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
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
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。
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
这对我很有效
SELECT schemaname,relname,n_live_tup FROM pg_stat_user_tables ORDER BY n_live_tup DESC;
我做了一个小的变化,包括所有的表,也是非公共的表。
CREATE TYPE table_count AS (table_schema TEXT,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 table_schema,table_name
FROM information_schema.tables
where table_schema !=''pg_catalog''
and table_schema !=''information_schema''
ORDER BY 1,2
LOOP
FOR the_count IN EXECUTE ''SELECT COUNT(*) AS "count" FROM '' || t_name.table_schema||''.''||t_name.table_name
LOOP
END LOOP;
r.table_schema := t_name.table_schema;
r.table_name := t_name.table_name;
r.num_rows := the_count.count;
RETURN NEXT r;
END LOOP;
RETURN;
END;
' LANGUAGE plpgsql;
使用select count_em_all();叫它。
希望这对你有用。 保罗