SUM values from two tables with GROUP BY and WHERE
I have two tables below named sent_table and received_table. I am attempting to mash them together in a query to achieve output_table. All my attempts so far result in a huge amount of duplicates and totally bogus sum values.
I am assuming I would need to use GROUP BY and WHERE to achieve this goal. I want to be able to filter based on the users name.
sent_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 100 | 1 |
| 2 | dave | 200 | 1 |
| 3 | dave | 300 | 2 |
+----+------+-------+----------+
received_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 400 | 1 |
| 2 | dave | 500 | 2 |
| 3 | dave | 600 | 2 |
+----+------+-------+----------+
output table
+------+----------+----------+
| sent | received | order_id |
+------+----------+----------+
| 300 | 400 | 1 |
| 300 | 1100 | 2 |
+------+----------+----------+
I tried the following with no joy. This does not impose any restrictions on how I would desire to solve this problem. It is just how I attempted to do it.
SELECT *
FROM
( select SUM(value) as sent, order_id FROM sent_table WHERE name='dave' GROUP BY order_id) A
CROSS JOIN
( select SUM(value) as received, order_id FROM received_table WHERE name='dave' GROUP BY order_id) B
Any help would be greatly appreciated.
postgresql
asked Nov 25 '18 at 9:38
DanDan
10812
add a comment |
I have two tables below named sent_table and received_table. I am attempting to mash them together in a query to achieve output_table. All my attempts so far result in a huge amount of duplicates and totally bogus sum values.
I am assuming I would need to use GROUP BY and WHERE to achieve this goal. I want to be able to filter based on the users name.
sent_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 100 | 1 |
| 2 | dave | 200 | 1 |
| 3 | dave | 300 | 2 |
+----+------+-------+----------+
received_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 400 | 1 |
| 2 | dave | 500 | 2 |
| 3 | dave | 600 | 2 |
+----+------+-------+----------+
output table
+------+----------+----------+
| sent | received | order_id |
+------+----------+----------+
| 300 | 400 | 1 |
| 300 | 1100 | 2 |
+------+----------+----------+
I tried the following with no joy. This does not impose any restrictions on how I would desire to solve this problem. It is just how I attempted to do it.
SELECT *
FROM
( select SUM(value) as sent, order_id FROM sent_table WHERE name='dave' GROUP BY order_id) A
CROSS JOIN
( select SUM(value) as received, order_id FROM received_table WHERE name='dave' GROUP BY order_id) B
Any help would be greatly appreciated.
postgresql
asked Nov 25 '18 at 9:38
DanDan
10812
add a comment |
I have two tables below named sent_table and received_table. I am attempting to mash them together in a query to achieve output_table. All my attempts so far result in a huge amount of duplicates and totally bogus sum values.
I am assuming I would need to use GROUP BY and WHERE to achieve this goal. I want to be able to filter based on the users name.
sent_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 100 | 1 |
| 2 | dave | 200 | 1 |
| 3 | dave | 300 | 2 |
+----+------+-------+----------+
received_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 400 | 1 |
| 2 | dave | 500 | 2 |
| 3 | dave | 600 | 2 |
+----+------+-------+----------+
output table
+------+----------+----------+
| sent | received | order_id |
+------+----------+----------+
| 300 | 400 | 1 |
| 300 | 1100 | 2 |
+------+----------+----------+
I tried the following with no joy. This does not impose any restrictions on how I would desire to solve this problem. It is just how I attempted to do it.
SELECT *
FROM
( select SUM(value) as sent, order_id FROM sent_table WHERE name='dave' GROUP BY order_id) A
CROSS JOIN
( select SUM(value) as received, order_id FROM received_table WHERE name='dave' GROUP BY order_id) B
Any help would be greatly appreciated.
postgresql
asked Nov 25 '18 at 9:38
DanDan
10812
I have two tables below named sent_table and received_table. I am attempting to mash them together in a query to achieve output_table. All my attempts so far result in a huge amount of duplicates and totally bogus sum values.
I am assuming I would need to use GROUP BY and WHERE to achieve this goal. I want to be able to filter based on the users name.
sent_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 100 | 1 |
| 2 | dave | 200 | 1 |
| 3 | dave | 300 | 2 |
+----+------+-------+----------+
received_table
+----+------+-------+----------+
| id | name | value | order_id |
+----+------+-------+----------+
| 1 | dave | 400 | 1 |
| 2 | dave | 500 | 2 |
| 3 | dave | 600 | 2 |
+----+------+-------+----------+
output table
+------+----------+----------+
| sent | received | order_id |
+------+----------+----------+
| 300 | 400 | 1 |
| 300 | 1100 | 2 |
+------+----------+----------+
I tried the following with no joy. This does not impose any restrictions on how I would desire to solve this problem. It is just how I attempted to do it.
