postgresql - 在 postgres 中，如何删除所有共享相同前缀的列

Question

我一直在使用以下代码删除所有共享相同前缀的表(在本例中删除名称以 'supenh_' 开头的所有表):

DO
$do$
DECLARE
   _tbl text;
BEGIN
FOR _tbl  IN
    SELECT quote_ident(table_schema) || '.'
        || quote_ident(table_name)      -- escape identifier and schema-qualify!
    FROM   information_schema.tables
    WHERE  table_name LIKE 'supenh_' || '%'  -- your table name prefix
    AND    table_schema NOT LIKE 'pg_%'     -- exclude system schemas
LOOP
  --  RAISE NOTICE '%',
   EXECUTE
  'DROP TABLE ' || _tbl;
END LOOP;
END
$do$;

有没有办法修改此代码/或使用不同的脚本以便从一个特定的表中删除所有以相同前缀开头的列(例如，'patient1_') ？

Answer 1

你可以把它写成 PL/pgSQL 函数:

CREATE OR REPLACE FUNCTION drop_columns_with_prefix(tbl_name TEXT, column_prefix TEXT)  RETURNS VOID AS
$BODY$
DECLARE
    _column TEXT;
BEGIN
    FOR _column  IN
        SELECT quote_ident(column_name)
        FROM   information_schema.columns
        WHERE  table_name = tbl_name
        AND    column_name LIKE column_prefix || '%'
        AND    table_schema NOT LIKE 'pg_%'
    LOOP
      --  RAISE NOTICE '%',
       EXECUTE
      'ALTER TABLE ' || tbl_name || ' DROP COLUMN ' || _column;
    END LOOP;
END
$BODY$
LANGUAGE plpgsql VOLATILE;

调用它使用:

SELECT drop_columns_with_prefix('tbl_name', 'prefix_');

或者如果您不想将其用作函数:

DO
$do$
DECLARE
    _column TEXT;
BEGIN
FOR _column  IN
    SELECT quote_ident(column_name)
    FROM   information_schema.columns
    WHERE  table_name = 'tbl_name'
    AND    column_name LIKE 'prefix_%'
    AND    table_schema NOT LIKE 'pg_%'
LOOP
  --  RAISE NOTICE '%',
   EXECUTE
  'ALTER TABLE tbl_name DROP COLUMN ' || _column;
END LOOP;
END
$do$