如何使用 laravel 数据库查询为 sql 中的分组项目命名?
How to give a name to the grouped items in sql with laravel db query?
使用这段代码我得到了这个输出:
$q = DB::table('payments')
->join('agency', 'agency.id', '=', 'payments.agency')
->join('paymenttype','paypemttype.id', '=', 'payments.paymenttype')
->select(DB::raw('sum(payment) as sum'), 'agency.agency', 'paymenttype.paymenttype')
->where('payments.school', '=', Sentry::getUser()->school)
->groupBy('payments.agency')
->groupBy('payments.paymenttype')
->get();
输出为:
[
{
sum: 200,
agency: "city1",
paymenttype: "credit card"
},
{
sum: 200,
agency: "city1",
paymenttype: "transfer"
},
{
sum: 200,
agency: "city2",
paymenttype: "credit card"
},
{
sum: 200,
agency: "city2",
paymenttype: "transfer"
}
]
但是我想要table左手边那个城市,所以我想要这种输出
- city1 [
{
sum: 200,
paymenttype: "credit card"
},
{
sum: 200,
paymenttype: "transfer"
}],
- city2 [
{
sum: 200,
paymenttype: "credit card"
},
{
sum: 200,
paymenttype: "transfer"
}
]
所以我的意思是对象中的对象我想得到这样的数据。
如果你愿意帮助我,我会很高兴,再次感谢你的帮助。
您可以使用keyBy
方法。
由于您没有使用 Eloquent,您可以自己创建一个 collection。
Laravel 5:
$result = collect($q)->keyBy('agency');
Laravel 4:
use Illuminate\Support\Collection;
$result = Collection::make($q)->keyBy('agency');
使用这段代码我得到了这个输出:
$q = DB::table('payments')
->join('agency', 'agency.id', '=', 'payments.agency')
->join('paymenttype','paypemttype.id', '=', 'payments.paymenttype')
->select(DB::raw('sum(payment) as sum'), 'agency.agency', 'paymenttype.paymenttype')
->where('payments.school', '=', Sentry::getUser()->school)
->groupBy('payments.agency')
->groupBy('payments.paymenttype')
->get();
输出为:
[
{
sum: 200,
agency: "city1",
paymenttype: "credit card"
},
{
sum: 200,
agency: "city1",
paymenttype: "transfer"
},
{
sum: 200,
agency: "city2",
paymenttype: "credit card"
},
{
sum: 200,
agency: "city2",
paymenttype: "transfer"
}
]
但是我想要table左手边那个城市,所以我想要这种输出
- city1 [
{
sum: 200,
paymenttype: "credit card"
},
{
sum: 200,
paymenttype: "transfer"
}],
- city2 [
{
sum: 200,
paymenttype: "credit card"
},
{
sum: 200,
paymenttype: "transfer"
}
]
所以我的意思是对象中的对象我想得到这样的数据。 如果你愿意帮助我,我会很高兴,再次感谢你的帮助。
您可以使用keyBy
方法。
由于您没有使用 Eloquent,您可以自己创建一个 collection。
Laravel 5:
$result = collect($q)->keyBy('agency');
Laravel 4:
use Illuminate\Support\Collection;
$result = Collection::make($q)->keyBy('agency');