Get the latest record or the first transaction (sale) BigQuery SQL standard









up vote
1
down vote

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I have a table on BigQuery with the following columns:



user_id visit_date referral transaction
1234 20180101 site2 0
1234 20180102 site3 1
1234 20180103 site2 1
4567 20180104 site4 0
4567 20180105 site5 0
5678 20180101 site2 0
5768 20180102 site3 1


My goal is to have a table in the following format as output:



path transactions
site2 > site3 2
site2 1
site4 > site5 0


What I don't understand is how to 'reset' the path for a user that have multiple conversions in the same period, as is the case for user_id = 1234.



So far I manage use the following query, but it is not the desired output.



SELECT
referral_path,
SUM(transactions) AS transactions
FROM (
SELECT
user_id,
STRING_AGG(DISTINCT(referral), ',') AS referral_path,
MAX(transactions) AS transactions
FROM (
SELECT
user_id,
referral,
transactions
FROM
table
ORDER BY
user_id )a
GROUP BY
user_id )b
GROUP BY
referral_path
ORDER BY
transactions DESC









share|improve this question





















  • what you mean by reset the path - please explain more. also do you allow only two referrals in the path?
    – Mikhail Berlyant
    Nov 9 at 18:06











  • It could have more than 2 referrals. By reset I mean, for the first user (id=1234) for the same period, it has 2 conversions, after the first conversion, the path should restart from site2. Instead of being site2 > site3 > site2.
    – Legos
    Nov 9 at 18:15










  • tnx. i see. got it
    – Mikhail Berlyant
    Nov 9 at 18:15














up vote
1
down vote

favorite












I have a table on BigQuery with the following columns:



user_id visit_date referral transaction
1234 20180101 site2 0
1234 20180102 site3 1
1234 20180103 site2 1
4567 20180104 site4 0
4567 20180105 site5 0
5678 20180101 site2 0
5768 20180102 site3 1


My goal is to have a table in the following format as output:



path transactions
site2 > site3 2
site2 1
site4 > site5 0


What I don't understand is how to 'reset' the path for a user that have multiple conversions in the same period, as is the case for user_id = 1234.



So far I manage use the following query, but it is not the desired output.



SELECT
referral_path,
SUM(transactions) AS transactions
FROM (
SELECT
user_id,
STRING_AGG(DISTINCT(referral), ',') AS referral_path,
MAX(transactions) AS transactions
FROM (
SELECT
user_id,
referral,
transactions
FROM
table
ORDER BY
user_id )a
GROUP BY
user_id )b
GROUP BY
referral_path
ORDER BY
transactions DESC









share|improve this question





















  • what you mean by reset the path - please explain more. also do you allow only two referrals in the path?
    – Mikhail Berlyant
    Nov 9 at 18:06











  • It could have more than 2 referrals. By reset I mean, for the first user (id=1234) for the same period, it has 2 conversions, after the first conversion, the path should restart from site2. Instead of being site2 > site3 > site2.
    – Legos
    Nov 9 at 18:15










  • tnx. i see. got it
    – Mikhail Berlyant
    Nov 9 at 18:15












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have a table on BigQuery with the following columns:



user_id visit_date referral transaction
1234 20180101 site2 0
1234 20180102 site3 1
1234 20180103 site2 1
4567 20180104 site4 0
4567 20180105 site5 0
5678 20180101 site2 0
5768 20180102 site3 1


My goal is to have a table in the following format as output:



path transactions
site2 > site3 2
site2 1
site4 > site5 0


What I don't understand is how to 'reset' the path for a user that have multiple conversions in the same period, as is the case for user_id = 1234.



So far I manage use the following query, but it is not the desired output.



SELECT
referral_path,
SUM(transactions) AS transactions
FROM (
SELECT
user_id,
STRING_AGG(DISTINCT(referral), ',') AS referral_path,
MAX(transactions) AS transactions
FROM (
SELECT
user_id,
referral,
transactions
FROM
table
ORDER BY
user_id )a
GROUP BY
user_id )b
GROUP BY
referral_path
ORDER BY
transactions DESC









share|improve this question













I have a table on BigQuery with the following columns:



user_id visit_date referral transaction
1234 20180101 site2 0
1234 20180102 site3 1
1234 20180103 site2 1
4567 20180104 site4 0
4567 20180105 site5 0
5678 20180101 site2 0
5768 20180102 site3 1


My goal is to have a table in the following format as output:



path transactions
site2 > site3 2
site2 1
site4 > site5 0


What I don't understand is how to 'reset' the path for a user that have multiple conversions in the same period, as is the case for user_id = 1234.



So far I manage use the following query, but it is not the desired output.



SELECT
referral_path,
SUM(transactions) AS transactions
FROM (
SELECT
user_id,
STRING_AGG(DISTINCT(referral), ',') AS referral_path,
MAX(transactions) AS transactions
FROM (
SELECT
user_id,
referral,
transactions
FROM
table
ORDER BY
user_id )a
GROUP BY
user_id )b
GROUP BY
referral_path
ORDER BY
transactions DESC






sql google-bigquery






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 9 at 18:03









Legos

4226




4226











  • what you mean by reset the path - please explain more. also do you allow only two referrals in the path?
    – Mikhail Berlyant
    Nov 9 at 18:06











  • It could have more than 2 referrals. By reset I mean, for the first user (id=1234) for the same period, it has 2 conversions, after the first conversion, the path should restart from site2. Instead of being site2 > site3 > site2.
    – Legos
    Nov 9 at 18:15










  • tnx. i see. got it
    – Mikhail Berlyant
    Nov 9 at 18:15
















  • what you mean by reset the path - please explain more. also do you allow only two referrals in the path?
    – Mikhail Berlyant
    Nov 9 at 18:06











  • It could have more than 2 referrals. By reset I mean, for the first user (id=1234) for the same period, it has 2 conversions, after the first conversion, the path should restart from site2. Instead of being site2 > site3 > site2.
    – Legos
    Nov 9 at 18:15










  • tnx. i see. got it
    – Mikhail Berlyant
    Nov 9 at 18:15















what you mean by reset the path - please explain more. also do you allow only two referrals in the path?
– Mikhail Berlyant
Nov 9 at 18:06





what you mean by reset the path - please explain more. also do you allow only two referrals in the path?
– Mikhail Berlyant
Nov 9 at 18:06













It could have more than 2 referrals. By reset I mean, for the first user (id=1234) for the same period, it has 2 conversions, after the first conversion, the path should restart from site2. Instead of being site2 > site3 > site2.
– Legos
Nov 9 at 18:15




It could have more than 2 referrals. By reset I mean, for the first user (id=1234) for the same period, it has 2 conversions, after the first conversion, the path should restart from site2. Instead of being site2 > site3 > site2.
– Legos
Nov 9 at 18:15












tnx. i see. got it
– Mikhail Berlyant
Nov 9 at 18:15




tnx. i see. got it
– Mikhail Berlyant
Nov 9 at 18:15












1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










Below is for BigQuery Standard SQL



#standardSQL
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path


You can test, play with above using dummy data from your question as in below



#standardSQL
WITH `project.dataset.table` AS (
SELECT 1234 user_id, '20180101' visit_date, 'site2' referral, 0 transaction UNION ALL
SELECT 1234, '20180102', 'site3', 1 UNION ALL
SELECT 1234, '20180103', 'site2', 1 UNION ALL
SELECT 4567, '20180104', 'site4', 0 UNION ALL
SELECT 4567, '20180105', 'site5', 0 UNION ALL
SELECT 5678, '20180101', 'site2', 0 UNION ALL
SELECT 5678, '20180102', 'site3', 1
)
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path
ORDER BY transactions DESC


with result



Row path transactions 
1 site2 > site3 2
2 site2 1
3 site4 > site5 0





share|improve this answer




















  • The solution works fine! I just don't understand "1 PRECEDING" command at the end of the over function. So within a specific user_id, It goes from unbounded preceding (the first one row) to the last?
    – Legos
    Nov 12 at 18:10






  • 1




    that's correct. this logic catches when group value is changed so then you can use grp in GROUP BY - I recommend you to run inner queries one by one to see how it works - so you will be able to use this approach by your own :o)
    – Mikhail Berlyant
    Nov 12 at 18:12










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










Below is for BigQuery Standard SQL



#standardSQL
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path


You can test, play with above using dummy data from your question as in below



#standardSQL
WITH `project.dataset.table` AS (
SELECT 1234 user_id, '20180101' visit_date, 'site2' referral, 0 transaction UNION ALL
SELECT 1234, '20180102', 'site3', 1 UNION ALL
SELECT 1234, '20180103', 'site2', 1 UNION ALL
SELECT 4567, '20180104', 'site4', 0 UNION ALL
SELECT 4567, '20180105', 'site5', 0 UNION ALL
SELECT 5678, '20180101', 'site2', 0 UNION ALL
SELECT 5678, '20180102', 'site3', 1
)
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path
ORDER BY transactions DESC


with result



Row path transactions 
1 site2 > site3 2
2 site2 1
3 site4 > site5 0





share|improve this answer




















  • The solution works fine! I just don't understand "1 PRECEDING" command at the end of the over function. So within a specific user_id, It goes from unbounded preceding (the first one row) to the last?
    – Legos
    Nov 12 at 18:10






  • 1




    that's correct. this logic catches when group value is changed so then you can use grp in GROUP BY - I recommend you to run inner queries one by one to see how it works - so you will be able to use this approach by your own :o)
    – Mikhail Berlyant
    Nov 12 at 18:12














up vote
2
down vote



accepted










Below is for BigQuery Standard SQL



#standardSQL
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path


You can test, play with above using dummy data from your question as in below



#standardSQL
WITH `project.dataset.table` AS (
SELECT 1234 user_id, '20180101' visit_date, 'site2' referral, 0 transaction UNION ALL
SELECT 1234, '20180102', 'site3', 1 UNION ALL
SELECT 1234, '20180103', 'site2', 1 UNION ALL
SELECT 4567, '20180104', 'site4', 0 UNION ALL
SELECT 4567, '20180105', 'site5', 0 UNION ALL
SELECT 5678, '20180101', 'site2', 0 UNION ALL
SELECT 5678, '20180102', 'site3', 1
)
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path
ORDER BY transactions DESC


with result



Row path transactions 
1 site2 > site3 2
2 site2 1
3 site4 > site5 0





share|improve this answer




















  • The solution works fine! I just don't understand "1 PRECEDING" command at the end of the over function. So within a specific user_id, It goes from unbounded preceding (the first one row) to the last?
    – Legos
    Nov 12 at 18:10






