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
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。
其他回答
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! !
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
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
这对我很有效
SELECT schemaname,relname,n_live_tup FROM pg_stat_user_tables ORDER BY n_live_tup DESC;
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