请帮帮我!
我遇到了这个错误。
Error: Cannot add or update a child row: a foreign key constraint fails (world.alarmes, CONSTRAINT fk_alarmes_registos1 FOREIGN KEY (idRegisto) REFERENCES registos (idRegisto) ON DELETE NO ACTION ON UPDATE NO ACTION)
我有这些表格。
CREATE TABLE `registos` (
`data_registo` char(10) NOT NULL,
`hora_registo` time NOT NULL,
`idSensor` varchar(8) NOT NULL,
`Temperatura` char(6) DEFAULT NULL,
`Humidade` char(6) DEFAULT NULL,
`pt_orvalho` char(6) DEFAULT NULL,
`idRegisto` int(11) NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`idRegisto`,`idSensor`,`data_registo`,`hora_registo`),
KEY `fk_registos_sensores1_idx` (`idSensor`),
CONSTRAINT `fk_registos_sensores1` FOREIGN KEY (`idSensor`) REFERENCES `sensores` (`idSensor`) ON DELETE NO ACTION ON UPDATE NO ACTION
)
CREATE TABLE `alarmes` (
`idAlarme` int(11) NOT NULL AUTO_INCREMENT,
`descricao_alarme` varchar(45) DEFAULT NULL,
`data_criacao` datetime DEFAULT CURRENT_TIMESTAMP,
`idRegisto` int(11) NOT NULL DEFAULT ''0'',
PRIMARY KEY (`idAlarme`,`idRegisto`),
KEY `fk_alarmes_registos1_idx` (`idRegisto`),
CONSTRAINT `fk_alarmes_registos1` FOREIGN KEY (`idRegisto`) REFERENCES `registos` (`idRegisto`) ON DELETE NO ACTION ON UPDATE NO ACTION
)
当我在表格中插入记录时会弹出错误。
insert into registos values ('2014-03-31', '14:03:32', 'BrgTH032', '22.3', '45.3', '9.9', '32');
如果我这样做:
SET FOREIGN_KEY_CHECKS=0
下一次插入已经被接受,但是当我再次尝试时。回来给出同样的错误。
我一直在研究并失败,因为registos表引用了sensores表中的外键。如果表中没有相应的条目,则无法直接插入关系表。
但我不知道如何解决这个问题。
请帮帮我。
-------编辑(我使用触发器填充表Alarmes)------------------------
DELIMITER $$
create TRIGGER alerta
BEFORE INSERT ON registos
FOR EACH ROW
begin
Set @tempmax=0;
Set @tempmin=0;
select lim_inf_temp, lim_sup_temp into @tempmin, @tempmax from sensores where idSensor=NEW.idSensor;
Set @maxidAlarme=0;
if (CAST(NEW.Temperatura AS UNSIGNED)<@tempmin) then
SELECT MAX(idAlarme) into @maxidAlarme FROM alarmes;
SET @maxidAlarme=@maxidAlarme+1;
INSERT INTO alarmes(idAlarme,descricao_alarme, idRegisto) VALUES (@maxidAlarme,"temperatura inserida inferior ao normal",New.idRegisto);
end if;
if (CAST(NEW.Temperatura AS UNSIGNED)>@tempmax) then
SELECT MAX(idAlarme) into @maxidAlarme FROM alarmes;
SET @maxidAlarme=@maxidAlarme+1;
INSERT INTO alarmes(idAlarme,descricao_alarme, idRegisto) VALUES (@maxidAlarme,"temperatura inserida superior ao normal",New.idRegisto);
end if;
end $$;
DELIMITER ;
答案 0 :(得分:2)
您试图在表格中插入的值多于允许的7(七)但预期的6(六)。
请始终在“插入”查询中包含您要插入的列。
此表格中有七列,但其中一列是'自动增量'列,因此应该有 6(六个)值插入查询。
CREATE TABLE `registos` (
`data_registo` char(10) NOT NULL,
`hora_registo` time NOT NULL,
`idSensor` varchar(8) NOT NULL,
`Temperatura` char(6) DEFAULT NULL,
`Humidade` char(6) DEFAULT NULL,
`pt_orvalho` char(6) DEFAULT NULL,
`idRegisto` int(11) NOT NULL AUTO_INCREMENT,
PRIMARY KEY (`idRegisto`,`idSensor`,`data_registo`,`hora_registo`),
KEY `fk_registos_sensores1_idx` (`idSensor`),
CONSTRAINT `fk_registos_sensores1
这是'insert'查询:
insert into registos values ('2014-03-31', '14:03:32', 'BrgTH032', '22.3', '45.3', '9.9', '32');
有七个值但您希望查询看起来像(添加了列):
insert into registos (data_registo, hora_registo, idSensor, Temperatura, Humidade, pt_orvalho)
values ('2014-03-31', '14:03:32', 'BrgTH032', '22.3', '45.3', '9.9', '32');
我建议查询应该是:
insert into registos (data_registo, hora_registo, idSensor, Temperatura, Humidade, pt_orvalho)
values ('2014-03-31', '14:03:32', 'BrgTH032', '22.3', '45.3', '9.9');
'registos'上的触发器必须是插入触发器才能获取新的'idRegisto'值。
create TRIGGER alerta
AFTER INSERT ON registos
FOR EACH ROW
begin