问题描述
我需要查询日期之间的订单,并按日期(如10月,11月)对它们进行分组。 并获得组中total_price和sales_count的总和。 我不需要查询的详细信息,我只需要每个组的结果。
我希望我的刀片桌看起来像这样:
$orderGroups = Order::withCount('orderItems')->filtered($request->all())->get()->groupBy(function($val) {
return Carbon::parse($val->created_at)->format('Y-M');
});
$orders = [];
foreach($orderGroups as $order => $value){
$orders[$order]['total_price'] = $value->sum('total_price');
$orders[$order]['total_order_items'] = $value->sum('order_items_count');
$orders[$order]['avrage_order_items'] = $value->avg('order_items_count');
}
return $orders;
我可以这样得到它:
from typing import List
def longest_chain(submatrix: List[int]) -> int:
"""
Given a list of integers,return the length of the longest chain of 1's
that start from the beginning.
You MUST use a while loop for this! We will check.
>>> longest_chain([1,1,0])
2
>>> longest_chain([0,1])
0
>>> longest_chain([1,1])
1
"""
count = 0
i = 0
while submatrix[i] == 1 and i < len(submatrix)-1:
count += 1
i += 1
return count
def largest_rectangle_at_position(matrix: List[List[int]],x: int,y: int) -> int:
"""
Returns the area of the largest rectangle whose top left corner is at
position <x>,<y> in <matrix>.
You MUST make use of <longest_chain> here as you loop through each row
of the matrix. Do not modify the input matrix.
>>> case1 = [[1,0],[1,1],0]]
>>> largest_rectangle_at_position(case1,0)
4
>>> largest_rectangle_at_position(case1,2,0)
5
>>> largest_rectangle_at_position(case1,2)
6
"""
我还需要按季节和10天将它们分组。您建议如何处理?
解决方法
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