下面的查询将给出每个表的行数和大小

选择table_schema, table_name， pg_relation_size('“' | | table_schema | |”“。”“| | table_name | |”“”)/ 1024/1024 size_MB, (xpath('/row/c/text()'， query_to_xml(format('select count(*) AS c from %I.)%I'， table_schema, table_name)， false, true， ")))[1]::text::int AS rows_n 从information_schema.tables order by size_MB

我想从所有表的总数+表的列表与他们的计数。有点像绩效表，显示大部分时间都花在了哪里

WITH results AS ( 
  SELECT nspname AS schemaname,relname,reltuples
    FROM pg_class C
    LEFT JOIN pg_namespace N ON (N.oid = C.relnamespace)
    WHERE 
      nspname NOT IN ('pg_catalog', 'information_schema') AND
      relkind='r'
     GROUP BY schemaname, relname, reltuples
)

SELECT * FROM results
UNION
SELECT 'all' AS schemaname, 'all' AS relname, SUM(reltuples) AS "reltuples" FROM results

ORDER BY reltuples DESC

当然，你也可以在这个版本的结果上加上一个LIMIT条款，这样你就可以得到最大的n个违例者以及总数。

需要注意的一点是，在大量进口后，您需要让它静置一段时间。我通过跨几个表向数据库中添加5000行(使用实际导入数据)来测试这一点。它显示了大约一分钟的1800条记录(可能是一个可配置的窗口)

这是基于https://stackoverflow.com/a/2611745/1548557的工作，所以感谢并认可在CTE中使用的查询

对于那些试图评估他们需要哪一个Heroku计划，又不能等待Heroku的慢行计数器刷新的人来说，一个简单实用的答案是:

基本上你想在psql中运行\dt，将结果复制到你最喜欢的文本编辑器中(它看起来像这样:

 public | auth_group                     | table | axrsosvelhutvw
 public | auth_group_permissions         | table | axrsosvelhutvw
 public | auth_permission                | table | axrsosvelhutvw
 public | auth_user                      | table | axrsosvelhutvw
 public | auth_user_groups               | table | axrsosvelhutvw
 public | auth_user_user_permissions     | table | axrsosvelhutvw
 public | background_task                | table | axrsosvelhutvw
 public | django_admin_log               | table | axrsosvelhutvw
 public | django_content_type            | table | axrsosvelhutvw
 public | django_migrations              | table | axrsosvelhutvw
 public | django_session                 | table | axrsosvelhutvw
 public | exercises_assignment           | table | axrsosvelhutvw

)，然后运行regex搜索并替换，如下所示:

^[^|]*\|\s+([^|]*?)\s+\| table \|.*$

to:

select '\1', count(*) from \1 union/g

这将会给你一个非常类似的结果:

select 'auth_group', count(*) from auth_group union
select 'auth_group_permissions', count(*) from auth_group_permissions union
select 'auth_permission', count(*) from auth_permission union
select 'auth_user', count(*) from auth_user union
select 'auth_user_groups', count(*) from auth_user_groups union
select 'auth_user_user_permissions', count(*) from auth_user_user_permissions union
select 'background_task', count(*) from background_task union
select 'django_admin_log', count(*) from django_admin_log union
select 'django_content_type', count(*) from django_content_type union
select 'django_migrations', count(*) from django_migrations union
select 'django_session', count(*) from django_session
;

(您需要删除最后一个联合，并手动在末尾添加分号)

在psql中运行它，就完成了。

            ?column?            | count
--------------------------------+-------
 auth_group_permissions         |     0
 auth_user_user_permissions     |     0
 django_session                 |  1306
 django_content_type            |    17
 auth_user_groups               |   162
 django_admin_log               |  9106
 django_migrations              |    19
[..]

您可以使用此查询生成所有表名及其计数

select ' select  '''|| tablename  ||''', count(*) from ' || tablename ||' 
union' from pg_tables where schemaname='public'; 

上述查询的结果将是

select  'dim_date', count(*) from dim_date union 
select  'dim_store', count(*) from dim_store union
select  'dim_product', count(*) from dim_product union
select  'dim_employee', count(*) from dim_employee union

您需要删除最后一个联合符，并在末尾添加分号!!

select  'dim_date', count(*) from dim_date union 
select  'dim_store', count(*) from dim_store union
select  'dim_product', count(*) from dim_product union
select  'dim_employee', count(*) from dim_employee  **;**

跑! !

摘自我在GregSmith的回答中的评论，使其更具可读性:

with tbl as (
  SELECT table_schema,table_name 
  FROM information_schema.tables
  WHERE table_name not like 'pg_%' AND table_schema IN ('public')
)
SELECT 
  table_schema, 
  table_name, 
  (xpath('/row/c/text()', 
    query_to_xml(format('select count(*) AS c from %I.%I', table_schema, table_name), 
    false, 
    true, 
    '')))[1]::text::int AS rows_n 
FROM tbl ORDER BY 3 DESC;

感谢@ a_horis_with_no_name

下面是一个解决方案，它不需要函数来获得每个表的精确计数:

select table_schema, 
       table_name, 
       (xpath('/row/cnt/text()', xml_count))[1]::text::int as row_count
from (
  select table_name, table_schema, 
         query_to_xml(format('select count(*) as cnt from %I.%I', table_schema, table_name), false, true, '') as xml_count
  from information_schema.tables
  where table_schema = 'public' --<< change here for the schema you want
) t

query_to_xml将运行传递的SQL查询并返回带有结果的XML(该表的行数)。外层xpath()将从该xml中提取计数信息并将其转换为数字

实际上并不需要派生表，但可以使xpath()更容易理解——否则整个query_to_xml()将需要传递给xpath()函数。