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，联系方式和维护者都是预先存在的表格。然后，我单独运行了前三个插入，并检查了数据是否已插入表中。当我运行整个块时，我得到一个非常普通的错误：

[查询73中的错误]您的SQL语法有错误；检查与您的MySQL服务器版本相对应的手册以获取正确的语法，以在第2行的'@maintainer_franks_maint_id'附近使用执行停止！

编辑-请参见下面的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);

并添加批准的流程：

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;

我希望这可以弄清楚。

Answer 1

对不起，每个人在某种程度上似乎是用户错误。存在两个问题：一个是INSERT INTO approved_process行开头的一个奇怪的不可见字符，而且由于某些原因，Sequel Pro无法分辨出@franks的电子邮件值和@franks之间的区别变量声明的开始。

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

1 个答案: