Question

我在MySQL中创建触发器时遇到问题。我的＆＃34; tb_pessoa＆＃34;表存储我的用户＆＃39;数据。我希望在制作新插件时，它会存储在＆＃34; tb_email＆＃34;列出用户的电子邮件，检查它是否已经存在，但是当用光标执行查询时，它表示光标没有打开，它就是。它能是什么？

CREATE DEFINER=`root`@`localhost` TRIGGER `db_agenda`.`tb_pessoa_AFTER_INSERT` AFTER INSERT ON `tb_pessoa` FOR EACH ROW
BEGIN
    declare id_email, id_tipo_email int(11);
    declare nm_email varchar(100);
    declare cr_email cursor for
    select id_email, nm_email, id_tipo_email
        from tb_email
            where nm_email like new.nm_email;
    declare continue handler for sqlstate '02000'
  open cr_email;
  lp_email: loop
    fetch cr_email into id_email, nm_email, id_tipo_email;
    if (id_email = null) then
      insert into tb_email set
        id_email = (select max(id_email)+1 from tb_email),
        nm_email = nm_email,
        id_tipo_email = 1;
        close cr_email;
        leave lp_email;
    end if;
  end loop lp_email;
  close cr_email;
END

Answer 1

您不需要使用游标，只需执行一个INSERT来获取SELECT的值。

CREATE TRIGGER tb_pessoa_AFTER_INSERT AFTER INSERT ON tb_pessoa
ON EACH ROW
    INSERT INTO tb_email (nm_email, id_tipo_email, id_email)
    SELECT e.nm_email, 1, @id_email := @id_email + 1
    FROM tb_email AS e
    CROSS JOIN (SELECT @id_email := MAX(id_email) FROM tb_email) AS i
    WHERE e.nm_email LIKE NEW.nm_email

此外，如果NEW.nm_email不包含任何通配符%或_，则应使用=代替LIKE。

触发错误1326 MySQL

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