mysql - 另一个外键问题 - #1451 - 无法删除或更新父行：外键约束失败

Question

在 MySQL 中执行此操作时：

UPDATE `client_therapist` SET `therapist_id` = 3 LIMIT 1 ;

我收到以下错误：

#1451 - Cannot delete or update a parent row: a foreign key constraint fails (`test_structure`.`client_group`, CONSTRAINT `client_group_therapist_id` FOREIGN KEY (`therapist_id`) REFERENCES `client_therapist` (`therapist_id`) ON DELETE NO ACTION ON UPDATE NO ACTION) 

我已经到处搜索并阅读了 StackOverflow 上提出的几个外键问题......无济于事。我认为我目前还没有完全理解外键。

这是我的数据库结构：

导致问题的外键位于 client_group 表上。我的理解是它的意思是“每当添加一行时，治疗师_id 必须与 client_treatment 表中的治疗师_id 匹配。”

只是输入这个让我意识到问题的一部分是当添加 client_group 时，它可能会绑定到 client_treatment 表中的某一行。因此，如果我尝试更新那个 client_treatment 行，它就会吓坏了。所以我想我的问题是 - 我该如何解决这个问题？我需要更新 client_therapy 表中某一行的 treatment_id，但只要 client_group 表中有一行，我似乎就不能。

这是我的表格，上面有一些虚拟信息：

CREATE SCHEMA IF NOT EXISTS `test_structure` DEFAULT CHARACTER SET utf8 ;
USE `test_structure` ;

-- -----------------------------------------------------
-- Table `test_structure`.`user`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `test_structure`.`user` (
  `user_id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT ,
  `email` VARCHAR(80) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NOT NULL ,
  `password` VARCHAR(40) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NOT NULL,
  `display_name` VARCHAR(60) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `first_name` VARCHAR(45) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NOT NULL ,
  `last_name` VARCHAR(45) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NOT NULL ,
  `street_address` VARCHAR(120) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `city` VARCHAR(45) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `state` CHAR(2) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `zip` VARCHAR(10) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `work_phone` VARCHAR(20) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `cell_phone` VARCHAR(20) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `home_phone` VARCHAR(20) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `date_created` TIMESTAMP NULL DEFAULT NULL ,
  `expired` TINYINT(1) NOT NULL DEFAULT '0' ,
  `date_last_login` TIMESTAMP NULL DEFAULT CURRENT_TIMESTAMP ,
  `login_count` INT(10) UNSIGNED NULL DEFAULT '0' ,
  PRIMARY KEY (`user_id`) ,
  UNIQUE INDEX `email` (`email` ASC) )
ENGINE = InnoDB
AUTO_INCREMENT = 0
DEFAULT CHARACTER SET = utf8
COLLATE = utf8_unicode_ci, 
COMMENT = 'Users table.  All Therapists, Clients, Admins and others wil' ;


-- -----------------------------------------------------
-- Table `test_structure`.`client_therapist`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `test_structure`.`client_therapist` (
  `client_therapist_id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT ,
  `client_id` INT(10) UNSIGNED NOT NULL ,
  `therapist_id` INT(10) UNSIGNED NOT NULL ,
  `start_date` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP ,
  `end_date` TIMESTAMP NULL DEFAULT NULL ,
  `therapist_approval` TINYINT(1) NOT NULL DEFAULT '0' ,
  `date_client_agreed_to_terms` TIMESTAMP NULL DEFAULT NULL ,
  PRIMARY KEY (`client_therapist_id`) ,
  INDEX `client_therapist_client_id` (`client_id` ASC) ,
  INDEX `client_therapist_therapist_id` (`therapist_id` ASC) ,
  CONSTRAINT `client_therapist_client_id`
    FOREIGN KEY (`client_id` )
    REFERENCES `test_structure`.`user` (`user_id` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `client_therapist_therapist_id`
    FOREIGN KEY (`therapist_id` )
    REFERENCES `test_structure`.`user` (`user_id` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB
AUTO_INCREMENT = 0
DEFAULT CHARACTER SET = utf8
COLLATE = utf8_unicode_ci, 
COMMENT = 'This defines the relationship between a client and therapist' ;



-- -----------------------------------------------------
-- Table `test_structure`.`client_group`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `test_structure`.`client_group` (
  `client_group_id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT ,
  `therapist_id` INT(10) UNSIGNED NOT NULL ,
  `title` VARCHAR(128) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NOT NULL ,
  `description` VARCHAR(255) CHARACTER SET 'utf8' COLLATE 'utf8_unicode_ci' NULL DEFAULT NULL ,
  `date_added` TIMESTAMP NULL DEFAULT CURRENT_TIMESTAMP ,
  PRIMARY KEY (`client_group_id`) ,
  INDEX `client_group_therapist_id` (`therapist_id` ASC) ,
  CONSTRAINT `client_group_therapist_id`
    FOREIGN KEY (`therapist_id` )
    REFERENCES `test_structure`.`client_therapist` (`therapist_id` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB
AUTO_INCREMENT = 0
DEFAULT CHARACTER SET = utf8
COLLATE = utf8_unicode_ci;


