Drop all tables in a Redshift schema - without dropping permissions - sql

Question

I would be interested to drop all tables in a Redshift schema. Even though this solution works
DROP SCHEMA public CASCADE;
CREATE SCHEMA public;
is NOT good for me since that it drops SCHEMA permissions as well.
A solution like
DO $$ DECLARE
r RECORD;
BEGIN
-- if the schema you operate on is not "current", you will want to
-- replace current_schema() in query with 'schematodeletetablesfrom'
-- *and* update the generate 'DROP...' accordingly.
FOR r IN (SELECT tablename FROM pg_tables WHERE schemaname = current_schema()) LOOP
EXECUTE 'DROP TABLE IF EXISTS ' || quote_ident(r.tablename) || ' CASCADE';
END LOOP;
END $$;
as reported in this thread How can I drop all the tables in a PostgreSQL database?
would be ideal. Unfortunately it doesn't work on Redshift (apparently there is no support for for loops).
Is there any other solution to achieve it?

Answer 1

Run this SQL and copy+paste the result on your SQL client.
If you want to do it programmatically you need to built little bit code around it.
SELECT 'DROP TABLE IF EXISTS ' || tablename || ' CASCADE;'
FROM pg_tables
WHERE schemaname = '<your_schema>'

Answer 2

I solved it through a procedure that deletes all records. Using this technique to truncate fails but deleting it works fine for my intents and purposes.
create or replace procedure sp_truncate_dwh() as $$
DECLARE
tables RECORD;
BEGIN
FOR tables in SELECT tablename
FROM pg_tables
WHERE schemaname = 'dwh'
order by tablename
LOOP
EXECUTE 'delete from dwh.' || quote_ident(tables.tablename) ;
END LOOP;
RETURN;
END;
$$ LANGUAGE plpgsql;
--call sp_truncate_dwh()

Answer 3

In addition to demircioglu's answer, I had to add Commit after every drop statement to drop all tables in my schema. SELECT 'DROP TABLE IF EXISTS ' || tablename || ' CASCADE; COMMIT;' FROM pg_tables WHERE schemaname = '<your_schema>'
P.S.: I do not have required reputation to add this note as a comment and had to add as an answer.

Answer 4

Using Python and pyscopg2 locally on my PC I came up with this script to delete all tables in schema:
import psycopg2
schema = "schema_to_be_deleted"
try:
conn = psycopg2.connect("dbname='{}' port='{}' host='{}' user='{}' password='{}'".format("DB_NAME", "DB_PORT", "DB_HOST", "DB_USER", "DB_PWD"))
cursor = conn.cursor()
cursor.execute("SELECT tablename FROM pg_tables WHERE schemaname = '%s'" % schema)
rows = cursor.fetchall()
for row in rows:
cursor.execute("DROP TABLE {}.{}".format(schema, row[0]))
cursor.close()
conn.commit()
except psycopg2.DatabaseError as error:
logger.error(error)
finally:
if conn is not None:
conn.close()
Replace correctly values for DB_NAME, DB_PORT, DB_HOST, DB_USER and DB_PWD to connect to the Redshift DB

Answer 5

The following recipe differs from other answers in the regard that it generates one SQL statement for all tables we're going to delete.
SELECT
'DROP TABLE ' ||
LISTAGG("table", ', ') ||
';'
FROM
svv_table_info
WHERE
"table" LIKE 'staging_%';
Example result:
DROP TABLE staging_077815128468462e9de8ca6fec22f284, staging_abc, staging_123;
As in other answers, you will need to copy the generated SQL and execute it separately.
References
|| operator concatenates strings
LISTAGG function concatenates every table name into a string with a separator
The table svv_table_info is used because LISTAGG doesn't want to work with pg_tables for me. Complaint:
One or more of the used functions must be applied on at least one user created tables. Examples of user table only functions are LISTAGG, MEDIAN, PERCENTILE_CONT, etc
UPD. I just now noticed that SVV_TABLE_INFO page says:
The SVV_TABLE_INFO view doesn't return any information for empty tables.
...which means empty tables will not be in the list returned by this query. I usually delete transient tables to save disk space, so this does not bother me much; but in general this factor should be considered.