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

不确定bash中的答案对您来说是否可以接受，但FWIW…

PGCOMMAND=" psql -h localhost -U fred -d mydb -At -c \"
            SELECT   table_name
            FROM     information_schema.tables
            WHERE    table_type='BASE TABLE'
            AND      table_schema='public'
            \""
TABLENAMES=$(export PGPASSWORD=test; eval "$PGCOMMAND")

for TABLENAME in $TABLENAMES; do
    PGCOMMAND=" psql -h localhost -U fred -d mydb -At -c \"
                SELECT   '$TABLENAME',
                         count(*) 
                FROM     $TABLENAME
                \""
    eval "$PGCOMMAND"
done

简单的两步:(注意:不需要改变任何东西-只是复制粘贴) 1. 创建函数

create function 
cnt_rows(schema text, tablename text) returns integer
as
$body$
declare
  result integer;
  query varchar;
begin
  query := 'SELECT count(1) FROM ' || schema || '.' || tablename;
  execute query into result;
  return result;
end;
$body$
language plpgsql;

2. 运行此查询获取所有表的行数

select sum(cnt_rows) as total_no_of_rows from (select 
  cnt_rows(table_schema, table_name)
from information_schema.tables
where 
  table_schema not in ('pg_catalog', 'information_schema') 
  and table_type='BASE TABLE') as subq;

或 按表获取行数

select
  table_schema,
  table_name, 
  cnt_rows(table_schema, table_name)
from information_schema.tables
where 
  table_schema not in ('pg_catalog', 'information_schema') 
  and table_type='BASE TABLE'
order by 3 desc;

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
[..]

如果您在psql shell中，使用\gexec允许您执行syed的答案和Aur的答案中描述的语法，而无需在外部文本编辑器中手动编辑。

with x (y) as (
    select
        'select count(*), '''||
        tablename||
        ''' as "tablename" from '||
        tablename||' '
    from pg_tables
    where schemaname='public'
)
select
    string_agg(y,' union all '||chr(10)) || ' order by tablename'
from x \gexec

注意，string_agg()既用于分隔所有语句之间的联合，也用于将分隔的数据箭头粉碎为一个单元，以便传递到缓冲区。

\ gexec 将当前查询缓冲区发送到服务器，然后将查询输出的每一行的每一列(如果有的话)视为要执行的SQL语句。

我不记得我收集这个的URL了。但希望这能帮助到你:

CREATE TYPE table_count AS (table_name TEXT, num_rows INTEGER); 

CREATE OR REPLACE FUNCTION count_em_all () RETURNS SETOF table_count  AS '
DECLARE 
    the_count RECORD; 
    t_name RECORD; 
    r table_count%ROWTYPE; 

BEGIN
    FOR t_name IN 
        SELECT 
            c.relname
        FROM
            pg_catalog.pg_class c LEFT JOIN pg_namespace n ON n.oid = c.relnamespace
        WHERE 
            c.relkind = ''r''
            AND n.nspname = ''public'' 
        ORDER BY 1 
        LOOP
            FOR the_count IN EXECUTE ''SELECT COUNT(*) AS "count" FROM '' || t_name.relname 
            LOOP 
            END LOOP; 

            r.table_name := t_name.relname; 
            r.num_rows := the_count.count; 
            RETURN NEXT r; 
        END LOOP; 
        RETURN; 
END;
' LANGUAGE plpgsql; 

执行select count_em_all();应该得到所有表的行数。