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需要一些帮助 我在数据库中有5张桌子,这里是MySQL代码:我如何使用mysql和php匹配数据?
CREATE TABLE IF NOT EXISTS `candidate` (
`CID` int(4) NOT NULL AUTO_INCREMENT,
`title` varchar(5) NOT NULL,
`fname` varchar(30) NOT NULL,
`lname` varchar(30) NOT NULL,
`dob` date NOT NULL,
`email` varchar(50) NOT NULL,
`address` varchar(255) NOT NULL,
`city` varchar(50) NOT NULL,
`postcode` varchar(10) NOT NULL,
`phone_num` varchar(11) NOT NULL,
`username` varchar(40) NOT NULL,
`password` varchar(40) NOT NULL,
`regdate` datetime NOT NULL,
`acc_type` enum('c','s') NOT NULL DEFAULT 'c',
`emailactivate` enum('0','1') NOT NULL DEFAULT '0',
`cv_name` varchar(60) NOT NULL,
`cv` blob NOT NULL,
PRIMARY KEY (`CID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='// this is the table for the candidates' AUTO_INCREMENT=175 ;
-- --------------------------------------------------------
--
-- Table structure for table `candidate_skill`
--
CREATE TABLE IF NOT EXISTS `candidate_skill` (
`CSID` int(4) NOT NULL AUTO_INCREMENT,
`CID` int(4) NOT NULL,
`S_CODE` int(4) NOT NULL,
PRIMARY KEY (`CSID`),
KEY `CID` (`CID`),
KEY `S_CODE` (`S_CODE`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='//match candidate and skill' AUTO_INCREMENT=102 ;
-- --------------------------------------------------------
--
-- Table structure for table `job`
--
CREATE TABLE IF NOT EXISTS `job` (
`JID` int(4) NOT NULL AUTO_INCREMENT,
`job_title` varchar(40) NOT NULL,
`job_desc` varchar(255) NOT NULL,
`start_date` date NOT NULL,
`end_date` date NOT NULL,
PRIMARY KEY (`JID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 COMMENT='// this is the table for the job vacancies' AUTO_INCREMENT=10 ;
-- --------------------------------------------------------
--
-- Table structure for table `skill`
--
CREATE TABLE IF NOT EXISTS `skill` (
`S_CODE` int(4) NOT NULL AUTO_INCREMENT COMMENT '// this is the skill primary key',
`skill_name` varchar(40) NOT NULL,
`skill_desc` varchar(255) NOT NULL,
PRIMARY KEY (`S_CODE`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=11 ;
-- --------------------------------------------------------
--
-- Table structure for table `skill_job`
--
CREATE TABLE IF NOT EXISTS `skill_job` (
`SJID` int(4) NOT NULL AUTO_INCREMENT,
`JID` int(4) NOT NULL,
`S_CODE` int(4) NOT NULL,
PRIMARY KEY (`SJID`),
KEY `S_CODE` (`S_CODE`),
KEY `JID` (`JID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=94 ;
这是查询给我的问题:
$query = ("SELECT * FROM candidate, candidate_skill, job, skill, skill_job WHERE candidate_skill.S_CODE=skill_job.S_CODE AND skill.S_CODE=candidate_skill.S_CODE AND candidate.CID=candidate_skill.CID AND job.JID = skill_job.JID");
$result = mysql_query ($query);
$candidate_id ='';
$canskill ='';
while ($result = mysql_fetch_array ($res)){
$candidate_id .= $r['CID'];
$canskill .= $r ['S_CODE'];
}
}
}
echo $candidate_id;
我想使用他们的技能匹配候选人和工作,并且我有成功的类型,但是即使他们只有一项技能并且工作具有三种技能,即使候选人与工作匹配。 有人可以请指点我在正确的方向,因为我已经搜索,测试并记录了几天但我无法移过这个问题
谢谢你兄弟,你救了我在那里: – 2012-07-15 08:38:55
你能让我知道我可以如何重组整个查询吗? – 2012-07-16 02:51:07