[SOLVED] Is it possible in snowflake to write a query that lists the columns that have all null values?

Is it possible in snowflake to write a query that lists the columns that have all null values?

In snowsight within snowflake, you can profile tables and see the % of null values in the UI, but is there an easy way to query for this data or export it from the UI? I just need to create a new table off a table with 1k+, but exclude columns that only have null vaules?

Input:

CREATE OR REPLACE TABLE TAB(Q INT, X INT, Y INT, Z INT)
AS
SELECT NULL, 0041X00002OEkZ5QAL,2, NULL UNION ALL
SELECT NULL, NULL, NULL, NULL UNION ALL
SELECT NULL, NULL, NULL, NULL UNION ALL
SELECT NULL, NULL, 2, NULL;

Solution

Using Snowpark to remove all-nullable columns:

import snowflake.snowpark as snowpark
from snowflake.snowpark.functions import col, count

def main(session: snowpark.Session): 
    # input
    df = session.table('PUBLIC.TAB')

    # columns with all nulls
    dropcols = [c for c in df.columns if df.agg(count(col(c))).collect()[0][0]==0]

    # save as new table without nulls columns
    df.drop(dropcols).write.save_as_table('PUBLIC.TAB_WITHOUT_NULL',mode='overwrite')
            
    return 'Success'

For input:

CREATE OR REPLACE TABLE TAB(Q INT, X INT, Y INT, Z INT)
AS
SELECT NULL, 1,2, NULL UNION ALL
SELECT NULL, NULL, NULL, NULL UNION ALL
SELECT NULL, 3, NULL, NULL UNION ALL
SELECT NULL, NULL, 2, NULL;

SELECT * FROM TAB;
/*
Q   X   Y   Z
    1   2   
        
    3       
        2   
*/

Output:

SELECT * FROM TAB_WITHOUT_NULL;
/*
X   Y
1   2
    
3   
    2
*/