如何在 redshift 中查找数据库、模式、表的大小答案

【问题标题】：how to find size of database, schema, table in redshift如何在 redshift 中查找数据库、模式、表的大小
【发布时间】：2014-03-13 03:08:06
【问题描述】：

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、模式大小和表大小？

但以下不适用于红移（以上版本）

SELECT pg_database_size('db_name');
SELECT pg_size_pretty( pg_relation_size('table_name') );

有没有像 oracle 一样的替代方法（来自 DBA_SEGMENTS）

对于表大小，我有以下查询，但不确定 MBYTES 的确切含义。对于第 3 行，MBYTES = 372。这意味着 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)

【问题讨论】：

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 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; ------------此查询中 mbytes 的确切含义是什么。 ?

标签： amazon-web-services amazon-redshift

【解决方案1】：

SVV_TABLE_INFO 是一个 Redshift 系统表，它显示有关 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，请在您的 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 为单位的表大小。
tbl_rows 列是表中的总行数，包括已标记为删除但尚未清空的行。

Source

查看SVV_TABLE_INFO Redshift 文档以获取其他有趣的列以从该系统表中检索。

【讨论】：

【解决方案2】：

这个查询要容易得多：

-- 列出集群中最大的 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;

【讨论】：

【解决方案3】：

上述答案并不总是对所使用的表空间给出正确答案。 AWS 支持已提供此查询以供使用：

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;

【讨论】：

这是我正在使用的
此查询是否有意过滤一个切片？ WHERE a.slice = 0
@imcdnzl：什么是 unsorted_mbytes？当你计算总内存时，你需要将 mbytes 和 unsorted_mybytes 相加吗？
当你计算总内存时，你需要将 mbytes 和 unsorted_mybytes 相加吗？

【解决方案4】：

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;

【讨论】：

【解决方案5】：

你可以查看这个存储库，我相信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，您可以使用以下视图： https://github.com/awslabs/amazon-redshift-utils/blob/master/src/AdminViews/v_space_used_per_tbl.sql

然后随意查询。例如：select * from admin.v_space_used_per_tbl;

【讨论】：

【解决方案6】：

这是我正在使用的（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;

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

【讨论】：

这不起作用。它会从不是最大的表中挑选一小部分样本。

【解决方案7】：

我不确定按数据库和方案分组，但这里有一个按表获取使用情况的简短方法，

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;

【讨论】：

【解决方案8】：

是的，您的示例中的 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;

【讨论】：

这似乎没有得到公共架构中的任何表
“pgn.nspowner > 1”过滤掉了公共模式，这可能是@SandipPingle 也没有取回行的原因。
最佳答案！！