Question

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。

Answer 1

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

Answer 2

我不知道具体的psycopg2 lib，但可以使用以下查询来检查表的存在：

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。

Answer 3

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

Answer 4

#!/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()

Answer 5

第一个答案对我不起作用。我发现成功检查了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

Answer 6

我知道您要提供psycopg2答案，但是我想我会添加一个基于pandas的实用程序功能（它在后台使用了psycopg2），只是因为pd.read_sql_query()使事情变得如此便捷，例如避免创建/关闭游标。

import pandas as pd

def db_table_exists(conn, tablename):
    # thanks to Peter Hansen's answer for this sql
    sql = f"select * from information_schema.tables where table_name='{tablename}'" 
    
    # return results of sql query from conn as a pandas dataframe
    results_df = pd.read_sql_query(sql, conn)

    # True if we got any results back, False if we didn't
    return bool(len(results_df))

与这里的其他答案类似，我仍然使用psycopg2创建数据库连接对象conn。

Answer 7

以下解决方案也在处理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])

Answer 8

扩展上面对EXISTS的使用，我需要一些东西来测试表的总体存在。我发现在select语句上使用访存测试结果会在空的现有表上产生结果“ None”-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')

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

8 个答案: