Bigquery - How to Calculate the sum of two continuous rows - google-bigquery

Question

How can I get the sum of two rows clubbed together for instance If I have 5 rows in total, I should get 3 rows a result.
Below is my table:
2020-08-01 1
2020-08-02 3
2020-08-03 4
2020-08-04 2
2020-08-05 4
I want to achive this:
4
6
4
August 1 and 2 = 4
August 3 and 4 = 6
August 5 = 4

Answer 1

You could use ROW_NUMBER here:
WITH cte AS (
SELECT dt, val, ROW_NUMBER() OVER (ORDER BY dt) rn
FROM yourTable
)
SELECT SUM(val)
FROM cte
GROUP BY FLOOR((rn - 1) / 2)
GROUP BY MIN(dt);
Here is a demo link, shown in SQL Server, but whose logic should also be working for BigQuery:
Demo

Answer 2

Below is for Bigquery Standard SQL
#standardSQL
SELECT SUM(value) AS value,
STRING_AGG(FORMAT_DATE('%B %d', day), ' and ') || ' = ' || CAST(SUM(value) AS STRING) AS calc
FROM (
SELECT day, value, DIV(ROW_NUMBER() OVER(ORDER BY day) - 1, 2) grp
FROM `project.dataset.table` t
)
GROUP BY grp
ORDER BY grp
You can test, play with above using sample data from your question as in below example
#standardSQL
WITH `project.dataset.table` AS (
SELECT DATE '2020-08-01' day, 1 value UNION ALL
SELECT '2020-08-02', 3 UNION ALL
SELECT '2020-08-03', 4 UNION ALL
SELECT '2020-08-04', 2 UNION ALL
SELECT '2020-08-05', 4
)
SELECT SUM(value) AS value,
STRING_AGG(FORMAT_DATE('%B %d', day), ' and ') || ' = ' || CAST(SUM(value) AS STRING) AS calc
FROM (
SELECT day, value, DIV(ROW_NUMBER() OVER(ORDER BY day) - 1, 2) grp
FROM `project.dataset.table` t
)
GROUP BY grp
ORDER BY grp
with output
Row value calc
1 4 August 01 and August 02 = 4
2 6 August 03 and August 04 = 6
3 4 August 05 = 4