How do I use flatten logic in Snowflake - snowflake-cloud-data-platform

Question

select * from
(SELECT ID,purchase_list,
CASE WHEN purchase_list LIKE '12432%' THEN replace(purchase_list,12432,'Amt_Saved: Mastercard(12432)')
END AS purchase_amt
FROM (
SELECT ID,index,d.value::string AS purchase_list
FROM (
SELECT ID,c.value::string AS purchase_list
FROM table_1,LATERAL flatten(INPUT=>split(purchase_order_list, '|')) c
), LATERAL flatten(INPUT=>split(purchase_list, ';')) d
)
)WHERE purchase_amt is not null
When I run the above query, I'm getting results as mentioned below:
ID PURCHASE_LIST PURCHASE_AMT
12810789 12432=3.00 Savings1: Mastercard(12432)=3.00
12810789 12432=0.99 Savings1: Mastercard(12432)=0.99
12810789 12432=0.49 Savings1: Mastercard(12432)=0.49
I want to sum up these amounts (3.00 + 0.99 + 0.49)=4.48 and display this result in same column or another COLUMN
The result should look like:
ID PURCHASE_LIST PURCHASE_AMT
12810789 12432=4.48 4.48
How to implement this using SUM function by using any other logic?
Kindly guide. Thanks in advance :)

Answer 1

Firstly your existing SQL can be rewritten to be:
SELECT
a.id,
d.value::string AS purchase_list,
IFF( purchase_list LIKE '12432%', REPLACE(purchase_list, 12432, 'Amt_Saved: Mastercard(12432)'), null) purchase_amt
FROM table_1 AS a
,LATERAL flatten(INPUT=>split(a.purchase_order_list, '|')) c
,LATERAL flatten(INPUT=>split(c.value::string, ';')) d
WHERE purchase_amt IS NOT NULL;
So we can add a SUM() OVER() clause to get a total per row, it you are want parts and total:
SELECT
a.id,
d.value::string AS purchase_list,
IFF( purchase_list LIKE '12432%', REPLACE(purchase_list, 12432, 'Amt_Saved: Mastercard(12432)'), null) purchase_amt,
SUM(REGEXP_REPLACE(purchase_list, '12432=([0-9.]+)', '\\1', 1,1,'e')::number(9,3)) over (partition by id) as sum_total
FROM table_1 AS a
,LATERAL flatten(INPUT=>split(a.purchase_order_list, '|')) c
,LATERAL flatten(INPUT=>split(c.value::string, ';')) d
WHERE purchase_amt IS NOT NULL;
ID
PURCHASE_LIST
PURCHASE_AMT
SUM_TOTAL
12810789
12432=3.00
Amt_Saved: Mastercard(12432)=3.00
4.48
12810789
12432=0.99
Amt_Saved: Mastercard(12432)=0.99
4.48
12810789
12432=0.49
Amt_Saved: Mastercard(12432)=0.49
4.48
OR you can group and not care about all the sub values:
SELECT
a.id,
SUM(REGEXP_REPLACE(d.value::string, '12432=([0-9.]+)', '\\1', 1,1,'e')::number(9,3)) AS sum_total
FROM table_1 AS a
,LATERAL flatten(INPUT=>split(a.purchase_order_list, '|')) c
,LATERAL flatten(INPUT=>split(c.value::string, ';')) d
GROUP BY 1;
gives:
ID
SUM_TOTAL
12810789
4.48