我有这张桌子:
Attendances
id | member_id | attended_at
1 | 1 | 01/01/2018
2 | 2 | 01/01/2018
3 | 3 | 01/01/2018
4 | 4 | 01/01/2018
5 | 5 | 01/01/2018
Members
id | church_id
1 | 1
2 | 1
3 | 3
4 | 2
5 | 3
Churches
id | church_location
1 | Olongapo
2 | Bataan
3 | Subic
我的问题来自“出勤率”表,我需要知道每个教堂位置有多少人参加,以下是我需要的结果:
Olongapo - 2
Bataan - 1
Subic - 2
您能帮我填写下面的代码吗?
$attendances = Attendance::
答案 0 :(得分:1)
给出该结果的原始SQL将如下所示:
select church_location, count(a.id) from churches as c
inner join members as m on c.id = m.church_id
inner join attendances as a on a.member_id = m.id
group by c.id;
现在要翻译一下,在Laravel中,我将从教会模型开始,而不是相反。
$attendances = Church::with(['members' => function($query) {
$query->join('attendances as a', 'a.member_id', '=', 'members.id');
}])
->select([
'church_location',
DB::raw('count(a.id) as attendance')
]
->groupBy('id')->get();
您可能需要对其进行一些修改,就像我在没有现有模型的情况下将其转换为laravel一样,但是它应该与您想要的接近。原始查询在我尝试时肯定可以正常工作。
-编辑
要注意的另一件事是,您必须同时在模型中设置关系,才能使上面的代码正常工作。
-使用查询生成器进行编辑
DB::table('churches as c')
->select([
'c.church_location',
DB::raw('count(a.id) as attendance')
])
->join('members as m', 'm.church_id', '=', 'c.id')
->join('attendances as a', 'm.member_id', '=', 'm.id')
->groupBy('c.id')
->get();