我有三个数据库表
CREATE TABLE `tblproject` (
`ProjectID` int(11) NOT NULL,
`ProjectStatusID` varchar(30) NOT NULL,
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
CREATE TABLE `tblSkills` (
`SkillID` int(11) NOT NULL,
`Skill` varchar(100) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
CREATE TABLE `tblprojectSkills` (
`ProjectSkillID` int(11) NOT NULL,
`ProjectID` int NOT NULL,
`SkillID` int NOT NULL,
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
在上表中。CCD_ 1在CCD_ 2和CCD_。ProjectID
在Project
和projectSkills
表中相关
我的项目模型如下
class Project_Model extends Model
{
protected $table = "tblproject";
protected $primaryKey = "ProjectID";
public $timestamps = false;
public function ProjectSkills() {
return $this->hasMany(''App'Models'ProjectSkill_Model', 'ProjectID');
}
}
下面是laravel 5.1中的数据库查询
'App'Models'Project'Project_Model
::with('ProjectSkills')
->where('ProjectID', '=', $ProjectID)->first();
问题
我可以获得技能ID,但是,如何从技能表中获得技能名称?
您可以使用闭包选择所需的字段:
'App'Models'Project'Project_Model
::with('ProjectSkills' => function($q)
{
$q->select('SkillID', 'Skill');
})
->where('ProjectID', '=', $ProjectID)->first();
或者,您可以直接在模型的关系中添加您想要的字段:
public function ProjectSkills() {
return $this->hasMany(''App'Models'ProjectSkill_Model', 'ProjectID')
->select('SkillID', 'Skill');
}