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

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

其他回答

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;

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

这对我很有效

SELECT schemaname,relname,n_live_tup FROM pg_stat_user_tables ORDER BY n_live_tup DESC;

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

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。