-- -----------------------------------------------------
-- Table `test_structure`.`client_group_member`
-- -----------------------------------------------------
CREATE  TABLE IF NOT EXISTS `test_structure`.`client_group_member` (
  `client_group_member_id` INT(10) UNSIGNED NOT NULL AUTO_INCREMENT ,
  `client_group_id` INT(10) UNSIGNED NOT NULL ,
  `client_id` INT(10) UNSIGNED NOT NULL ,
  PRIMARY KEY (`client_group_member_id`) ,
  INDEX `client_group_id` (`client_group_id` ASC) ,
  INDEX `client_group_member_id` (`client_id` ASC) ,
  CONSTRAINT `client_group_id`
    FOREIGN KEY (`client_group_id` )
    REFERENCES `test_structure`.`client_group` (`client_group_id` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION,
  CONSTRAINT `client_group_member_id`
    FOREIGN KEY (`client_id` )
    REFERENCES `test_structure`.`client_therapist` (`client_id` )
    ON DELETE NO ACTION
    ON UPDATE NO ACTION)
ENGINE = InnoDB
AUTO_INCREMENT = 0
DEFAULT CHARACTER SET = utf8
COLLATE = utf8_unicode_ci;


-- -----------------------------------------------------
-- INSERT TEST DATA
-- -----------------------------------------------------
INSERT INTO `test_structure`.`user` (`user_id`, `email`, `password`, `display_name`, `first_name`, `last_name`, `street_address`, `city`, `state`, `zip`, `work_phone`, `cell_phone`, `home_phone`, `date_created`, `expired`, `date_last_login`, `login_count`) VALUES (NULL, 'dr@dr.com', SHA1('12345'), 'Dr. Test', 'Dr.', 'Test', NULL, NULL, NULL, NULL, NULL, NULL, NULL, NOW(), '0', CURRENT_TIMESTAMP, '0');

SET @therapist_id = LAST_INSERT_ID();

INSERT INTO `test_structure`.`user` (`user_id`, `email`, `password`, `display_name`, `first_name`, `last_name`, `street_address`, `city`, `state`, `zip`, `work_phone`, `cell_phone`, `home_phone`, `date_created`, `expired`, `date_last_login`, `login_count`) VALUES (NULL, 'client@client.com', '12345', 'Client', 'Client', 'Client', NULL, NULL, NULL, NULL, NULL, NULL, NULL, NOW(), '0', CURRENT_TIMESTAMP, '0');

SET @client_id = LAST_INSERT_ID();

INSERT INTO `test_structure`.`client_therapist` (`client_therapist_id`, `client_id`, `therapist_id`, `start_date`, `end_date`, `therapist_approval`, `date_client_agreed_to_terms`) VALUES (NULL, @client_id, @therapist_id, CURRENT_TIMESTAMP, NOW(), '0', NOW());

INSERT INTO `test_structure`.`client_group` (`client_group_id`, `therapist_id`, `title`, `description`, `date_added`) VALUES (NULL, @therapist_id, 'Test', NULL, CURRENT_TIMESTAMP);

SET @group_id = LAST_INSERT_ID();

INSERT INTO `test_structure`.`client_group_member` (`client_group_member_id`, `client_group_id`, `client_id`) VALUES (NULL, @group_id, @client_id);

然后运行这段代码，看看foreign_key问题：

INSERT INTO `test_structure`.`user` (`user_id`, `email`, `password`, `display_name`, `first_name`, `last_name`, `street_address`, `city`, `state`, `zip`, `work_phone`, `cell_phone`, `home_phone`, `date_created`, `expired`, `date_last_login`, `login_count`) VALUES (NULL, 'newdr@newdr.com', SHA1('12345'), 'New Dr', 'Dr.', 'New', NULL, NULL, NULL, NULL, NULL, NULL, NULL, NOW(), '0', CURRENT_TIMESTAMP, '0');

SET @new_therapist_id = LAST_INSERT_ID();

UPDATE `test_structure`.`client_therapist` SET `therapist_id` = @new_therapist_id LIMIT 1;

非常感谢所有帮助/说明/指导！

更新

我遇到的问题是因为 client_group 表中的 therapy_id 绑定到 client_treatment 表中的 treatment_id 而不是 user 表。该 therapy_id 可以在 client_treatment 和 client_group 表中多次存在，但在 user 表中只存在一次，因此只要 client_group 表中仍有一行引用该 therapy_id，我就永远无法更新 client_treatment 表中的该列。

我应该做的是将 therapy_id 绑定到 user 表而不是 client_treatment 表，因为最好让外键引用每个表上的主 ID（这是真的吗？）。

现在我必须学习如何删除/编辑/更新表上的外键... :)

Answer 1

您需要使用该选项声明的外键ON UPDATE CASCADE。

以下是如何在表上删除和添加外键的示例：

ALTER TABLE client_group drop FOREIGN KEY `client_group_therapist_id`, 
  ADD FOREIGN KEY `client_group_therapist_id` (`therapist_id`) 
  REFERENCES `client_therapist` (`therapist_id`) ON DELETE NO ACTION ON UPDATE CASCADE;

然后，如果您更改 中的值client_therapist.therapist_id，它将自动更新 中的值client_group.therapist_id。

这种变化将是同时的和原子的——也就是说，没有其他同时运行的查询能够看到一个表中的数据发生了变化，而另一个表中的数据没有发生变化。

Answer 2

外键不正确，因为它引用了另一个表中的非唯一列。

我应该做的是将 therapy_id 绑定到 user 表而不是 client_treatment 表，因为最好让外键引用每个表上的主 ID。

ALTER TABLE client_group drop FOREIGN KEY `client_group_therapist_id`, 
ADD FOREIGN KEY `client_group_therapist_id` (`therapist_id`) 
REFERENCES `user` (`user_id`) ON DELETE NO ACTION ON UPDATE NO ACTION;