【发布时间】:2017-08-11 19:21:21
【问题描述】:
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 `tblprojectSkills` (
`ProjectSkillID` int(11) NOT NULL,
`ProjectID` int NOT NULL,
`SkillID` int 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;
在上面的表格中。 SkillID 与 tblSkills 和 tblprojectSkills 相关。
ProjectID 与 Project 和 projectSkills 表相关
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');
}
}
class ProjectSkill_Model extends Model
{
protected $table = "tblprojectskill";
protected $primaryKey = "ProjectSkillID";
public $timestamps = false;
}
class Skill_Model extends Model
{
protected $table = "tblskill";
protected $primaryKey = "SkillID";
public $timestamps = false;
}
laravel 5.1 中的数据库查询如下。
\App\Models\Project\Project_Model
::with('ProjectSkills')
->where('ProjectID', '=', $ProjectID)->first();
问题
我可以获得技能 ID,但是,我如何从技能表中获取技能名称?
【问题讨论】:
-
您错误地定义了关系 - 查看多对多关系:laravel.com/docs/5.1/eloquent-relationships#many-to-many
标签: php laravel-5 laravel-5.1