How to find count of transaction of type1 and type2 for each customer in mysql
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13-09-2019 - |
Question
I have a customer table:
id name
1 customer1
2 customer2
3 customer3
and a transaction table:
id customer amount type
1 1 10 type1
2 1 15 type1
3 1 15 type2
4 2 60 type2
5 3 23 type1
What I want my query to return is the following table
name type1 type2
customer1 2 1
customer2 0 1
customer3 1 0
Which shows that customer1 has made two transactions of type1 and 1 transaction of type2 and so forth.
Is there a query which I can use to obtain this result or do I have to use procedural code.
Solution
You could try
select c.id as customer_id
, c.name as customer_name
, sum(case when t.`type` = 'type1' then 1 else 0 end) as count_of_type1
, sum(case when t.`type` = 'type2' then 1 else 0 end) as count_of_type2
from customer c
left join `transaction` t
on c.id = t.customer
group by c.id, c.name
This query needs to iterate only once over the join.
OTHER TIPS
dirk beat me to it ;)
Similar but will work in mysql 4.1 too.
Select c.name,
sum(if(type == 1,1,0)) as `type_1_total`,
sum(if(type == 2,1,0)) as `type_2_total`,
from
customer c
join transaction t on (c.id = t.customer)
;
SELECT name,
(
SELECT COUNT(*)
FROM transaction t
WHERE t.customer = c.id
AND t.type = 'type1'
) AS type1,
(
SELECT COUNT(*)
FROM transaction t
WHERE t.customer = c.id
AND t.type = 'type2'
) AS type2
FROM customer c
To apply WHERE
conditions to these columns use this:
SELECT name
FROM (
SELECT name,
(
SELECT COUNT(*)
FROM transaction t
WHERE t.customer = c.id
AND t.type = 'type1'
) AS type1,
(
SELECT COUNT(*)
FROM transaction t
WHERE t.customer = c.id
AND t.type = 'type2'
) AS type2
FROM customer c
) q
WHERE type1 > 3
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