Mysql InnoDB外键定义无法解析列名

Mysql InnoDB外键定义无法解析列名,mysql,foreign-keys,innodb,Mysql,Foreign Keys,Innodb,在两个表之间创建外键约束时遇到问题 这是第一张表: CREATE TABLE `agews_rifiuti_cer` ( `id_cer` int(10) unsigned NOT NULL AUTO_INCREMENT, `agews_id` int(10) unsigned DEFAULT '0', `livello` tinyint(4) DEFAULT '1', `codice` varchar(10) DEFAULT NULL, `descrizione` varc

在两个表之间创建外键约束时遇到问题

这是第一张表:

CREATE TABLE `agews_rifiuti_cer` (
  `id_cer` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `agews_id` int(10) unsigned DEFAULT '0',
  `livello` tinyint(4) DEFAULT '1',
  `codice` varchar(10) DEFAULT NULL,
  `descrizione` varchar(255) DEFAULT NULL,
  `note` text,
  `flag_pericoloso` tinyint(1) DEFAULT '0',
  `id_cliente` int(10) unsigned DEFAULT '1',
  `flag_modificato` char(1) DEFAULT 'N',
  PRIMARY KEY (`id_cer`),
  KEY `fk_id_cliente_agews_sgs_codici_cer` (`id_cliente`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;
这是第二个:

CREATE TABLE `lin_98_47_rifiuti` (
  `id_rifiuto` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `id_azienda` int(10) unsigned NOT NULL DEFAULT '1',
  `id_sede` int(10) unsigned NOT NULL DEFAULT '1',
  `revisione_documento` int(10) unsigned NOT NULL DEFAULT '0',
  `rifiuto` varchar(255) DEFAULT NULL,
  `codice_cer` varchar(10) DEFAULT NULL,
  `nome_interno` varchar(255) DEFAULT NULL,
  `descrizione` text,
  `materie_prime` text,
  `contenitore` text,
  `deposito` text,
  `data_ultima_analisi` date DEFAULT NULL,
  `stato_fisico` enum('Solido pulverulento','Solido non pulverulento','Fangoso palabile','Liquido') DEFAULT 'Solido non pulverulento',
  `quantita` float(9,1) DEFAULT '0.0',
  `unita_misura` enum('Kg','l','mc') DEFAULT 'Kg',
  `id_pericolo` int(10) unsigned DEFAULT NULL,
  `destino` enum('Recupero','Smaltimento') DEFAULT NULL,
  `id_recupero` int(10) unsigned DEFAULT NULL,
  `id_smaltimento` int(10) unsigned DEFAULT NULL,
  `id_cer` int(10) unsigned DEFAULT NULL,
  `immagine` varchar(255) DEFAULT NULL,
  `image_type` varchar(20) DEFAULT NULL,
  `image_content` mediumblob,
  `image_size_x` smallint(5) unsigned DEFAULT '0',
  `image_size_y` smallint(5) unsigned DEFAULT '0',
  `flag_storico` tinyint(1) DEFAULT '0',
  `id_responsabile` int(10) unsigned DEFAULT '0',
  `nome_responsabile` varchar(255) DEFAULT '0',
  `id_ultima_modifica` int(10) unsigned DEFAULT '0',
  `create_log` tinyint(1) DEFAULT '1',
  PRIMARY KEY (`id_rifiuto`,`id_azienda`,`id_sede`,`revisione_documento`),
  KEY `fk_main_lin_98_47_rifiuti` (`id_azienda`,`id_sede`,`revisione_documento`),
  KEY `fk_id_responsabile_lin_98_47_rifiuti` (`id_responsabile`,`id_azienda`,`id_sede`,`revisione_documento`,`nome_responsabile`),
  KEY `fk_id_pericolo_lin_98_47_rifiuti` (`id_pericolo`),
  KEY `fk_id_recupero_lin_98_47_rifiuti` (`id_recupero`),
  KEY `fk_id_smaltimento_lin_98_47_rifiuti` (`id_smaltimento`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;
问题发生在我尝试这样做时:

