Question

我想在laravel中建立三个表之间的关系。目前我有三个型号

教室，

public function subjects(){
return $this->belongstoMany('Subject','subject_section_classroom');
}

栏目下，
的主题

我的表格

classrooms(id, name)
sections(id, name)
subjects(id, name)
subject_section_classroom( id, classroom_id, section_id, subject_id)

在我的课堂控制器中我有

public function assignsubjects($class_id, $section_id){
        $classroom = Classroom::find($class_id);
        $section = Section::find($section_id); 
        $subjects = Subject::lists('name','id'); 
        $selected_subjects = $classroom->subjects()->where('section_id', '=', 1);
        $subjects = Subject::lists('name','id'); 
        return view('assignedit', compact('classroom','section','subjects', 'selected_subjects')); 
}

但是我无法从上面的关系中获取selected_subjects。当我试图获取上述查询的sql（使用 - ＆gt; toSQL（））时，我得到了

`"select * from `myschool_subjects` inner join `myschool_subject_section_classroom` on `myschool_subjects`.`id` = `myschool_subject_section_classroom`.`subject_id` where `myschool_subject_section_classroom`.`classroom_id` = ? and `section_id` = ?"`

我无法想象我在这里做错了什么。请帮助。

Answer 1

我认为你的问题来自于 php $selected_subjects = $classroom->subjects()->where('section_id', '=', 1);

将其更改为： php $selected_subjects = $classroom->subjects()->where('section_id', '=', 1)->get();

无法在Laravel数据透视表中获得关系

1 个答案: