4 回答

TA貢獻(xiàn)1785條經(jīng)驗(yàn) 獲得超8個(gè)贊
您可以使用 laravel sum 函數(shù)
$deliveryCostTotal = Order::where('place_id',1)->sum('delivery_fees');
https://laravel.com/docs/6.x/queries#aggregates

TA貢獻(xiàn)1865條經(jīng)驗(yàn) 獲得超7個(gè)贊
使用 Mysql SUM
$deliveryCostTotal = Order::where('place_id',1)->selectRaw('SUM(delivery_fees) AS cost_total')->value('cost_total');

TA貢獻(xiàn)1898條經(jīng)驗(yàn) 獲得超8個(gè)贊
請(qǐng)嘗試這樣
$deliveryFees = Order::where('place_id',1)->sum('delivery_fees');

TA貢獻(xiàn)1895條經(jīng)驗(yàn) 獲得超3個(gè)贊
隨著您的問題得到解決,但我認(rèn)為我的答案是消除您的困惑array_sum并在數(shù)組中添加數(shù)據(jù)。因?yàn)槟皇菍⒅翟O(shè)置為數(shù)組并覆蓋它,所以用于array_push在數(shù)組中添加數(shù)據(jù)。
$orders = Order::where('place_id',1)->select('delivery_fees')->get()->toArray();
$orderArray = [];
foreach ($orders as $order)
{
array_push($orderArray,$order['delivery_fees']);
}
$deliveryCostTotal = array_sum($orderArray);
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