  • 1




    that's correct. this logic catches when group value is changed so then you can use grp in GROUP BY - I recommend you to run inner queries one by one to see how it works - so you will be able to use this approach by your own :o)
    – Mikhail Berlyant
    Nov 12 at 18:12












up vote
2
down vote



accepted







up vote
2
down vote



accepted






Below is for BigQuery Standard SQL



#standardSQL
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path


You can test, play with above using dummy data from your question as in below



#standardSQL
WITH `project.dataset.table` AS (
SELECT 1234 user_id, '20180101' visit_date, 'site2' referral, 0 transaction UNION ALL
SELECT 1234, '20180102', 'site3', 1 UNION ALL
SELECT 1234, '20180103', 'site2', 1 UNION ALL
SELECT 4567, '20180104', 'site4', 0 UNION ALL
SELECT 4567, '20180105', 'site5', 0 UNION ALL
SELECT 5678, '20180101', 'site2', 0 UNION ALL
SELECT 5678, '20180102', 'site3', 1
)
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path
ORDER BY transactions DESC


with result



Row path transactions 
1 site2 > site3 2
2 site2 1
3 site4 > site5 0





share|improve this answer












Below is for BigQuery Standard SQL



#standardSQL
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path


You can test, play with above using dummy data from your question as in below



#standardSQL
WITH `project.dataset.table` AS (
SELECT 1234 user_id, '20180101' visit_date, 'site2' referral, 0 transaction UNION ALL
SELECT 1234, '20180102', 'site3', 1 UNION ALL
SELECT 1234, '20180103', 'site2', 1 UNION ALL
SELECT 4567, '20180104', 'site4', 0 UNION ALL
SELECT 4567, '20180105', 'site5', 0 UNION ALL
SELECT 5678, '20180101', 'site2', 0 UNION ALL
SELECT 5678, '20180102', 'site3', 1
)
SELECT
path,
SUM(transaction) transactions
FROM (
SELECT
STRING_AGG(referral, ' > ') path,
SUM(transaction) transaction
FROM (
SELECT
user_id, visit_date, referral, transaction,
IFNULL(SUM(transaction) OVER(PARTITION BY user_id ORDER BY visit_date ROWS BETWEEN UNBOUNDED PRECEDING AND 1 PRECEDING), 0) grp
FROM `project.dataset.table`
)
GROUP BY user_id, grp
)
GROUP BY path
ORDER BY transactions DESC


with result



Row path transactions 
1 site2 > site3 2
2 site2 1
3 site4 > site5 0






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 9 at 18:24









Mikhail Berlyant

52.8k42964




52.8k42964











  • The solution works fine! I just don't understand "1 PRECEDING" command at the end of the over function. So within a specific user_id, It goes from unbounded preceding (the first one row) to the last?
    – Legos
    Nov 12 at 18:10






  • 1




    that's correct. this logic catches when group value is changed so then you can use grp in GROUP BY - I recommend you to run inner queries one by one to see how it works - so you will be able to use this approach by your own :o)
    – Mikhail Berlyant
    Nov 12 at 18:12
















  • The solution works fine! I just don't understand "1 PRECEDING" command at the end of the over function. So within a specific user_id, It goes from unbounded preceding (the first one row) to the last?
    – Legos
    Nov 12 at 18:10






  • 1




    that's correct. this logic catches when group value is changed so then you can use grp in GROUP BY - I recommend you to run inner queries one by one to see how it works - so you will be able to use this approach by your own :o)
    – Mikhail Berlyant
    Nov 12 at 18:12















The solution works fine! I just don't understand "1 PRECEDING" command at the end of the over function. So within a specific user_id, It goes from unbounded preceding (the first one row) to the last?
– Legos
Nov 12 at 18:10




The solution works fine! I just don't understand "1 PRECEDING" command at the end of the over function. So within a specific user_id, It goes from unbounded preceding (the first one row) to the last?
– Legos
Nov 12 at 18:10




1




1




that's correct. this logic catches when group value is changed so then you can use grp in GROUP BY - I recommend you to run inner queries one by one to see how it works - so you will be able to use this approach by your own :o)
– Mikhail Berlyant
Nov 12 at 18:12




that's correct. this logic catches when group value is changed so then you can use grp in GROUP BY - I recommend you to run inner queries one by one to see how it works - so you will be able to use this approach by your own :o)
– Mikhail Berlyant
Nov 12 at 18:12

















 

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