Question

我在使用函数启动程序时遇到无法连接到数据库的问题。如果我没有用函数启动它，它可以正常工作。我的程序从serverlist.txt中获取计算机名称并在数据库中查找它。然后它给了我一个＆＃34;位置ID＆＃34;那台电脑。

此版本有效：

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。它转向了例外情况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()

关于导致这种情况的任何想法？

Answer 1

当你在start函数中放入连接时它变成了一个本地对象而其他函数无法获得连接！

如果他们使用相同的连接，你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):

这样改变它应该没问题（把抓取功能放在试用部分内）

如何与功能连接Pyodbc？

1 个答案: