LAST_INSERT_ID() 变量声明的 mysql 通用语法错误 - 难以分析

Question

我正在尝试根据插入的插入 ID 设置变量。所以我写道:

INSERT INTO person (first_name, last_name, middle_names, suffix, title)
VALUES ("Frank", "Thring", "", "", "Mr");
SET @person_frank_thring_id = LAST_INSERT_ID();

INSERT INTO address (line2, line3, postcode, state, suburb)
VALUES ("Dock 3", "22 Boundary Rd", 2088, "NSW", "Mascot");
SET @address_franks_aircraft_maintenance_id = LAST_INSERT_ID();

INSERT INTO contact (email, phone)
VALUES ("info@franks.com", "0245732552");
SET @contact_franks_aircraft_maintenance_id = LAST_INSERT_ID();

INSERT INTO maintainer (abn, address_id, contact_id, nk, name, person_id)
VALUES (73507986550, @address_franks_aircraft_maintenance_id, @contact_franks_aircraft_maintenance_id,
"c49439a4-a24a-4e1b-bc92-ebad6caf5e74", "Frank's Airplane Repairs", @person_frank_thring_id);
SET @maintainer_franks_maint_id = LAST_INSERT_ID();

INSERT INTO approved_process (nk, approval_number, description, expires, maintainer_id)
VALUES ("ee114aab-d201-498d-85a4-2b95e9df8b3d", "13226A", "Welding and Heat Treatment", "2021-05-02", @maintainer_franks_maint_id);

人员、地址、联系人和维护者都是预先存在的表。我单独运行了前三个插入，并检查了数据是否已插入到表中。当我运行整个 block 时，我收到一个非常普遍的错误:

[ERROR in query 73] You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@maintainer_franks_maint_id' at line 2 Execution stopped!

编辑 - 请参阅下面的 DROP 和 CREATE 语句:

DROP TABLE IF EXISTS address;
DROP TABLE IF EXISTS contact;
DROP TABLE IF EXISTS person;
DROP TABLE IF EXISTS maintainer;

CREATE TABLE person (
    id BIGINT NOT NULL AUTO_INCREMENT,
    first_name varchar(255),
    last_name varchar(255),
    middle_names varchar(255),
    suffix varchar(255),
    title varchar(255),
    PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

CREATE TABLE address (
    id BIGINT NOT NULL AUTO_INCREMENT,
    line1 varchar(255),
    line2 varchar(255),
    line3 varchar(255),
    postcode integer,
    state varchar(255),
    suburb varchar(255),
    PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

CREATE TABLE contact (
    id BIGINT NOT NULL AUTO_INCREMENT,
    email varchar(255),
    fax varchar(255),
    mobile varchar(255),
    phone varchar(255),
    address_id bigint,
    PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

CREATE TABLE maintainer (
    id BIGINT NOT NULL AUTO_INCREMENT,
    abn bigint,
    image varchar(255),
    nk varchar(255),
    name varchar(255),
    address_id bigint,
    contact_id bigint,
    person_id bigint,
    PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

ALTER TABLE maintainer 
ADD CONSTRAINT UK_maintainer_nk unique (nk);

ALTER TABLE maintainer 
ADD CONSTRAINT FKs7jo395jusgm3631g7w845wy4 FOREIGN KEY (address_id) REFERENCES address (id);

ALTER TABLE maintainer 
ADD CONSTRAINT FKgixmfq21peg70qtff3q4ktq1 FOREIGN KEY (contact_id) REFERENCES contact (id);

ALTER TABLE maintainer 
ADD CONSTRAINT FK86boj3163qysduc7x3a2m84mh FOREIGN KEY (person_id) REFERENCES person (id);

并添加批准+流程:

CREATE TABLE approved_process (
    id BIGINT NOT NULL AUTO_INCREMENT,
    nk varchar(255),
    approval_number varchar(255),
    description varchar(255),
    expires date,
    maintainer_id bigint,
    subcontractor_id bigint,
    PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8; 

我希望这能澄清事情。

Answer 1

抱歉大家，这似乎是某种程度的用户错误。有两个问题:一个是 INSERT INTOroved_process 行开头有一个奇怪的不可见字符，而且由于某种原因，Sequel Pro 无法区分 @franks 之间的区别> 在电子邮件值中，@franks 在变量声明的开头。