团队,
我的红移版本是:
PostgreSQL 8.0.2 on i686-pc-linux-gnu, compiled by GCC gcc (GCC) 3.4.2 20041017 (Red Hat 3.4.2-6.fc3), Redshift 1.0.735
如何找出数据库大小,表空间,架构大小&桌子大小?
但下面的版本不适用于redshift(适用于以上版本)
SELECT pg_database_size('db_name');
SELECT pg_size_pretty( pg_relation_size('table_name') );
是否有任何替代品可以找到像oracle(来自DBA_SEGMENTS)
对于tble size,我有以下查询,但不确定MBYTES的确切分析。对于第3行,MBYTES = 372.它意味着372 MB?
select 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
join pg_namespace as pgn on pgn.oid = pgc.relnamespace
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 a.db_id, a.name;
database | schema | table | mbytes | rows
---------------+--------------+------------------+--------+----------
postgres | public | company | 8 | 1
postgres | public | table_data1_1 | 7 | 1
postgres | proj_schema1 | table_data1 | 372 | 33867540
postgres | public | table_data1_2 | 40 | 2000001
(4 rows)
答案 0 :(得分:59)
以上答案并不总是为使用的表空间提供正确的答案。 AWS支持已将此查询用于:
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;
答案 1 :(得分:18)
是的,你的例子中的mbytes是372Mb。这是我一直在使用的:
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;
答案 2 :(得分:11)
我不确定按数据库和方案进行分组,但这是一种按表格使用的简短方法,
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;
答案 3 :(得分:7)
你可以查看这个存储库,我确定你会在那里找到有用的东西。
https://github.com/awslabs/amazon-redshift-utils
要回答您的问题,您可以使用此视图: https://github.com/awslabs/amazon-redshift-utils/blob/master/src/AdminViews/v_space_used_per_tbl.sql
然后根据需要进行查询。
例如:select * from admin.v_space_used_per_tbl;
答案 4 :(得分:7)
其他答案之一的修改版本。这包括数据库名称,模式名称,表名称,总行数,磁盘大小和未排序大小:
-- 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;
答案 5 :(得分:2)
此查询更容易:
- 列出群集中最大的30个表
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;
答案 6 :(得分:0)
SVV_TABLE_INFO
是Redshift系统表,显示有关Redshift数据库中用户定义表(而非其他系统表)的信息。该表仅对超级用户可见。
要获取每个表的大小,请在Redshift集群上运行以下命令:
SELECT "table", size, tbl_rows
FROM SVV_TABLE_INFO
table
列是表名。size
列是以MB为单位的表格大小。tbl_rows
列是表中的总行数,包括已标记为删除但尚未清除的行。 请参阅SVV_TABLE_INFO
Redshift文档,以获取其他有趣的列以从该系统表中检索。
答案 7 :(得分:-1)
这就是我正在使用的(请将数据库名称从'mydb'更改为您的数据库名称):
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/