Debuggen des MySQL-Fehlers "# 1064 - Sie haben einen Fehler in Ihrer SQL-Syntax"

Question

Ich verstehe, dass diese Frage mehrmals gestellt wurde, aber ich bin ein großer Anfänger, ich übertrage die Datenbank von SQL Workbench nach phpmyadmin. Ich erhalte diese Fehlermeldung:

SQL-Abfrage:

CREATE TABLE IF NOT EXISTS `ApplicationSystem`.`staff` (
    `StaffID` TEXT(6) NOT NULL, 
    `Name` VARCHAR(45) NULL, 
    `Email` VARCHAR(254) NULL, 
    `Password` VARCHAR(45) NULL, 
    PRIMARY KEY (`StaffID`), 
    CONSTRAINT `StaffID` 
    FOREIGN KEY() 
    REFERENCES `ApplicationSystem`.`UserType`() 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB 
MySQL said: Documentation 

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ') 
    REFERENCES `ApplicationSystem`.`UserType`() 
    ON DELETE NO ACTION 
    ' at line 8 

hier ist die eigentliche SQL-Code:

-- MySQL Script generated by MySQL Workbench 
-- 04/04/16 23:56:12 
-- Model: New Model Version: 1.0 
-- MySQL Workbench Forward Engineering 

SET @OLD_UNIQUE_CHECKS=@@UNIQUE_CHECKS, UNIQUE_CHECKS=0; 
SET @OLD_FOREIGN_KEY_CHECKS=@@FOREIGN_KEY_CHECKS, FOREIGN_KEY_CHECKS=0; 
SET @OLD_SQL_MODE=@@SQL_MODE, SQL_MODE='TRADITIONAL,ALLOW_INVALID_DATES'; 

-- ----------------------------------------------------- 
-- Schema ApplicationSystem 
-- ----------------------------------------------------- 

-- ----------------------------------------------------- 
-- Schema ApplicationSystem 
-- ----------------------------------------------------- 
CREATE SCHEMA IF NOT EXISTS `ApplicationSystem` DEFAULT CHARACTER SET utf8 ; 
USE `ApplicationSystem` ; 

-- ----------------------------------------------------- 
-- Table `ApplicationSystem`.`staff` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `ApplicationSystem`.`staff` (
    `StaffID` TEXT(6) NOT NULL, 
    `Name` VARCHAR(45) NULL, 
    `Email` VARCHAR(254) NULL, 
    `Password` VARCHAR(45) NULL, 
    PRIMARY KEY (`StaffID`), 
    CONSTRAINT `StaffID` 
    FOREIGN KEY() 
    REFERENCES `ApplicationSystem`.`UserType`() 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `ApplicationSystem`.`UserType` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `ApplicationSystem`.`UserType` (
    `Type` ENUM('A', 'HA', 'AT') NULL, 
    `ID` TEXT(6) NOT NULL, 
    INDEX `ID_idx` (`ID` ASC), 
    PRIMARY KEY (`ID`), 
    UNIQUE INDEX `ID_UNIQUE` (`ID` ASC), 
    CONSTRAINT `ID` 
    FOREIGN KEY (`ID`) 
    REFERENCES `ApplicationSystem`.`Applicants` (`ApplicantID`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION, 
    CONSTRAINT `ID` 
    FOREIGN KEY (`ID`) 
    REFERENCES `ApplicationSystem`.`staff` (`StaffID`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `ApplicationSystem`.`Applicants` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `ApplicationSystem`.`Applicants` (
    `ApplicantID` TEXT(6) NULL, 
    `Firstname` VARCHAR(40) NULL, 
    `surname` VARCHAR(40) NULL, 
    `DOB` DATE NULL, 
    `Gender` ENUM('yes', 'no') NULL, 
    `Address` VARCHAR(45) NULL, 
    `Postcode` VARCHAR(9) NULL, 
    `ApplicantMobile` TEXT(11) NULL, 
    `ApplicantHome` TEXT(11) NULL, 
    `ApplicantEmail` VARCHAR(255) NULL, 
    `ParentName` VARCHAR(40) NULL, 
    `Relationtoapplicant` VARCHAR(15) NULL, 
    `ParentEmail` VARCHAR(255) NULL, 
    `ParentAddress` VARCHAR(40) NULL, 
    `ParentPostcode` VARCHAR(9) NULL, 
    `ParentHome` TEXT(11) NULL, 
    `ParentMobile` TEXT(11) NULL, 
    `Currentschool` VARCHAR(40) NULL, 
    `ApplicantInYr11` ENUM('yes', 'no') NULL, 
    `Sibbling` ENUM('yes', 'no') NULL, 
    `SibblingName` VARCHAR(40) NULL, 
    `SibblingYear` TEXT(2) NULL, 
    `MedicalConditions` VARCHAR(30) NULL, 
    `AccessArrangementsAvailable` ENUM('yes', 'no') NULL, 
    `AccessArrangments` VARCHAR(20) NULL, 
    `Ethnic` VARCHAR(40) NULL, 
    `FirstLanguage` VARCHAR(15) NULL, 
    `HomeLanguage` VARCHAR(15) NULL, 
    `Religion` VARCHAR(15) NULL, 
    `Nationality` VARCHAR(20) NULL, 
    `ArmedForces` ENUM('yes', 'no') NULL, 
    `PreviouslyLookedAfter` ENUM('yes', 'no') NULL, 
    `EUResident` ENUM('yes', 'no') NULL, 
    `RestrictionResidence` ENUM('yes', 'no') NULL, 
    `school1` VARCHAR(45) NULL, 
    `school2` VARCHAR(45) NULL, 
    `school3` VARCHAR(45) NULL, 
    `school4` VARCHAR(45) NULL, 
    `UPN` TEXT(13) NULL, 
    `UCL` TEXT(13) NULL, 
    `ULN` TEXT(10) NULL, 
    `Password` VARCHAR(40) NULL, 
    `Course1` VARCHAR(30) NULL, 
    `Course2` VARCHAR(30) NULL, 
    `Course3` VARCHAR(30) NULL, 
    `Course4` VARCHAR(30) NULL, 
    `Course5` VARCHAR(30) NULL, 
    `Complete` TINYINT(1) NULL, 
    `Outcome` ENUM('R', 'A') NULL, 
    PRIMARY KEY (`ApplicantID`), 
    UNIQUE INDEX `idtable1_UNIQUE` (`ApplicantID` ASC), 
    UNIQUE INDEX `ApplicantEmail_UNIQUE` (`ApplicantEmail` ASC), 
    CONSTRAINT `ApplicantID` 
    FOREIGN KEY() 
    REFERENCES `ApplicationSystem`.`UserType`() 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `ApplicationSystem`.`table1` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `ApplicationSystem`.`table1` (
    `ID` TEXT(6) NOT NULL, 
    `outcome` TEXT(6) NULL, 
    PRIMARY KEY (`ID`)) 
ENGINE = InnoDB; 


