SQL Joining query using Selection - mysql

Question

I have one table of tbl_order where columns are :
id_order
username
delivery_date etc...
and other table tbl_order_foods where columns are :
id_order
id_food
food_quantity etc
Here the sample pic of tbl_order table
and tbl_order_foods table
Here i am want to select all the foods with quantity with price and unit based on delivery_date.
Ex: Suppose On 2020-04-18 there are 3 orders where foods: tomato's are ordered total 15 kg (3 orders * 5 quantity * food_min_unit_amount * unit) and so on others foods.
how i can get the foodlist of total ordered quantity based on delivery_date

Answer 1

The method you need is the GROUP BY keyword.
SQL grouping works on the given attributes. According to your question this would be the delivery_date. In the projection (the attributes following the SELECT keyword) you then can use attributes you state after the GROUP BY keyword and aggregation functions e.g. SUM and MAX.
Based on your question you could get the total price for all orders on the given date like this:
SELECT order.delivery_date, SUM(food.food_quantity) as amount, SUM(food.food_total_price) as revenue
FROM tbl_order as order
INNER JOIN tbl_order_foods as food ON order.id_order = food.id_order
GROUP BY order.delivery_date
This would result in a list like this:
date | amount | revenue
------------------------
05.01| 10 | 800.00
I don't know whether this is what you wish. If you also want to split it into order positions then you would GROUP BY the food_name as well and add it to the projection which then results in the SUM grouped by the orders on a given day of a given product.
SELECT order.delivery_date, food.food_name, SUM(food.food_quantity) as amount, SUM(food.food_total_price) as revenue
FROM tbl_order as order
INNER JOIN tbl_order_foods as food ON order.id_order = food.id_order
GROUP BY order.delivery_date, food.food_name
Which would result in something like this.
date | food_name | amount | revenue
-----------------------------------
05.01| Tomato | 10 | 800.00
05.01| Apple | 5 | 400.00

Answer 2

Check out GROUP BY with the SUM function, a good article from Tutorialspoint.com is here. The only difference is that you are performing a JOIN to create the resulting table.
SELECT DISTINCT [f.id_food], [o.delivery_date], SUM(f.food_quantity) as [delivered_sum_qty], SUM(f.food_total_price) as [sum_food_total_price]
FROM tbl_order o LEFT JOIN tbl_order_foods f
ON [o.id_order] = [f.id_order]
GROUP BY [o.delivery_date];
If you need it, a pretty clear explanation of how to use JOIN is here as well.