SELECT *
FROM
( select SUM(value) as sent, order_id FROM sent_table WHERE name='dave' GROUP BY order_id) A
CROSS JOIN
( select SUM(value) as received, order_id FROM received_table WHERE name='dave' GROUP BY order_id) B
Any help would be greatly appreciated.
postgresql
postgresql
asked Nov 25 '18 at 9:38
DanDan
10812
asked Nov 25 '18 at 9:38
DanDan
10812
asked Nov 25 '18 at 9:38
DanDan
10812
asked Nov 25 '18 at 9:38
DanDan
10812
asked Nov 25 '18 at 9:38
DanDan
10812
10812
add a comment |
add a comment |
1 Answer
1
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Do the sums on each table, grouping by order_id, then join the results. To get the rows even if one side is missing, do a FULL OUTER JOIN:
SELECT COALESCE(s.order_id, r.order_id) AS order_id, s.sent, r.received
FROM (
SELECT order_id, SUM(value) AS sent
FROM sent
GROUP BY order_id
) s
FULL OUTER JOIN (
SELECT order_id, SUM(value) AS received
FROM received
GROUP BY order_id
) r
USING (order_id)
ORDER BY 1
Result:
| order_id | sent | received |
| -------- | ---- | -------- |
| 1 | 300 | 400 |
| 2 | | 1100 |
Note the COALESCE on the order_id, so that if it's missing from sent it will be taken from recevied, so that that value will never be NULL.
If you want to have 0 in place of NULL (when e.g. there is no record for that order_id in either sent or received), you would do COALESCE(s.sent, 0) AS sent, COALESCE(r.received, 0) AS received.
https://www.db-fiddle.com/f/nq3xYrcys16eUrBRHT6xLL/2
answered Nov 25 '18 at 9:46
404404
3,0851726
That is awesome! :) One question, ifsentdoes not have an order_id 2, thereceiveddata for order_id 2 does not appear. Is there any quick way to still get the order_id 2 data?
– Dan
Nov 25 '18 at 9:53
@Dan Updated example for that scenario in answer.
– 404
Nov 25 '18 at 10:03
You my friend are amazing. Thank you sooooo much. Works a treat. I will spend some time learning what is exactly going on here and develop my own knowledge now that I know this stuff is possible. Thanks again!
– Dan
Nov 25 '18 at 10:07
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
Do the sums on each table, grouping by order_id, then join the results. To get the rows even if one side is missing, do a FULL OUTER JOIN:
SELECT COALESCE(s.order_id, r.order_id) AS order_id, s.sent, r.received
FROM (
SELECT order_id, SUM(value) AS sent
FROM sent
GROUP BY order_id
) s
FULL OUTER JOIN (
SELECT order_id, SUM(value) AS received
FROM received
GROUP BY order_id
) r
USING (order_id)
ORDER BY 1
Result:
| order_id | sent | received |
| -------- | ---- | -------- |
| 1 | 300 | 400 |
| 2 | | 1100 |
Note the COALESCE on the order_id, so that if it's missing from sent it will be taken from recevied, so that that value will never be NULL.
If you want to have 0 in place of NULL (when e.g. there is no record for that order_id in either sent or received), you would do COALESCE(s.sent, 0) AS sent, COALESCE(r.received, 0) AS received.
https://www.db-fiddle.com/f/nq3xYrcys16eUrBRHT6xLL/2
answered Nov 25 '18 at 9:46
404404
3,0851726
That is awesome! :) One question, ifsentdoes not have an order_id 2, thereceiveddata for order_id 2 does not appear. Is there any quick way to still get the order_id 2 data?
– Dan
Nov 25 '18 at 9:53
@Dan Updated example for that scenario in answer.
– 404
Nov 25 '18 at 10:03
You my friend are amazing. Thank you sooooo much. Works a treat. I will spend some time learning what is exactly going on here and develop my own knowledge now that I know this stuff is possible. Thanks again!
– Dan
Nov 25 '18 at 10:07
add a comment |
Do the sums on each table, grouping by order_id, then join the results. To get the rows even if one side is missing, do a FULL OUTER JOIN:
SELECT COALESCE(s.order_id, r.order_id) AS order_id, s.sent, r.received
FROM (
SELECT order_id, SUM(value) AS sent
FROM sent
GROUP BY order_id
) s
FULL OUTER JOIN (
SELECT order_id, SUM(value) AS received
FROM received
GROUP BY order_id
) r
USING (order_id)
ORDER BY 1
Result:
| order_id | sent | received |
| -------- | ---- | -------- |
| 1 | 300 | 400 |
| 2 | | 1100 |
Note the COALESCE on the order_id, so that if it's missing from sent it will be taken from recevied, so that that value will never be NULL.
If you want to have 0 in place of NULL (when e.g. there is no record for that order_id in either sent or received), you would do COALESCE(s.sent, 0) AS sent, COALESCE(r.received, 0) AS received.
https://www.db-fiddle.com/f/nq3xYrcys16eUrBRHT6xLL/2
answered Nov 25 '18 at 9:46
404404
3,0851726
That is awesome! :) One question, ifsentdoes not have an order_id 2, thereceiveddata for order_id 2 does not appear. Is there any quick way to still get the order_id 2 data?
