MySQL:为什么这个数据库创建脚本不起作用?

时间:2010-02-25 01:42:59

标签: sql mysql database mysql-error-1005

出于某种原因,这个MySQL失败了:

CREATE SCHEMA IF NOT EXISTS `partB` DEFAULT CHARACTER SET latin1 COLLATE latin1_swedish_ci ;
USE `partB`;

CREATE TABLE Employees ( ssn CHAR(11),
Name CHAR(30),
mlot INTEGER,
PRIMARY KEY(ssn))
ENGINE = InnoDB;

CREATE TABLE Dept_Mgr ( did INTEGER,
dname CHAR(20),
ssn CHAR(11) NOT NULL,
PRIMARY KEY (did),
FOREIGN KEY (ssn) REFERENCES Employees
ON DELETE NO ACTION)
ENGINE = InnoDB;

它给出错误:

  

错误1005(HY000):无法创建表格   partb.dept_mgr(错误:150)

导致这种情况的原因是什么?

3 个答案:

答案 0 :(得分:2)

您必须在外表中为键指定列:

FOREIGN KEY (ssn) REFERENCES Employees (ssn) ...

答案 1 :(得分:1)

此命令:

SHOW ENGINE INNODB STATUS;
当您在创建外键时遇到问题,

是您的朋友。输出(删节)

------------------------
LATEST FOREIGN KEY ERROR
------------------------
100225  2:51:42 Error in foreign key constraint of table test/dept_mgr:
FOREIGN KEY (ssn) REFERENCES Employees
ON DELETE NO ACTION)
ENGINE = InnoDB:
Syntax error close to:

ON DELETE NO ACTION)
ENGINE = InnoDB

如果您将陈述更改为:

CREATE TABLE Dept_Mgr ( 
    did INTEGER,
    dname CHAR(20),
    ssn CHAR(11) NOT NULL,
    PRIMARY KEY (did),
    FOREIGN KEY (ssn) REFERENCES Employees(ssn)
) engine = innodb;

它确实有用。

答案 2 :(得分:0)

您只有引用表的外键,而不是列。

尝试:引用Employee.ssn