CODEWITHSUNDEEP
mysqlsqlgoogle-bigquery
BeanWyatt
BeanWyatt

Reputation: 23

SQL Query: Indicate if a row had a value of 0

If I have a table that logs how many products were sold each day - and does not put in a log if 0 items were sold - how would I go about identifying all of my products had a day of 0 sales?

CafeTable sample

Date    Item    QuantitySold
Jan 1   coffee  4
Jan 1   tea     1
Jan 2   tea     3
Jan 3   coffee  2
Jan 3   tea     4

Desired output would be something like this. So far I can get to a version that lists each day and quantity sold, but can't figure out if there's a way to also "tag" the product.

Item    TotalSold   SoldDaily
coffee     6           FALSE
tea        8           TRUE

This is my rough starting point - the JOIN is to try to force all dates to show, but it's still skipping the row for coffee on Jan 2. And from there, I can't figure out how I would potentially then perform the next step - I've tried a few CASE WHENs but not coming together.

with dates as (select distinct date from cafetable)
 
 select date, 
 product,
 sum(quantitysold) as quant_sold,
 min(quantitysold) as min_sold
 from dates c
 left join cafetable d on d.date=c.date
 group by 1,2

I don't think I'm even articulating this well. I work with this table a dozen times a day and I feel like this query has fried my brain.

Upvotes: 2

Views: 42

Answers (3)

Sergey Geron
Sergey Geron

Reputation: 10172

If there are days when there are no sales at all or there are zero sales at the last day of the report then you can use GENERATE_DATE_ARRAY:

WITH test_table AS (
  SELECT DATE '2020-01-01' AS Date, 'coffee' AS Item, 4 AS QuantitySold UNION ALL
  SELECT DATE '2020-01-01', 'tea', 1 UNION ALL
  SELECT DATE '2020-01-02', 'tea', 3 UNION ALL
  SELECT DATE '2020-01-03', 'coffee', 2 UNION ALL
  SELECT DATE '2020-01-03', 'tea', 4
)  
, all_items_table AS (
  SELECT DISTINCT Item
  FROM test_table
)
SELECT *
FROM all_items_table 
CROSS JOIN UNNEST(GENERATE_DATE_ARRAY('2020-01-01', '2020-01-03')) AS Date
LEFT JOIN test_table USING (Date, Item)

enter image description here

WITH test_table AS (
  SELECT DATE '2020-01-01' AS Date, 'coffee' AS Item, 4 AS QuantitySold UNION ALL
  SELECT DATE '2020-01-01', 'tea', 1 UNION ALL
  SELECT DATE '2020-01-02', 'tea', 3 UNION ALL
  SELECT DATE '2020-01-03', 'coffee', 2 UNION ALL
  SELECT DATE '2020-01-03', 'tea', 4
)  
, all_items_table AS (
  SELECT DISTINCT Item
  FROM test_table
)
SELECT 
  Item,
  LOGICAL_AND(QuantitySold IS NOT NULL) AS SoldDaily
FROM all_items_table 
CROSS JOIN UNNEST(GENERATE_DATE_ARRAY('2020-01-01', '2020-01-03')) AS Date
LEFT JOIN test_table USING (Date, Item)
GROUP BY Item

enter image description here

Upvotes: 0

Mikhail Berlyant
Mikhail Berlyant

Reputation: 172993

Below is for BigQuery Standard SQL

#standardSQL
select item, 
  sum(QuantitySold) as TotalSold,
  count(*) = date_diff(max(date), min(date), day) + 1 as SoldDaily
from `project.dataset.CafeTable`
group by item

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

#standardSQL
with `project.dataset.CafeTable` as (
  select date '2020-01-01' date, 'coffee' item, 4 QuantitySold union all
  select '2020-01-01', 'tea', 1 union all
  select '2020-01-02', 'tea', 3 union all
  select '2020-01-03', 'coffee', 2 union all
  select '2020-01-03', 'tea', 4 
)
select item, 
  sum(QuantitySold) as TotalSold,
  count(*) = date_diff(max(date), min(date), day) + 1 as SoldDaily
from `project.dataset.CafeTable`
group by item

with output

enter image description here

Upvotes: 1

Tim Biegeleisen
Tim Biegeleisen

Reputation: 521249

You could use logic here which asserts that all dates be present or not:

Below is for BigQuery Standard SQL:

select
    c.date, 
    d.product,
    sum(quantitysold) as quant_sold,
    min(quantitysold) as min_sold,
    case when count(distinct c.date) = (select count(distinct date) from dates)
         then TRUE else FALSE end AS SoldDaily
from dates c
left join cafetable d
    on d.date = c.date
group by
    1, 2;

Upvotes: 1

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