以下Postgres SQL查询将列出所有模式及其大小和索引大小的所有表。如果表只是一个索引表,它将显示为100%索引。
SELECT schema,
name,
pg_size_pretty(CASE WHEN is_index THEN 0 ELSE s END) AS size,
pg_size_pretty(CASE WHEN is_index THEN s ELSE st - s END) AS index,
CASE WHEN st = 0 THEN 0
WHEN is_index THEN 100
ELSE 100 - ((s*100) / st) END || '%' as ratio,
pg_size_pretty(st) as total
FROM (SELECT *,
st = s AS is_index
FROM (SELECT nspname as schema,
relname as name,
pg_relation_size(nspname || '.' || relname) as s,
pg_total_relation_size(nspname || '.' || relname) as st
FROM pg_class
JOIN pg_namespace ON (relnamespace = pg_namespace.oid)) AS p)
AS pp
ORDER BY st DESC LIMIT 30;
它将给出以下结果:
schema | name | size | index | ratio | total
----------------+------------------------+---------+---------+-------+---------
public | conf | 4072 kB | 4360 kB | 52% | 8432 kB
archive | product_param | 4048 kB | 3968 kB | 50% | 8016 kB
public | conf_pkey | 0 bytes | 4320 kB | 100% | 4320 kB
archive | product_value | 1568 kB | 1136 kB | 43% | 2704 kB
public | param_mapping | 1472 kB | 832 kB | 37% | 2304 kB
archive | supplie_price | 944 kB | 896 kB | 49% | 1840 kB
public | product_param_param_id | 0 bytes | 1552 kB | 100% | 1552 kB
archive | product_param_id | 0 bytes | 1536 kB | 100% | 1536 kB
我已经到了一个地方,我看不到所有树木的森林,而且它开始变得有点笨拙。
我想知道其中有什么东西可以简化或冗余吗?如果可以使查询更简单,则列不一定保持不变。
答案 0 :(得分:3)
我使用此查询获得可比较的结果(具有不同的格式):
select
nspname as schema,
relname as name,
pg_relation_size(pg_class.oid) as size,
pg_indexes_size(pg_class.oid) as index,
pg_total_relation_size(pg_class.oid) as total,
100 * case when relkind = 'i' then pg_relation_size(pg_class.oid)
else pg_indexes_size(pg_class.oid) end
/ pg_total_relation_size(pg_class.oid) as i_ratio
from
pg_class
join pg_namespace on relnamespace = pg_namespace.oid
order by 5 desc
答案 1 :(得分:1)
首先,为什么不使用CTE,它们会让你的代码更具可读性。 然后你不返回is_index所以它似乎是冗余
with p as (
SELECT nspname as schema,
relname as name,
pg_relation_size(nspname || '.' || relname) as s,
pg_total_relation_size(nspname || '.' || relname) as st
FROM pg_class
JOIN pg_namespace
ON (relnamespace = pg_namespace.oid)
),
pp as (
SELECT *,
case when st = s then 0 else s end as size,
case when st = s then s else st-s end as index
FROM p
)
select schema,
name,
pg_size_pretty(size) as size,
pg_size_pretty(index) as index,
(case st
when 0 then 0
else index*100 / st
end) || '%' ratio,
st total
from pp
order by st desc limit 30;
答案 2 :(得分:1)
我真正想做的就是指出quzary的响应应该是使用oid而不是创建无法解析回oid的字符串。
现在我必须写一篇合适的帖子(也许这是阻止新手评论的重点?)这是另一个清洁和漂亮的版本:
WITH p AS (
SELECT n.nspname AS schema,
c.relname AS name,
pg_relation_size(c.oid) AS s,
pg_total_relation_size(c.oid) AS st
FROM pg_class c, pg_namespace n
WHERE c.relnamespace = n.oid
)
SELECT schema, name,
pg_size_pretty(s) AS size,
pg_size_pretty(st - s) AS index,
(100.0 * s / NULLIF(st, 0))::numeric(10,1) AS "% data of total",
st AS total
FROM p
ORDER BY st DESC
LIMIT 30;
请注意,添加以下行可能很有用:
AND c.relkind = 'r'
WHERE
p
条款中的。这将仅限于关系/表,并使代码对表大小的一般摘要有用。
答案 3 :(得分:0)
不要忘记pg_relation_size
和pg_total_relation_size
不区分大小写!
pg_relation_size(nspname || '.' || relname)
实际应该是:
pg_relation_size('"' || nspname || '.' || relname || '"')
所以它也适用于大写。 (花了一些时间才弄明白这个)