我有三个表:项目,技能和project_skills。在projects表中,我保存了项目的一般数据。第二个表格技能,我持有技能id和技能名称,我也有projects_skills表,这是持有项目的技能关系。以下是表格计划:
CREATE TABLE IF NOT EXISTS `project_skills` (
`project_id` int(11) NOT NULL,
`skill_id` int(11) NOT NULL,
KEY `project_id` (`project_id`),
KEY `skill_id` (`skill_id`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_turkish_ci;
CREATE TABLE IF NOT EXISTS `projects` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`employer_id` int(11) NOT NULL,
`project_title` varchar(100) COLLATE utf8_turkish_ci NOT NULL,
`project_description` text COLLATE utf8_turkish_ci NOT NULL,
`project_budget` int(11) NOT NULL,
`project_allowedtime` int(11) NOT NULL,
`project_deadline` datetime NOT NULL,
`total_bids` int(11) NOT NULL,
`average_bid` int(11) NOT NULL,
`created` datetime NOT NULL,
`active` tinyint(1) NOT NULL,
PRIMARY KEY (`id`),
KEY `created` (`created`),
KEY `employer_id` (`employer_id`),
KEY `active` (`active`),
FULLTEXT KEY `project_title` (`project_title`,`project_description`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_turkish_ci AUTO_INCREMENT=3 ;
CREATE TABLE IF NOT EXISTS `skills` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`category` int(11) NOT NULL,
`name` varchar(100) COLLATE utf8_turkish_ci NOT NULL,
`seo_name` varchar(100) COLLATE utf8_turkish_ci NOT NULL,
`total_projects` int(11) NOT NULL,
PRIMARY KEY (`id`),
KEY `seo_name` (`seo_name`)
) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_turkish_ci AUTO_INCREMENT=224 ;我想选择具有相关技能名称的项目。我想我必须使用联接,但我不知道该怎么做。谢谢
发布于 2010-06-05 14:43:43
听起来像一个JOIN是完全正确的:
SELECT ...
FROM projects
INNER JOIN project_skills ON (project_skills.project_id = projects.id)
INNER JOIN skills ON (skills.id = project_skills.skill_id)发布于 2010-06-05 14:42:43
select * from projects
left join project_skills on projects.id = project_skills.project_id
left join skills on project_skills.skills_id = skills .id注意:您不需要所有的列,但是这将让您在选择所需的列之前看到发生了什么。
https://stackoverflow.com/questions/2980769
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