如何在RedShift/ParAccel中测量磁盘上的表空间

Question

我在RedShift中有一个表。 如何查看它使用了多少磁盘空间？

Answer 1

从该演示使用查询：http://www.slideshare.net/AmazonWebServices/amazon-redshift-best-practices

分析磁盘空间使用情况集群：

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 mbytes desc, a.db_id, a.name; 

分析节点之间的表分配：一个模式过滤器

select slice, col, num_values, minvalue, maxvalue 
from svv_diskusage 
where name = '__INSERT__TABLE__NAME__HERE__' and col = 0 
order by slice, col; 

Answer 2

添加所有者和到以上查询：

select 
cast(use.usename as varchar(50)) as owner, 
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 use on (pgc.relowner = use.usesysid) 
join pg_namespace as pgn on pgn.oid = pgc.relnamespace 
    -- leave out system schemas 
    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; 

Answer 3

刚想过我会扩大这个，因为我面临着一个不均匀分布的问题。我添加了一些链接和字段，以便按节点和切片分析空间。同时添加的最大/最小值，并且每片价值数0列

select 
cast(use.usename as varchar(50)) as owner, 
trim(pgdb.datname) as Database, 
trim(pgn.nspname) as Schema, 
trim(a.name) as Table, 
a.node, 
a.slice, 
b.mbytes, 
a.rows, 
a.num_values, 
a.minvalue, 
a.maxvalue 
from 
(select 
    a.db_id, 
    a.id, 
    s.node, 
    s.slice, 
    a.name, 
    d.num_values, 
    d.minvalue, 
    d.maxvalue, 
    sum(rows) as rows 
    from stv_tbl_perm a 
    inner join stv_slices s on a.slice = s.slice 
    inner join (
    select tbl, slice, sum(num_values) as num_values, min(minvalue) as minvalue, max(maxvalue) as maxvalue 
    from svv_diskusage 
    where col = 0 
    group by 1, 2) d on a.id = d.tbl and a.slice = d.slice 
    group by 1, 2, 3, 4, 5, 6, 7, 8 
) as a 
join pg_class as pgc on pgc.oid = a.id 
left join pg_user use on (pgc.relowner = use.usesysid) 
join pg_namespace as pgn on pgn.oid = pgc.relnamespace 
    -- leave out system schemas 
    and pgn.nspowner > 1 
join pg_database as pgdb on pgdb.oid = a.db_id 
join 
    (select 
    tbl, 
    slice, 
    count(*) as mbytes 
    from stv_blocklist 
    group by tbl, slice 
) b on a.id = b.tbl 
    and a.slice = b.slice 
order by mbytes desc, a.db_id, a.name, a.node; 

Answer 4

我知道这个问题是旧的，已经接受了答案，但我必须指出的是，答案是错误的。 查询以“mb”输出的内容实际上是“块数”。只有块大小为1MB（这是默认值），答案才是正确的。

如果块大小不同（在我的情况下，例如是256K），则必须将块数乘以其大小（以字节为单位）。我建议如下修改您的查询，我的块大小相乘以字节为单位（262144个字节）块的数量，然后通过（1024 * 1024），以MB为单位的总分为输出：

select 
    trim(pgdb.datname) as Database, 
    trim(pgn.nspname) as Schema, 
    trim(a.name) as Table, 
    b.mbytes as previous_wrong_value, 
    (b.mbytes * 262144)::bigint/(1024*1024) as "Total 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(blocknum) as mbytes 
    from stv_blocklist 
    group by tbl 
) b on a.id = b.tbl 
order by mbytes desc, a.db_id, a.name;