我通常不依赖于统计数据，尤其是在PostgreSQL中。

SELECT table_name, dsql2('select count(*) from '||table_name) as rownum
FROM information_schema.tables
WHERE table_type='BASE TABLE'
    AND table_schema='livescreen'
ORDER BY 2 DESC;

CREATE OR REPLACE FUNCTION dsql2(i_text text)
  RETURNS int AS
$BODY$
Declare
  v_val int;
BEGIN
  execute i_text into v_val;
  return v_val;
END; 
$BODY$
  LANGUAGE plpgsql VOLATILE
  COST 100;

我喜欢Daniel Vérité的回答。 但是当你不能使用CREATE语句时，你可以使用bash解决方案，如果你是windows用户，可以使用powershell解决方案:

# You don't need this if you have pgpass.conf
$env:PGPASSWORD = "userpass"

# Get table list
$tables = & 'C:\Program Files\PostgreSQL\9.4\bin\psql.exe' -U user -w -d dbname -At -c "select table_name from information_schema.tables where table_type='BASE TABLE' AND table_schema='schema1'"

foreach ($table in $tables) {
    & 'C:\path_to_postresql\bin\psql.exe' -U root -w -d dbname -At -c "select '$table', count(*) from $table"
}

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! !

如果您不介意可能过时的数据，您可以访问查询优化器使用的相同统计信息。

喜欢的东西:

SELECT relname, n_tup_ins - n_tup_del as rowcount FROM pg_stat_all_tables;

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
[..]

要获得估计，请参阅格雷格·史密斯的答案。

为了得到确切的数字，到目前为止，其他答案都受到一些问题的困扰，其中一些问题很严重(见下文)。这里有一个版本，希望更好:

CREATE FUNCTION rowcount_all(schema_name text default 'public')
  RETURNS table(table_name text, cnt bigint) as
$$
declare
 table_name text;
begin
  for table_name in SELECT c.relname FROM pg_class c
    JOIN pg_namespace s ON (c.relnamespace=s.oid)
    WHERE c.relkind = 'r' AND s.nspname=schema_name
  LOOP
    RETURN QUERY EXECUTE format('select cast(%L as text),count(*) from %I.%I',
       table_name, schema_name, table_name);
  END LOOP;
end
$$ language plpgsql;

它接受模式名作为参数，如果没有给出参数，则接受public。

要使用特定的模式列表或来自查询的列表而不修改函数，可以从查询中调用它，如下所示:

WITH rc(schema_name,tbl) AS (
  select s.n,rowcount_all(s.n) from (values ('schema1'),('schema2')) as s(n)
)
SELECT schema_name,(tbl).* FROM rc;

这将生成一个包含模式、表和行计数的3列输出。

下面是这个函数避免的其他答案中的一些问题:

Table and schema names shouldn't be injected into executable SQL without being quoted, either with quote_ident or with the more modern format() function with its %I format string. Otherwise some malicious person may name their table tablename;DROP TABLE other_table which is perfectly valid as a table name. Even without the SQL injection and funny characters problems, table name may exist in variants differing by case. If a table is named ABCD and another one abcd, the SELECT count(*) FROM... must use a quoted name otherwise it will skip ABCD and count abcd twice. The %I of format does this automatically. information_schema.tables lists custom composite types in addition to tables, even when table_type is 'BASE TABLE' (!). As a consequence, we can't iterate oninformation_schema.tables, otherwise we risk having select count(*) from name_of_composite_type and that would fail. OTOH pg_class where relkind='r' should always work fine. The type of COUNT() is bigint, not int. Tables with more than 2.15 billion rows may exist (running a count(*) on them is a bad idea, though). A permanent type need not to be created for a function to return a resultset with several columns. RETURNS TABLE(definition...) is a better alternative.