This community-built FAQ covers the “Product Sum 2” exercise from the lesson “Advanced Aggregates”.
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This exercise can be found in the following Codecademy content:
SQL: Analyzing Business Metrics
FAQs on the exercise Product Sum 2
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hello to everybody,
i need heipl. can someone tell me, why in the second values in select we don’t need to specify the name. How do we get this percentage grouped by name?
the select quere uses only (sum(amount_paid)/(select sum(amount_paid) from order_items)*100.
for me it semms like every answer sould be 100.
sorry i can’t get it
many thanks in advice
I am just getting to this exercise, but the way I understand it only the numerator is affected by the GROUP BY.
In that way the numerator is the sum by item (name) and the denominator is the grand total, which includes all the items.
In the main query we calculate the revenue per name. This is then divided by the subquery, which simply sums up the total revenue across the rows. This is then multiplied by 100.0 to get a percentage.
You’re essentially combining these two tables:
SELECT name, SUM(amount_paid)
GROUP BY 1
ORDER BY 2 DESC;
--> revenue per food item
--> total revenue across food items (178311.5)
By combining them in one query:
SELECT name, ROUND(SUM(amount_paid)/(SELECT sum(amount_paid) FROM order_items)*100, 2)
GROUP BY 1
ORDER BY 2 DESC;
The first query
SELECT name, SUM(amount_paid) becomes the numerator, and
SELECT sum(amount_paid) FROM order_items becomes the denominator. Within the ROUND() function, this division is multiplied by 100 to get a percentage, and then limited to two decimals. Swapping out the subquery for the value it calculates, 178311.5, would return the same table.
In regards to your questions 'Every answer should be 100 ': this would be the case if the denominator has the same value, however using the subquery makes sure that the denominator is the total revenue. Using the full query
SELECT name, ROUND(SUM(amount_paid)/SUM(amount_paid)*100, 2) would accomplish that, however it wouldn’t be very useful in this context.