Big query find data that could be in multiple columns - sql

Question

I have a table with the following data
id|task1_name|task1_date|task2_name|task2_date
1,breakfast,1/1/20,,
2,null,null,breakfast,,1/1/20
3,null,null,lunch,,1/1/20
4,dinner,1/1/20,lunch,1/1/10
I'd like to build a view that always displayed the task names in the same column or null if it could not be found in any of the columns e.g.
id|dinner_date|lunch_date|breakfast_date
1,1/1/20, null, null
2,null, null, 1/1/20
2,1/1/20, 1/1/10, null
I've tried using a nested IF statement e.g.
SELECT *
IF(task_1_name = 'dinner', task1_date, IF(task2_date = 'dinner', task2_date, NULL)) as `dinner_date`
FROM t
But as there are 50 or so columns in the real dataset, this seems like a stupid solution and would get complex very quickly, is there a smarter way here?

Answer 1

One method uses case expressions:
select t.*,
(case when task1_name = 'dinner' then task1_date
when task2_name = 'dinner' then task2_date
when task3_name = 'dinner' then task3_date
end) as dinner_date
from t;

Answer 2

Below is for BigQuery Standard SQL and generic enough to addresses concerns expressed in question. You don't need to know in advance number of columns and tasks names (although they should not have , or : which should not be a big limitation here and can be addressed if needed)
#standardSQL
CREATE TEMP TABLE ttt AS
SELECT id,
SPLIT(k, '_')[OFFSET(0)] task,
MAX(IF(SPLIT(k, '_')[OFFSET(1)] = 'name', v, NULL)) AS name,
MAX(IF(SPLIT(k, '_')[OFFSET(1)] = 'date', v, NULL)) AS DAY
FROM (
SELECT id,
TRIM(SPLIT(kv, ':')[OFFSET(0)], '"') k,
TRIM(SPLIT(kv, ':')[OFFSET(1)], '"') v
FROM `project.dataset.table` t,
UNNEST(SPLIT(TRIM(TO_JSON_STRING(t), '{}'))) kv
WHERE TRIM(SPLIT(kv, ':')[OFFSET(0)], '"') != 'id'
AND TRIM(SPLIT(kv, ':')[OFFSET(1)], '"') != 'null'
)
GROUP BY id, task;
EXECUTE IMMEDIATE '''
SELECT id, ''' || (
SELECT STRING_AGG(DISTINCT "MAX(IF(name = '" || name || "', day, NULL)) AS " || name || "_date")
FROM ttt
) || '''
FROM ttt
GROUP BY 1
ORDER BY 1
'''
Note; the assumption here is only about columns name to be task<N>_name and task<N>_date
If to apply to sample data (similar) to yours in question
WITH `project.dataset.table` AS (
SELECT 1 id, 'breakfast' task1_name, '1/1/21' task1_date, NULL task2_name, NULL task2_date UNION ALL
SELECT 2, NULL, NULL, 'breakfast', '1/1/22' UNION ALL
SELECT 3, NULL, NULL, 'lunch', '1/1/23' UNION ALL
SELECT 4, 'dinner', '1/1/24', 'lunch', '1/1/10'
)
output is
Row id breakfast_date lunch_date dinner_date
1 1 1/1/21 null null
2 2 1/1/22 null null
3 3 null 1/1/23 null
4 4 null 1/1/10 1/1/24

Answer 3

Here is another solution which doesn't use dynamic SQL, doesn't rely on specific column names and works with arbitrary number of columns:
WITH table AS (
SELECT 1 id, 'breakfast' task1_name, '1/1/21' task1_date, NULL task2_name, NULL task2_date UNION ALL
SELECT 2, NULL, NULL, 'breakfast', '1/1/22' UNION ALL
SELECT 3, NULL, NULL, 'lunch', '1/1/23' UNION ALL
SELECT 4, 'dinner', '1/1/24', 'lunch', '1/1/10'
)
SELECT
REGEXP_EXTRACT(f, r'breakfast\, ([^\,\)]*)'),
REGEXP_EXTRACT(f, r'lunch\, ([^\,\)]*)'),
REGEXP_EXTRACT(f, r'dinner\, ([^\,\)]*)')
FROM (
SELECT FORMAT("%t", t) f FROM table t
)