返回非零退出状态 123

Question

我正在编写一些代码，假设从文件夹中获取权限列表，并将此列表与另一个列表之间的 users\group 列表进行比较，如果它们之间合适，他会编写文件夹路径 so 文件。 这是代码：

import subprocess
import sys
import os, pprint, sys
from sys import argv

script, FILENAME = argv  ##get folder\file
## copy permissions list to temporary file function
def check_per(path):   ##"check_per" copy permission list to string and compares
        test=open('tempG','w')
    dir = 'GroupList'
    output = subprocess.check_output(['cacls', path]).splitlines()
    test.write(output)
    while (True):
        d=readline()
        if not d: break
        d=d[0:len(d)-1:]
        d=str(d)
        while (True):
            line=groups.readline()
            line=line[0:len(line)-1:]
            if not line: break
            temp=d.find(line, 0, len(d))
            if temp!=-1: 
                dest=open(dir+'/'+line+'.txt','a')
                dest.write(path)
                dest.close()
        groups.seek(0)



temp= open('temp.txt','w')  ##temporary folder from sub-folders path list
file=open('folders_list.txt', 'w') ##arrange path in
##making list of all sub folders under the chosen path
temp.write("%s" %[x[0] for x in os.walk(os.path.abspath(FILENAME))])
temp.close()
temp=open('temp.txt', 'r')
##arrange every path in separate line
while (True):
    a=temp.read(1)
    if not a: break
    if a==',':file.write("\n")
    else:file.write(a)
file.close()
file=open('folders_list.txt', 'r')
temp.close()
os.remove('temp.txt') ##remove temp file    
groups=open('GroupList.txt','r') ###open the file that contain the groups list
dir = 'GroupList'
os.mkdir(dir) ##make new dir for groups files
## making text file for every group in new folder "grouplist" 
while(True):          ##make text file for every group
    a=groups.readline()
    if not a:break
    a=a[0:len(a)-1:]+'.txt'
    b=open(dir+'/'+a,'w')
    b.close()
while (True):  ##taking path from folders list and call "check_per" 
    path=file.readline()
    if not path: break
    path=path[0:len(path)-1:]
    check_per(path)

当我测试代码时，我收到以下错误消息： 输出 = subprocess.check_output(['cacls', path]).splitlines() 文件“C:\Python33\lib\subprocess.py”，第 586 行，在 check_output 中 引发 CalledProcessError(retcode, process.args, output=output) subprocess.CalledProcessError: 命令 '['cacls', "['D:\\Scripting\\permissi on project'"]' 返回非零退出状态 123

有什么想法吗？

Answer 1

消息是说命令返回退出状态 123，表示错误。

根据http://msdn.microsoft.com/en-us/library/windows/desktop/ms681382(v=vs.85).aspx状态123表示

文件名、目录名或卷标语法不正确。

也许你的目录名有问题？空间似乎太大了。