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。
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
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。
其他回答
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;
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
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。
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! !