-- ----------------------------------------------------- 
-- Table `ApplicationSystem`.`outcome` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `ApplicationSystem`.`outcome` (
    `ID` TEXT(6) NOT NULL, 
    `Outcome` ENUM('R', 'A') NULL, 
    PRIMARY KEY (`ID`), 
    CONSTRAINT `ID` 
    FOREIGN KEY (`ID`) 
    REFERENCES `ApplicationSystem`.`Applicants` (`ApplicantID`) 
    ON DELETE NO ACTION 
    ON UPDATE NO ACTION) 
ENGINE = InnoDB; 

USE `ApplicationSystem` ; 

-- ----------------------------------------------------- 
-- Placeholder table for view `ApplicationSystem`.`view1` 
-- ----------------------------------------------------- 
CREATE TABLE IF NOT EXISTS `ApplicationSystem`.`view1` (`id` INT); 

-- ----------------------------------------------------- 
-- View `ApplicationSystem`.`view1` 
-- ----------------------------------------------------- 
DROP TABLE IF EXISTS `ApplicationSystem`.`view1`; 
USE `ApplicationSystem`; 


SET SQL_MODE=@OLD_SQL_MODE; 
SET FOREIGN_KEY_CHECKS=@OLD_FOREIGN_KEY_CHECKS; 
SET UNIQUE_CHECKS=@OLD_UNIQUE_CHECKS 

Answer 1

CONSTRAINT `StaffID` 
    FOREIGN KEY() 
    REFERENCES `ApplicationSystem`.`UserType`() 

einen Fremdschlüssel zu deklarieren, müssen Sie Geben Sie die spezifischen Spalten an, die an der Einschränkung teilnehmen. Du hast das hier nicht getan. Die korrekte Syntax ist

CONSTRAINT `StaffID` 
    FOREIGN KEY (<name of a column in this table>) 
    REFERENCES `ApplicationSystem`.`UserType` (<name of the matching column in the referenced table>) 

Füllung im <placeholders> mit den richtigen tatsächlichen Spaltennamen.

Sebas hat einen guten Punkt, obwohl; die Daten sind weder in Workbench noch in phpMyAdmin "in"; Dies sind nur Verwaltungsclients, die eine Verbindung zum Datenbankprozess herstellen. Sie sollten daher nur eine neue Verbindung zur vorhandenen Datenbank einrichten können.

Answer 2

FOREIGN KEY (`here shoud be the column name the key belongs to`)