Question

我正在尝试实现this，但我得到一个循环或多个级联路径，无法查看问题所在以及如何解决问题。

CREATE TABLE Staff(
    Id INT NOT NULL PRIMARY KEY
)

CREATE TABLE Supevision (
    SupervisorId INT NOT NULL FOREIGN KEY REFERENCES Staff(Id) ON UPDATE CASCADE ON DELETE CASCADE,
    SupervisedId INT NOT NULL FOREIGN KEY REFERENCES Staff(Id) ON UPDATE CASCADE ON DELETE CASCADE,
    PRIMARY KEY (SupervisorId, SupervisedId)
)

Answer 1

试试这个：

表：员工

CREATE TABLE IF NOT EXISTS `staff` (
  `id` int(5) NOT NULL AUTO_INCREMENT,
  `name` varchar(15) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT

表：超级视觉

CREATE TABLE IF NOT EXISTS `supevision` (
  `details_id` int(5) NOT NULL AUTO_INCREMENT,
  `SupervisorId` int(5) NOT NULL,
  `SupervisedId` int(5) NOT NULL,
  `is` int(1) NOT NULL DEFAULT '1',
  PRIMARY KEY (`details_id`),
  KEY `SupervisorId` (`SupervisorId`),
  KEY `SupervisedId` (`SupervisedId`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT

查询：

ALTER TABLE `supevision`
  ADD CONSTRAINT `supevision_ibfk_2` FOREIGN KEY (`SupervisedId`) REFERENCES `staff` (`id`) ON DELETE CASCADE ON UPDATE CASCADE,
  ADD CONSTRAINT `supevision_ibfk_1` FOREIGN KEY (`SupervisorId`) REFERENCES `staff` (`id`) ON DELETE CASCADE ON UPDATE CASCADE;

Answer 2

您可以删除其中一个外部约束，但可以用触发器替换它。

DROP TABLE Supevision
DROP TABLE Staff


CREATE TABLE Staff(
    Id INT NOT NULL PRIMARY KEY
)
GO

CREATE TABLE Supevision (
    SupervisorId INT NOT NULL FOREIGN KEY REFERENCES Staff(Id) ON UPDATE CASCADE ON DELETE CASCADE,
    SupervisedId INT NOT NULL --FOREIGN KEY REFERENCES Staff(Id) ON UPDATE CASCADE ON DELETE CASCADE,
    PRIMARY KEY (SupervisorId, SupervisedId)
)
GO

CREATE TRIGGER DELETEStuff
   ON  Staff 
   FOR DELETE
AS 
BEGIN

    DELETE sv
    FROM deleted d
    INNER JOIN Supevision sv
        ON d.Id = sv.SupervisedId 

    DELETE sv
    FROM deleted d
    INNER JOIN Supevision sv
        ON d.Id = sv.SupervisorId 

END
GO

CREATE TRIGGER UPDATEStuff
   ON  Staff 
   FOR UPDATE
AS
BEGIN
    UPDATE Supevision
    SET SupervisedId = inserted.id
    FROM inserted, deleted
    WHERE Supevision.SupervisedId = deleted.Id 

    UPDATE Supevision
    SET SupervisorId = inserted.id
    FROM inserted, deleted
    WHERE Supevision.SupervisorId = deleted.Id 

END
GO

INSERT INTO Staff (ID)
VALUES (1), (2), (3), (4), (5), (6), (7)

INSERT INTO Supevision (SupervisorId, SupervisedId)
VALUES (1, 2), (2, 3), (3, 4), (5, 6), (6, 7)

DELETE FROM Staff WHERE ID = 3

UPDATE Staff SET Id = 60 WHERE Id=6

SELECT * FROM Supevision
SELECT * FROM Staff

递归的多对多关系

2 个答案: