I need a solution regarding the Fine the difference between the last two entries difference in MySQL
SELECT DATE_FORMAT(order_datetime,'%m/%Y') as date,
SUM(order_total_after_tax) as number
FROM tbl_order
WHERE status = 'Confirmed'
GROUP BY DATE_FORMAT(order_datetime,'%Y/%m')
Limit 2
OutPut :
date number
02/2019 2345.01
03/2019 103751.05
But i need an exact solution is i need to find the difference between the last 2 result
2345.01 -103751.05
Result : -101406.04
how to do in MySQL What is the Query?
You could try a cross join
select a.date, a.number, b.number, a.number - b.number my_diff
from (
SELECT DATE_FORMAT(order_datetime,'%m/%Y') as date
, SUM(order_total_after_tax) as number
FROM tbl_order
where status='Confirmed'
GROUP BY DATE_FORMAT(order_datetime,'%Y/%m')
Limit 2
) a
CROSS JOIN (
SELECT DATE_FORMAT(order_datetime,'%m/%Y') as date
, SUM(order_total_after_tax) as number
FROM tbl_order
where status='Confirmed'
GROUP BY DATE_FORMAT(order_datetime,'%Y/%m')
) b ON a.date = b.date and a.number <> b.number
One method is a cross join
, but it looks like this:
SELECT last.number - last_1.number
FROM (SELECT DATE_FORMAT(order_datetime,'%m/%Y') as date,
SUM(order_total_after_tax) as number
FROM tbl_order
WHERE status = 'Confirmed'
GROUP BY DATE_FORMAT(order_datetime,'%Y/%m')
ORDER BY date
LIMIT 1
) as last CROSS JOIN
(SELECT DATE_FORMAT(order_datetime,'%m/%Y') as date,
SUM(order_total_after_tax) as number
FROM tbl_order
WHERE status = 'Confirmed'
GROUP BY DATE_FORMAT(order_datetime,'%Y/%m')
ORDER BY date
LIMIT 1 OFFSET 1
) last_1
SELECT last.number - last_1.number FROM (SELECT DATE_FORMAT(order_datetime,'%m/%Y') as date, SUM(order_total_after_tax) as number FROM tbl_order WHERE status = 'Confirmed' GROUP BY DATE_FORMAT(order_datetime,'%Y/%m') ORDER BY date DESC LIMIT 1 ) as last CROSS JOIN (SELECT DATE_FORMAT(order_datetime,'%m/%Y') as date, SUM(order_total_after_tax) as number FROM tbl_order WHERE status = 'Confirmed' GROUP BY DATE_FORMAT(order_datetime,'%Y/%m') ORDER BY date DESC LIMIT 1 OFFSET 1 ) last_1
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