Как найти размер базы данных, схему, таблицу в redshift

Ответ 1

Приведенные выше ответы не всегда дают правильные ответы на используемое табличное пространство. Поддержка AWS предоставила этот запрос:

SELECT   TRIM(pgdb.datname) AS Database,
         TRIM(a.name) AS Table,
         ((b.mbytes/part.total::decimal)*100)::decimal(5,2) AS pct_of_total,
         b.mbytes,
         b.unsorted_mbytes
FROM     stv_tbl_perm a
JOIN     pg_database AS pgdb
  ON     pgdb.oid = a.db_id
JOIN     ( SELECT   tbl,
                    SUM( DECODE(unsorted, 1, 1, 0)) AS unsorted_mbytes,
                    COUNT(*) AS mbytes
           FROM     stv_blocklist
           GROUP BY tbl ) AS b
       ON a.id = b.tbl
JOIN     ( SELECT SUM(capacity) AS total
           FROM   stv_partitions
           WHERE  part_begin = 0 ) AS part
      ON 1 = 1
WHERE    a.slice = 0
ORDER BY 4 desc, db_id, name;

Ответ 2

Да, mbytes в вашем примере - 372Mb. Вот что я использовал:

select
  cast(use2.usename as varchar(50)) as owner, 
  pgc.oid,
  trim(pgdb.datname) as Database,
  trim(pgn.nspname) as Schema,
  trim(a.name) as Table,
  b.mbytes,
  a.rows
from 
 (select db_id, id, name, sum(rows) as rows
  from stv_tbl_perm a
  group by db_id, id, name
  ) as a
 join pg_class as pgc on pgc.oid = a.id
 left join pg_user use2 on (pgc.relowner = use2.usesysid)
 join pg_namespace as pgn on pgn.oid = pgc.relnamespace 
    and pgn.nspowner > 1
 join pg_database as pgdb on pgdb.oid = a.db_id
 join 
   (select tbl, count(*) as mbytes
    from stv_blocklist
    group by tbl
   ) b on a.id = b.tbl
 order by mbytes desc, a.db_id, a.name; 

Ответ 3

Я не уверен в группировке по базе данных и схеме, но здесь короткий способ получить использование по таблице,

SELECT tbl, name, size_mb FROM
(
  SELECT tbl, count(*) AS size_mb
  FROM stv_blocklist
  GROUP BY tbl
)
LEFT JOIN
(select distinct id, name FROM stv_tbl_perm)
ON id = tbl
ORDER BY size_mb DESC
LIMIT 10;

Ответ 4

Модифицированные версии одного из других ответов. Это включает имя базы данных, имя схемы, имя таблицы, общее количество строк, размер на диске и несортированный размер:

-- sort by row count
select trim(pgdb.datname) as Database, trim(pgns.nspname) as Schema, trim(a.name) as Table,
    c.rows, ((b.mbytes/part.total::decimal)*100)::decimal(5,3) as pct_of_total, b.mbytes, b.unsorted_mbytes
    from stv_tbl_perm a
    join pg_class as pgtbl on pgtbl.oid = a.id
    join pg_namespace as pgns on pgns.oid = pgtbl.relnamespace
    join pg_database as pgdb on pgdb.oid = a.db_id
    join (select tbl, sum(decode(unsorted, 1, 1, 0)) as unsorted_mbytes, count(*) as mbytes from stv_blocklist group by tbl) b on a.id=b.tbl
    join (select id, sum(rows) as rows from stv_tbl_perm group by id) c on a.id=c.id
    join (select sum(capacity) as total from stv_partitions where part_begin=0) as part on 1=1
    where a.slice=0
    order by 4 desc, db_id, name;


-- sort by space used
select trim(pgdb.datname) as Database, trim(pgns.nspname) as Schema, trim(a.name) as Table,
    c.rows, ((b.mbytes/part.total::decimal)*100)::decimal(5,3) as pct_of_total, b.mbytes, b.unsorted_mbytes
    from stv_tbl_perm a
    join pg_class as pgtbl on pgtbl.oid = a.id
    join pg_namespace as pgns on pgns.oid = pgtbl.relnamespace
    join pg_database as pgdb on pgdb.oid = a.db_id
    join (select tbl, sum(decode(unsorted, 1, 1, 0)) as unsorted_mbytes, count(*) as mbytes from stv_blocklist group by tbl) b on a.id=b.tbl
    join (select id, sum(rows) as rows from stv_tbl_perm group by id) c on a.id=c.id
    join (select sum(capacity) as total from stv_partitions where part_begin=0) as part on 1=1
    where a.slice=0
    order by 6 desc, db_id, name;

Ответ 5

вы можете проверить этот репозиторий, я уверен, что вы найдете там полезный материал.

https://github.com/awslabs/amazon-redshift-utils

для ответа на ваш вопрос вы можете использовать это представление: https://github.com/awslabs/amazon-redshift-utils/blob/master/src/AdminViews/v_space_used_per_tbl.sql

и затем запросите, как вам нравится. например: select * from admin.v_space_used_per_tbl;

Ответ 6

Этот запрос намного проще:

- Список 30 самых больших таблиц в вашем кластере

SELECT 
 "schema"
,"table"  AS table_name
,ROUND((size/1024.0),2) AS "Size in Gigabytes"
,pct_used AS "Physical Disk Used by This Table"
FROM svv_table_info
ORDER BY pct_used DESC
LIMIT 30;

Ответ 7

Это то, что я использую (пожалуйста, измените имя базы данных из "mydb" на ваше имя базы данных):

SELECT CAST(use2.usename AS VARCHAR(50)) AS OWNER
 ,TRIM(pgdb.datname) AS DATABASE
 ,TRIM(pgn.nspname) AS SCHEMA
 ,TRIM(a.NAME) AS TABLE
 ,(b.mbytes) / 1024 AS Gigabytes
 ,a.ROWS
FROM (
 SELECT db_id
 ,id
 ,NAME
 ,SUM(ROWS) AS ROWS
 FROM stv_tbl_perm a
 GROUP BY db_id
 ,id
 ,NAME
 ) AS a
JOIN pg_class AS pgc ON pgc.oid = a.id
LEFT JOIN pg_user use2 ON (pgc.relowner = use2.usesysid)
JOIN pg_namespace AS pgn ON pgn.oid = pgc.relnamespace
 AND pgn.nspowner > 1
JOIN pg_database AS pgdb ON pgdb.oid = a.db_id
JOIN (
 SELECT tbl
 ,COUNT(*) AS mbytes
 FROM stv_blocklist
 GROUP BY tbl
 ) b ON a.id = b.tbl
WHERE pgdb.datname = 'mydb'
ORDER BY mbytes DESC
 ,a.db_id
 ,a.NAME;

src: https://aboutdatabases.wordpress.com/2015/01/24/amazon-redshift-how-to-get-the-sizes-of-all-tables/