MySQL - 一对多关系无法正常工作

我有2个表，一个用于员工，另一个用于部门。一个部门可以有多个员工，但一个员工只能在一个部门工作。他们的关系是[1：很多]。MySQL - 一对多关系无法正常工作

我想在MySQL中做到这一点，但我遇到了一个问题。如果我有8个不同的部门，我尝试添加超过800名员工总数，谁在不同的部门工作，我得到以下错误：

Cannot add or update a child row: a foreign key constraint fails (`testdb`.`employee`, CONSTRAINT `employee_ibfk_1` FOREIGN KEY (`id`) REFERENCES `department` (`id`))

如果我有8名或更少的员工，一切都很正常。添加第9名员工后，我收到上述错误。

部门表：

CREATE TABLE IF NOT EXISTS department(
    id INT(20) NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    name VARCHAR(255) NOT NULL UNIQUE 
)ENGINE=INNODB;

处插入：

INSERT INTO department(name) VALUES ('Athens'); 
INSERT INTO department(name) VALUES ('Patras'); 
INSERT INTO department(name) VALUES ('Kalamata'); 
INSERT INTO department(name) VALUES ('Heraklion'); 
INSERT INTO department(name) VALUES ('Thessaloniki'); 
INSERT INTO department(name) VALUES ('Xanthi'); 
INSERT INTO department(name) VALUES ('Larisa'); 
INSERT INTO department(name) VALUES ('Alexandroupoli');

Employee表：

CREATE TABLE IF NOT EXISTS employee(
    id INT(20) NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    first_name VARCHAR(255) NOT NULL, 
    last_name VARCHAR(255) NOT NULL, 
    email VARCHAR(255) NOT NULL, 
    born INT(20) NOT NULL, 
    country VARCHAR(255) NOT NULL, 
    department_name VARCHAR(255) NOT NULL, 
    FOREIGN KEY (id) references department(id) 
)ENGINE=INNODB;

个员工插入：

INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('Vaggelis','Michos','[email protected]','1995','Greece','Athens'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('James','Gunn','[email protected]','1970','USA','Athens'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('George','McMahon','[email protected]','1978','Usa','Patras'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('John','Jones','[email protected]','1992','England','Patras'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('Marinos','Kuriakopoulos','[email protected]','1986','Greece','Alexandroupoli'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('Dimitris','Nikolaou','[email protected]','1984','Greece','Larisa'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('Soufiane','El Kaddouri','[email protected]','1974','France','Xanthi'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('Maria','Apostolou','[email protected]','1997','Greece','Larisa'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('Ioannis','Marinou','[email protected]','1982','Greece','Kalamata'); 
INSERT INTO employee(first_name,last_name,email,born,country,department_name) VALUES('Thanasis','Athanasiou','[email protected]','1989','Cyprus','Heraklion');

这是什么样子后CREATE-INSERT操作：

正如你所看到的，在第9插入员工表，插入失败，我得到我描述的错误，这是：

Cannot add or update a child row: a foreign key constraint fails (`testdb`.`employee`, CONSTRAINT `employee_ibfk_1` FOREIGN KEY (`id`) REFERENCES `department` (`id`))

来源

2017-08-30 ceid-vg

返回部门的外键被设置为员工表中的id是employee_id;它需要引用department_id。我将department_id添加到您的员工表中，并将外键引用更改为department_id。我删除了department_name，因为它是冗余数据。

CREATE TABLE IF NOT EXISTS employee(
    id INT(20) NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    first_name VARCHAR(255) NOT NULL, 
    last_name VARCHAR(255) NOT NULL, 
    email VARCHAR(255) NOT NULL, 
    born INT(20) NOT NULL, 
    country VARCHAR(255) NOT NULL, 
    department_id INT(20) 
    FOREIGN KEY (department_id) references department(id)

）ENGINE = INNODB;

来源

2017-08-30 23:54:14 LAS

在员工插入，我需要分给DEPARTMENT_ID我在部门插入了同样的ID？ –

@ ceid-vg是的，插入员工时使用的department_id与员工所在部门的部门表中的ID相匹配。当您想要显示员工的部门名称时，请加入这两个表格。我同意Jacques的说法，这些ID最好命名为dept_id和emp_id，以使您的代码更具人类可读性。 – LAS

您在employee的外键中使用了错误的字段。为了清楚起见，您应该对每个id字段命名不同。 dept_id和employee_id 然后employee表应该有一个字段称为dept_id（未DEPARTMENT_NAME），这将验证对department：

CREATE TABLE IF NOT EXISTS department(
    dept_id INT(20) NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    name VARCHAR(255) NOT NULL UNIQUE 
)ENGINE=INNODB; 

CREATE TABLE IF NOT EXISTS employee(
    empl_id INT(20) NOT NULL AUTO_INCREMENT PRIMARY KEY, 
    first_name VARCHAR(255) NOT NULL, 
    last_name VARCHAR(255) NOT NULL, 
    email VARCHAR(255) NOT NULL, 
    born INT(20) NOT NULL, 
    country VARCHAR(255) NOT NULL, 
    dept_id INT(20) NOT NULL, 
    FOREIGN KEY (dept_id) references department(dept_id)

）ENGINE = INNODB;

然后，您可以得到department_name通过JOIN

来源

2017-08-30 23:58:08

MySQL - 一对多关系无法正常工作

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