varchar(80) NOT NULL,
`niveauEtude` varchar(80) NOT NULL,
`niveauExperience` varchar(80) NOT NULL,
`langue` varchar(50) DEFAULT NULL,
`poste` varchar(50) NOT NULL,
`pays` varchar(50) NOT NULL,
`ville` varchar(50) NOT NULL,
`adresse` varchar(80) NOT NULL,
`datePublication` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
`etat` varchar(50) NOT NULL,
`photo` varchar(255)  NULL,
`nometr` text NOT NULL,
PRIMARY KEY (`id`),
CONSTRAINT `fk_idEntrepriseAnn` FOREIGN KEY (idEntreprise) REFERENCES tbl_entreprise(id)ON DELETE CASCADE ON UPDATE CASCADE
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=6 ;