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Home/ Questions/Q 3785024
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T11:27:56+00:00 2026-05-19T11:27:56+00:00

The MySQL workbench is a great tool, but you have to keep your head

  • 0

The MySQL workbench is a great tool, but you have to keep your head on as it does no error checking whatsoever.

So, it generated this SQL syntax to create a table and it won’t work.
And I can’t figure out WHY

It gives me this error:

#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
') NULL DEFAULT NULL , `maaltijd` TINYINT(1) NOT NULL ,
`cheques` TINYINT(1'
at line 14

This is the query:

    CREATE  TABLE IF NOT EXISTS `nascholing`.`aanvraag` (
      `aanvraagId` INT(11) NOT NULL AUTO_INCREMENT ,
      `userId` BIGINT(20) NOT NULL ,
      `creationDate` DATETIME NULL DEFAULT NULL ,
      `vakGroep` INT(11) NULL DEFAULT NULL ,
      `personeelType` INT(11) NOT NULL ,
      `leeftijd` INT(11) NULL DEFAULT NULL ,
      `thema` VARCHAR(255) NOT NULL ,
      `intern` TINYINT(1) NULL DEFAULT -1 ,
      `duur` INT(11) NOT NULL ,
      `tijdens` INT(11) NULL DEFAULT NULL ,
      `type` INT(11) NULL DEFAULT NULL ,
      `voor` INT(11) NOT NULL ,
      `kost` DOUBLE(11) NULL DEFAULT NULL ,
      `maaltijd` TINYINT(1) NOT NULL ,
      `cheques` TINYINT(1) NOT NULL ,
      `betalingsmethode` INT(11) NOT NULL ,
      `aanschafBoek` TINYINT(1) NOT NULL ,
      `kostBoek` DOUBLE(11) NULL DEFAULT NULL ,
      PRIMARY KEY (`aanvraagId`) ,
      UNIQUE INDEX `aanvraagId_UNIQUE` (`aanvraagId` ASC) ,
      INDEX `fk_aanvraag_user1` (`userId` ASC) ,
      INDEX `fk_aanvraag_duur1` (`duur` ASC) ,
      INDEX `fk_aanvraag_tijdens1` (`tijdens` ASC) ,
      INDEX `fk_aanvraag_type1` (`type` ASC) ,
      INDEX `fk_aanvraag_vakGroep1` (`vakGroep` ASC) ,
      INDEX `fk_aanvraag_voorEmail1` (`voor` ASC) ,
      INDEX `fk_aanvraag_betalingsMethode1` (`betalingsmethode` ASC) ,
      CONSTRAINT `fk_aanvraag_user1`
        FOREIGN KEY (`userId` )
        REFERENCES `nascholing`.`user` (`userId` )
        ON DELETE NO ACTION
        ON UPDATE NO ACTION,
      CONSTRAINT `fk_aanvraag_duur1`
        FOREIGN KEY (`duur` )
        REFERENCES `nascholing`.`duur` (`duurId` )
        ON DELETE NO ACTION
        ON UPDATE NO ACTION,
      CONSTRAINT `fk_aanvraag_tijdens1`
        FOREIGN KEY (`tijdens` )
        REFERENCES `nascholing`.`tijdens` (`tijdensId` )
        ON DELETE NO ACTION
        ON UPDATE NO ACTION,
      CONSTRAINT `fk_aanvraag_type1`
        FOREIGN KEY (`type` )
        REFERENCES `nascholing`.`type` (`typeId` )
        ON DELETE NO ACTION
        ON UPDATE NO ACTION,
      CONSTRAINT `fk_aanvraag_vakGroep1`
        FOREIGN KEY (`vakGroep` )
        REFERENCES `nascholing`.`vakGroep` (`vakGroepId` )
        ON DELETE NO ACTION
        ON UPDATE NO ACTION,
      CONSTRAINT `fk_aanvraag_voorEmail1`
        FOREIGN KEY (`voor` )
        REFERENCES `nascholing`.`voorEmail` (`voorEmailId` )
        ON DELETE NO ACTION
        ON UPDATE NO ACTION,
      CONSTRAINT `fk_aanvraag_betalingsMethode1`
        FOREIGN KEY (`betalingsmethode` )
        REFERENCES `nascholing`.`betalingsMethode` (`betalingsMethodeId` )
        ON DELETE NO ACTION
        ON UPDATE NO ACTION)
    ENGINE = InnoDB
    DEFAULT CHARACTER SET = utf8
    COLLATE = utf8_general_ci;
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-19T11:27:57+00:00Added an answer on May 19, 2026 at 11:27 am

    There is a need to specify decimal point for type double

    example,

    `kost` DOUBLE(11, 2) NULL DEFAULT NULL, ...
    

    so on for all the type double

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