检查python下是否存在postgresql表(可能还有Psycopg2)

问题描述 投票:40回答:7

如何使用Psycopg2 Python库确定表是否存在?我想要一个真或假的布尔值。

python postgresql psycopg2
7个回答
66
投票

怎么样:

>>> import psycopg2
>>> conn = psycopg2.connect("dbname='mydb' user='username' host='localhost' password='foobar'")
>>> cur = conn.cursor()
>>> cur.execute("select * from information_schema.tables where table_name=%s", ('mytable',))
>>> bool(cur.rowcount)
True

使用EXISTS的替代方案更好,因为它不需要检索所有行,而只需要存在至少一个这样的行:

>>> cur.execute("select exists(select * from information_schema.tables where table_name=%s)", ('mytable',))
>>> cur.fetchone()[0]
True

19
投票

我具体不知道psycopg2 lib,但可以使用以下查询来检查表是否存在:

SELECT EXISTS(SELECT 1 FROM information_schema.tables 
              WHERE table_catalog='DB_NAME' AND 
                    table_schema='public' AND 
                    table_name='TABLE_NAME');

使用information_schema而不是直接从pg_ *表中进行选择的优点是查询的某种程度的可移植性。


3
投票
select exists(select relname from pg_class 
where relname = 'mytablename' and relkind='r');

2
投票
#!/usr/bin/python
# -*- coding: utf-8 -*-

import psycopg2
import sys


con = None

try:

    con = psycopg2.connect(database='testdb', user='janbodnar') 
    cur = con.cursor()
    cur.execute('SELECT 1 from mytable')          
    ver = cur.fetchone()
    print ver    //здесь наш код при успехе


except psycopg2.DatabaseError, e:
    print 'Error %s' % e    
    sys.exit(1)


finally:

    if con:
        con.close()

1
投票

第一个答案对我不起作用。我发现成功检查pg_class中的关系:

def table_exists(con, table_str):
    exists = False
    try:
        cur = con.cursor()
        cur.execute("select exists(select relname from pg_class where relname='" + table_str + "')")
        exists = cur.fetchone()[0]
        print exists
        cur.close()
    except psycopg2.Error as e:
        print e
    return exists

0
投票

以下解决方案也是处理schema

import psycopg2

with psycopg2.connect("dbname='dbname' user='user' host='host' port='port' password='password'") as conn:
    cur = conn.cursor()
    query = "select to_regclass(%s)"
    cur.execute(query, ['{}.{}'.format('schema', 'table')])

exists = bool(cur.fetchone()[0])

0
投票

扩展上面对EXISTS的使用,我需要一些东西来测试表的存在。我发现在select语句上使用fetch测试结果会在空的现有表上产生结果“None” - 这并不理想。

这是我想出的:

import psycopg2

def exist_test(tabletotest):

    schema=tabletotest.split('.')[0]
    table=tabletotest.split('.')[1]
    existtest="SELECT EXISTS (SELECT 1 FROM information_schema.tables WHERE table_schema = '"+schema+"' AND table_name = '"+table+"' );"

    print('existtest',existtest)

    cur.execute(existtest) # assumes youve already got your connection and cursor established

    # print('exists',cur.fetchall()[0])
    return ur.fetchall()[0] # returns true/false depending on whether table exists


exist_test('someschema.sometable')
© www.soinside.com 2019 - 2024. All rights reserved.