python之如何使用函数连接Pyodbc

我遇到了一个问题，即在使用函数启动程序时无法连接到数据库。如果我不使用函数启动它，它就可以正常工作。 我的程序从 serverlist.txt 中获取一个计算机名称并在数据库中查找它。然后它会为我提供该计算机的“位置 ID”。

这个版本有效:

import os 
import shutil 
import fileinput 
import pypyodbc 
 
def replaceid(servername): 
    try: 
        cursor = connection.cursor() 
 
        SQLCommand = ("SELECT Name, Location_ID " 
            "FROM dbo.I_Location "   # table name 
            "with (nolock)" 
            "WHERE Name = ?") 
        Values = [servername] 
        cursor.execute(SQLCommand,Values) 
        results = cursor.fetchone() 
        if results: 
 
            print (" Name: " + results[0] + " Location ID: " + str(results[1])) 
            print (" ") 
        else: 
            print (" Location ID for " + servername + " does not exist.") 
            print (" ") 
            connection.close() 
    except: 
        print("Database is down or you are not connected to network.") 
        exit() 
 
def grab(servername): 
# copy config from remote computer 
 
    source = r'//' + servername + '/c$/Administrator/' 
    dest = "." 
    file = "Admin.config" 
    if os.path.isfile(os.path.join(source, file)) 
        try: 
            shutil.copyfile(os.path.join(source, file), os.path.join(dest, file)) 
 
        except: 
            print (" Local directory you are copying to does not exist.") 
    else:  
        pass 
 
    replaceid(servername) 
 
 
 
os.system('cls' if os.name == 'nt' else 'clear') 
array = [] 
with open("serverlist.txt", "r") as f: 
    for servername in f: 
 
        try: 
            connection = pypyodbc.connect('Driver={SQL Server};Server=mydbx;Database=WinOasis;Trusted_Connection=yes;')  
        except pypyodbc.Error as ex: 
            sqlstate = ex.args[0] 
            if sqlstate == '28000': 
                print ("You do not have access.") 
        grab(servername.strip()) 

当我在底部添加 start() 函数时，它不起作用。它移动到说的异常(exception) 数据库已关闭或您未连接到网络。

import os 
import shutil 
import fileinput 
import pypyodbc 
 
def replaceid(servername): 
    try: 
        cursor = connection.cursor() 
 
        SQLCommand = ("SELECT Name, Location_ID " 
            "FROM dbo.I_Location "   # table name 
            "with (nolock)" 
            "WHERE Name = ?") 
        Values = [servername] 
        cursor.execute(SQLCommand,Values) 
        results = cursor.fetchone() 
        if results: 
 
            print (" Name: " + results[0] + " Location ID: " + str(results[1])) 
            print (" ") 
        else: 
            print (" Location ID for " + servername + " does not exist.") 
            print (" ") 
            connection.close() 
    except: 
        print("Database is down or you are not connected to network.") 
        exit() 
 
def grab(servername): 
# copy config from remote computer 
 
    source = r'//' + servername + '/c$/Administrator/' 
    dest = "." 
    file = "Admin.config" 
    if os.path.isfile(os.path.join(source, file)) 
        try: 
            shutil.copyfile(os.path.join(source, file), os.path.join(dest, file)) 
 
        except: 
            print (" Local directory you are copying to does not exist.") 
    else:  
        pass 
 
    replaceid(servername) 
 
def start(): 
    # Option 1 
    os.system('cls' if os.name == 'nt' else 'clear') 
    array = [] 
    with open("serverlist.txt", "r") as f: 
        for servername in f: 
 
            try: 
                connection = pypyodbc.connect('Driver={SQL Server};Server=mydbx;Database=WinOasis;Trusted_Connection=yes;')  
            except pypyodbc.Error as ex: 
                sqlstate = ex.args[0] 
                if sqlstate == '28000': 
                    print ("You do not have access.") 
            grab(servername.strip()) 
 
start() 

关于造成这种情况的原因有什么想法吗？

请您参考如下方法:

当你将连接放在启动函数中时，它变成了一个本地对象，其他函数无法获得连接!!!

如果它们使用相同的连接，则必须将连接作为对象传递给每个函数!!!

grab(servername.strip(),connection) 
def grab(servername ,connection): 
def replaceid(servername,connection): 

这样改应该没问题(把grab函数放在try部分)