ALTER TABLE lin_98_47_rifiuti ADD CONSTRAINT fk_id_cer_lin_98_47_rifiuti FOREIGN KEY (id_cer) REFERENCES agews_rifiuti_cer(id_cer) ON UPDATE CASCADE ON DELETE CASCADE;
我得到了这个错误:

#1005 - Can't create table 'db_626suite.#sql-71c_13d5' (errno: 150)
而SHOW INNODB STATUS命令显示:

Error in foreign key constraint of table db_626suite/#sql-71c_13d5:
 FOREIGN KEY (id_cer) REFERENCES agews_rifiuti_cer(id_cer) ON UPDATE CASCADE ON DELETE CASCADE:
Cannot resolve column name close to:
) ON UPDATE CASCADE ON DELETE CASCADE
但是语法和字段定义对我来说似乎是正确的。我做错了什么

编辑:


<强> Marc B 表明,这可能是由于<代码> IDOCER 在<代码> AgWWSH RIFIUTUSTER CE/<代码>中被定义为<代码> NULL ,但我不认为是这样,事实上请考虑另一个表:

CREATE TABLE `agews_rifiuti_pericolo` (
  `id_pericolo` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `agews_id` int(10) unsigned DEFAULT '0',
  `codice` varchar(255) DEFAULT NULL,
  `pericolo` varchar(255) DEFAULT NULL,
  `note` text,
  `id_cliente` int(10) unsigned DEFAULT '1',
  `flag_modificato` char(1) DEFAULT 'N',
  PRIMARY KEY (`id_pericolo`),
  KEY `codice_rifiuti_recupero` (`codice`),
  KEY `fk_id_cliente_rifiuti_pericolo` (`id_cliente`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
在这里,
id\u pericolo
被定义为上面的
notnull
as
id\u cer
,但这是完美的:

ALTER TABLE lin_98_47_rifiuti ADD CONSTRAINT fk_id_pericolo_lin_98_47_rifiuti FOREIGN KEY (id_pericolo) REFERENCES agews_rifiuti_pericolo(id_pericolo) ON UPDATE CASCADE ON DELETE SET NULL;
即使在
lin\u 98\u 47\u rifiuti
中,字段
id\u pericolo
被定义为
默认空值

编辑2:

我刚刚尝试了这个最小设置:

CREATE TABLE `cer` (
  `id_cer` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `codice` varchar(10) DEFAULT NULL,
  PRIMARY KEY (`id_cer`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

CREATE TABLE `rifiuti` (
  `id_rifiuto` int(10) unsigned NOT NULL AUTO_INCREMENT,
  `id_cer` int(10) unsigned NULL DEFAULT NULL,
  `rifiuto` varchar(10) DEFAULT NULL,
  PRIMARY KEY (`id_rifiuto`)
) ENGINE=InnoDB  DEFAULT CHARSET=utf8;

ALTER TABLE rifiuti ADD FOREIGN KEY (id_cer) REFERENCES cer(id_cer) ON UPDATE CASCADE ON DELETE CASCADE;
这一切都正常工作,但我不明白我最初的查询中可能存在什么问题


我还认为,这证明了在lin_98_47表中将
id\u cer
定义为
DEFAULT NULL
不是将字段定义为
not NULL
DEFAULT NULL

的问题,而是将另一个表中匹配的FK字段定义为
not NULL

字段定义必须与要建立的外键关系完全匹配,这包括可空性:

`id_cer` int(10) unsigned NOT NULL AUTO_INCREMENT,

`id_cer` int(10) unsigned DEFAULT NULL,

我真的不明白为什么,但将所有表中的
id\u cer
重命名为
id\u codice
,同时保持其定义不变,解决了问题


因此,MySQL似乎对特定的字段名有一些问题。

您需要什么样的最小样本?这里不是插入数据的问题,只是表定义,我已经给出了所有的规范。在一个只有引起问题所需的列的表上进行类似的插入。您甚至可以在设置时自己找到解决方案。我不认为是这种情况,请参阅我的编辑。还有,我想我已经做了几百次了。。。