如何排序具有关系的记录在查询结果中排在最后

Question

我有两个表，名为 calls 和 call_histories。例如下面是我的查询。

$calls = Call::select([
    'calls.id as id', 'calls.reserved_at', 'calls.app_id',
    'calls.call_type_id', 'call_histories.call_id'
])->leftJoin('call_histories', 'calls.id', '=', 'call_histories.call_id')
    ->whereIn('call_type_id', $callTypesIds)
    ->where(DB::raw("(DATE_FORMAT(reserved_at,'%Y-%m-%d'))"), '<=', 
        Carbon::now()->format('Y-m-d'))
    ->where('app_id', $callAppId)
    ->free()
    ->orderBy('reserved_at')
    ->take(100);

我想知道一个电话是否有任何call_histories订单作为查询结果中的最后记录。我该怎么做？

Answer 1

因为如果找不到 call_histories.call_id 将为 null，您可以按此排序：

$calls = Call::select([
            'calls.id as id', 'calls.reserved_at', 'calls.app_id', 'calls.call_type_id', 'call_histories.call_id'
        ])->leftJoin('call_histories', 'calls.id', '=', 'call_histories.call_id')
            ->whereIn('call_type_id', $callTypesIds)
            ->where(DB::raw("(DATE_FORMAT(reserved_at,'%Y-%m-%d'))"), '<=', Carbon::now()->format('Y-m-d'))
            ->where('app_id', $callAppId)
            ->free()
            ->orderBy('reserved_at')
            ->orderBy('call_id')
            ->take(100);

如何排序具有关系的记录在查询结果中排在最后

How to order records with having a relationship comes last in query result

php

mysql

laravel

laravel-query-builder

eloquent-relationship