我正在寻找一种方法,为我在Postgres中所有的表找到行数。我知道我可以一次做一张表:

SELECT count(*) FROM table_name;

但我想看看所有表的行数,然后按它排序,以了解所有表的大小。


当前回答

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

其他回答

这里有一个更简单的方法。

tables="$(echo '\dt' | psql -U "${PGUSER}" | tail -n +4 | head -n-2 | tr -d ' ' | cut -d '|' -f2)"
for table in $tables; do
printf "%s: %s\n" "$table" "$(echo "SELECT COUNT(*) FROM $table;" | psql -U "${PGUSER}" | tail -n +3 | head -n-2 | tr -d ' ')"
done

输出应该如下所示

auth_group: 0
auth_group_permissions: 0
auth_permission: 36
auth_user: 2
auth_user_groups: 0
auth_user_user_permissions: 0
authtoken_token: 2
django_admin_log: 0
django_content_type: 9
django_migrations: 22
django_session: 0
mydata_table1: 9011
mydata_table2: 3499

你可以根据需要更新psql -U "${PGUSER}"部分来访问你的数据库

注意,head -n-2语法可能在macOS中不起作用,你可以使用不同的实现

在CentOS 7下的psql (PostgreSQL) 11.2上测试


如果你想按表排序,那就用sort来包装它

for table in $tables; do
printf "%s: %s\n" "$table" "$(echo "SELECT COUNT(*) FROM $table;" | psql -U "${PGUSER}" | tail -n +3 | head -n-2 | tr -d ' ')"
done | sort -k 2,2nr

输出;

mydata_table1: 9011
mydata_table2: 3499
auth_permission: 36
django_migrations: 22
django_content_type: 9
authtoken_token: 2
auth_user: 2
auth_group: 0
auth_group_permissions: 0
auth_user_groups: 0
auth_user_user_permissions: 0
django_admin_log: 0
django_session: 0

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;

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。

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。

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;