– Dan
Nov 25 '18 at 9:53
@Dan Updated example for that scenario in answer.
– 404
Nov 25 '18 at 10:03
You my friend are amazing. Thank you sooooo much. Works a treat. I will spend some time learning what is exactly going on here and develop my own knowledge now that I know this stuff is possible. Thanks again!
– Dan
Nov 25 '18 at 10:07
add a comment |
Do the sums on each table, grouping by order_id, then join the results. To get the rows even if one side is missing, do a FULL OUTER JOIN:
SELECT COALESCE(s.order_id, r.order_id) AS order_id, s.sent, r.received
FROM (
SELECT order_id, SUM(value) AS sent
FROM sent
GROUP BY order_id
) s
FULL OUTER JOIN (
SELECT order_id, SUM(value) AS received
FROM received
GROUP BY order_id
) r
USING (order_id)
ORDER BY 1
Result:
| order_id | sent | received |
| -------- | ---- | -------- |
| 1 | 300 | 400 |
| 2 | | 1100 |
Note the COALESCE on the order_id, so that if it's missing from sent it will be taken from recevied, so that that value will never be NULL.
If you want to have 0 in place of NULL (when e.g. there is no record for that order_id in either sent or received), you would do COALESCE(s.sent, 0) AS sent, COALESCE(r.received, 0) AS received.
https://www.db-fiddle.com/f/nq3xYrcys16eUrBRHT6xLL/2
answered Nov 25 '18 at 9:46
404404
3,0851726
Do the sums on each table, grouping by order_id, then join the results. To get the rows even if one side is missing, do a FULL OUTER JOIN:
SELECT COALESCE(s.order_id, r.order_id) AS order_id, s.sent, r.received
FROM (
SELECT order_id, SUM(value) AS sent
FROM sent
GROUP BY order_id
) s
FULL OUTER JOIN (
SELECT order_id, SUM(value) AS received
FROM received
GROUP BY order_id
) r
USING (order_id)
ORDER BY 1
Result:
| order_id | sent | received |
| -------- | ---- | -------- |
| 1 | 300 | 400 |
| 2 | | 1100 |
Note the COALESCE on the order_id, so that if it's missing from sent it will be taken from recevied, so that that value will never be NULL.
If you want to have 0 in place of NULL (when e.g. there is no record for that order_id in either sent or received), you would do COALESCE(s.sent, 0) AS sent, COALESCE(r.received, 0) AS received.
https://www.db-fiddle.com/f/nq3xYrcys16eUrBRHT6xLL/2
answered Nov 25 '18 at 9:46
404404
3,0851726
edited Nov 25 '18 at 10:07
answered Nov 25 '18 at 9:46
404404
3,0851726
answered Nov 25 '18 at 9:46
404404
3,0851726
answered Nov 25 '18 at 9:46
404404
3,0851726
3,0851726
That is awesome! :) One question, ifsentdoes not have an order_id 2, thereceiveddata for order_id 2 does not appear. Is there any quick way to still get the order_id 2 data?
– Dan
Nov 25 '18 at 9:53
@Dan Updated example for that scenario in answer.
– 404
Nov 25 '18 at 10:03
You my friend are amazing. Thank you sooooo much. Works a treat. I will spend some time learning what is exactly going on here and develop my own knowledge now that I know this stuff is possible. Thanks again!
– Dan
Nov 25 '18 at 10:07
add a comment |
That is awesome! :) One question, ifsentdoes not have an order_id 2, thereceiveddata for order_id 2 does not appear. Is there any quick way to still get the order_id 2 data?
– Dan
Nov 25 '18 at 9:53
@Dan Updated example for that scenario in answer.
– 404
Nov 25 '18 at 10:03
You my friend are amazing. Thank you sooooo much. Works a treat. I will spend some time learning what is exactly going on here and develop my own knowledge now that I know this stuff is possible. Thanks again!
– Dan
Nov 25 '18 at 10:07
That is awesome! :) One question, if
sent does not have an order_id 2, the received data for order_id 2 does not appear. Is there any quick way to still get the order_id 2 data?– Dan
Nov 25 '18 at 9:53
That is awesome! :) One question, if
sent does not have an order_id 2, the received data for order_id 2 does not appear. Is there any quick way to still get the order_id 2 data?– Dan
Nov 25 '18 at 9:53
@Dan Updated example for that scenario in answer.
– 404
Nov 25 '18 at 10:03
@Dan Updated example for that scenario in answer.
– 404
Nov 25 '18 at 10:03
You my friend are amazing. Thank you sooooo much. Works a treat. I will spend some time learning what is exactly going on here and develop my own knowledge now that I know this stuff is possible. Thanks again!
– Dan
Nov 25 '18 at 10:07
You my friend are amazing. Thank you sooooo much. Works a treat. I will spend some time learning what is exactly going on here and develop my own knowledge now that I know this stuff is possible. Thanks again!
– Dan
Nov 25 '18 at 10:07
add a comment |
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