問題描述

我正在嘗試將我的新架構(gòu)轉(zhuǎn)發(fā)到我的數(shù)據(jù)庫服務(wù)器上，但我不知道為什么我會(huì)收到這個(gè)錯(cuò)誤.

I am trying to forward engineer my new schema onto my database server, but I can't figure out why I am getting this error.

我試圖在這里搜索答案，但我發(fā)現(xiàn)的所有內(nèi)容都說要么將數(shù)據(jù)庫引擎設(shè)置為 InnoDB，要么確保我嘗試用作外鍵的鍵是主鍵他們自己的桌子.如果我沒記錯(cuò)的話，這兩件事我都做過.我還能做什么?

I've tried to search for the answer here, but everything I've found has said to either set the database engine to InnoDB or to make sure the keys I'm trying to use as a foreign key are primary keys in their own tables. I have done both of these things, if I'm not mistaken. What else can I do?

Executing SQL script in server

ERROR: Error 1215: Cannot add foreign key constraint

-- -----------------------------------------------------
-- Table `Alternative_Pathways`.`Clients_has_Staff`
-- -----------------------------------------------------

CREATE  TABLE IF NOT EXISTS `Alternative_Pathways`.`Clients_has_Staff` (
  `Clients_Case_Number` INT NOT NULL ,
  `Staff_Emp_ID` INT NOT NULL ,
  PRIMARY KEY (`Clients_Case_Number`, `Staff_Emp_ID`) ,
  INDEX `fk_Clients_has_Staff_Staff1_idx` (`Staff_Emp_ID` ASC) ,
  INDEX `fk_Clients_has_Staff_Clients_idx` (`Clients_Case_Number` ASC) ,
  CONSTRAINT `fk_Clients_has_Staff_Clients`
    FOREIGN KEY (`Clients_Case_Number` )
    REFERENCES `Alternative_Pathways`.`Clients` (`Case_Number` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `fk_Clients_has_Staff_Staff1`
    FOREIGN KEY (`Staff_Emp_ID` )
    REFERENCES `Alternative_Pathways`.`Staff` (`Emp_ID` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB

SQL 腳本執(zhí)行完成:語句:7 條成功，1 條失敗

SQL script execution finished: statements: 7 succeeded, 1 failed

這是父表的 SQL.

CREATE  TABLE IF NOT EXISTS `Alternative_Pathways`.`Clients` (
  `Case_Number` INT NOT NULL ,
  `First_Name` CHAR(10) NULL ,
  `Middle_Name` CHAR(10) NULL ,
  `Last_Name` CHAR(10) NULL ,
  `Address` CHAR(50) NULL ,
  `Phone_Number` INT(10) NULL ,
  PRIMARY KEY (`Case_Number`) )
ENGINE = InnoDB

CREATE  TABLE IF NOT EXISTS `Alternative_Pathways`.`Staff` (
  `Emp_ID` INT NOT NULL ,
  `First_Name` CHAR(10) NULL ,
  `Middle_Name` CHAR(10) NULL ,
  `Last_Name` CHAR(10) NULL ,
  PRIMARY KEY (`Emp_ID`) )
ENGINE = InnoDB

推薦答案

我猜 Clients.Case_Number 和/或 Staff.Emp_ID 并不完全相同數(shù)據(jù)類型為 Clients_has_Staff.Clients_Case_Number 和 Clients_has_Staff.Staff_Emp_ID.

I'm guessing that Clients.Case_Number and/or Staff.Emp_ID are not exactly the same data type as Clients_has_Staff.Clients_Case_Number and Clients_has_Staff.Staff_Emp_ID.

也許父表中的列是INT UNSIGNED?

它們在兩個(gè)表中的數(shù)據(jù)類型必須完全相同.

They need to be exactly the same data type in both tables.

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