2011-10-10 63 views
0

这里的数据:SELECT返回的数据,但UPDATE给受影响的行:0

CREATE TABLE `Charlies_Sierra_Papa` (
    `Mike` int(5) default NULL, 
    `cpf` char(11) default NULL, 
    `idFoxtrot` int(3) default NULL, 

) ENGINE=MyISAM AUTO_INCREMENT=254098 DEFAULT CHARSET=utf8; 

INSERT INTO `Charlies_Sierra_Papa` VALUES ('1', '12345678910', '12'); 
INSERT INTO `Charlies_Sierra_Papa` VALUES ('2', '11121314157', '12'); 
INSERT INTO `Charlies_Sierra_Papa` VALUES ('3', '57585960610', '12'); 


CREATE TABLE `Charlies` (
    `idCharlie` int(20) NOT NULL auto_increment, 
    `cpf` varchar(11) NOT NULL default '' 
    PRIMARY KEY (`idCharlie`), 
    UNIQUE KEY `cpf` (`cpf`), 
    UNIQUE KEY `idCharlie` (`idCharlie`), 
) ENGINE=MyISAM AUTO_INCREMENT=264670 DEFAULT CHARSET=latin1; 

INSERT INTO `Charlies` VALUES ('1', '12345678910'); 
INSERT INTO `Charlies` VALUES ('2', '11121314157'); 
INSERT INTO `Charlies` VALUES ('3', '57585960610'); 


CREATE TABLE `Mike` (
    `Mike` int(5) unsigned zerofill NOT NULL auto_increment, 
    `idCharlie` int(11) NOT NULL default '0', 
    PRIMARY KEY (`Mike`), 
    UNIQUE KEY `idCharlie` (`idCharlie`) 
) ENGINE=MyISAM AUTO_INCREMENT=12043 DEFAULT CHARSET=latin1; 

INSERT INTO `Mike` VALUES ('00001', '51214'); 
INSERT INTO `Mike` VALUES ('00002', '174135'); 
INSERT INTO `Mike` VALUES ('00003', '203553'); 

这里是SELECT和UPDATE:

UPDATE Charlies_Sierra_Papa AS csp, Charlies AS Cha, Mike_oc AS Mik 
SET csp.cpf = cast(Cha.cpf AS char(11)) 
WHERE csp.cpf = Cha.cpf 
AND Cha.idCharlie = Mik.idCharlie 
AND csp.Mike = Mik.Mike 
AND csp.idFoxtrot = 16 


SELECT * FROM Charlies_Sierra_Papa AS csp, Charlies AS Cha, Mike_oc AS Mik 
/*SET csp.cpf = cast(Cha.cpf AS char(11))*/ 
WHERE csp.cpf = Cha.cpf 
AND Cha.idCharlie = Mik.idCharlie 
AND csp.Mike = Mik.Mike 
AND csp.idFoxtrot = 16 

我的问题是:SELECT返回预期值,但是当我运行它时,UPDATE给了我一个烦人的“Affected rows:0”。

任何线索?

+0

“UPDATE”中*实际*改变了多少行,即在'UPDATE'语句出现后有不同的值?根据[文档](http://dev.mysql.com/doc/refman/5.0/en/update.html),“UPDATE返回实际更改的行数。” – mellamokb

+0

请勿使用隐式SQL连接,使用显式连接语法:'UPDATE Charlies_Sierra_Papa csp INNER JOIN Charlies Cha ON(csp.cpf = Cha.cpf)INNER JOIN Mike_oc Mik ON(Cha.idCharlie = Mik.idCharlie AND csp.Mike = Mik.Mike) SET csp.cpf = cast(Cha.cpf AS char(11)) WHERE csp.idFoxtrot = 16' – Johan

回答

0

这是一个微不足道的错误:

我试图更新csp.cpf = Cha.cpf 但我还搭售csp.cpf = Cha.cpf 因此,没有行更新,并没有雪茄。

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