Python 从ls-l命令的输出中提取模式

我想从ls-1hl命令中提取信息

我必须在不同的远程服务器上运行这个命令，从每个服务器上我得到不同类型的输出，因为其中一些是solaris，一些是GNU linux

下面是我要格式化的ls-1hl命令片段：-

sshuser@server'ls-1hl/path/to/directory'

ls_命令：-

-rw-r-----   1 root groupA     0 2014-08-04 10:42 a.log
-rw-------   1 root  groupA  48132720 Aug  1 23:45 core
-rwxr-xr-x   1 root  groupA      208 Mar 13 17:18 restart.sh
lrwxrwxrwx 1 root groupA   49 2014-06-06 09:32 MerSem -> ../combined_ABC/scripts
drwxrwxrwx   3 root groupA     1.0K Apr 22 16:12 logs

我想从上述片段中提取以下信息：-

第一线

permission = -rw-r-----  
links = 1
owner = rooot
group = groupA
size = 0
date = 2014-08-04 10:42
File/Directory Name = a.log

我的代码：-

Final_list = []
result = ls_command.split('\n')
for line in result:
   coldata = []
   a = re.search(r"^(.+?)\s{1,3}.*$",line)
   b = re.search(r"^.+?\s{1,3}(\d{1,5}).*$",line)
   c = re.search(r"^.+?\s{1,3}\d{1,5}(.+?)\s{1,3}.*$",line)
   d = re.search(r"^.+?\s{1,3}\d{1,5}.+?\s(.+?)\s{1,3}.*$",line)
   e = re.search(r"^.+?\s{1,3}\d{1,5}.+?\s.+?\s{1,3}([0-9.A-Za-z]+?)\s{1,3}.*$",line)
   f = re.search(r"^.+?\s{1,3}\d{1,5}.+?\s.+?\s{1,3}[0-9.A-Za-z]+?\s{1,3}(.+?)\s{1,3}.*$",line)
   g = re.search(r"^.*\s{1,3}(.+)$",line)
   if a:
           permissions = a.group(1)
   if b:
           links = b.group(1)
   if c:
           owner = c.group(1)
   if d:
           group = d.group(1)
   if e:
           size = e.group(1)
   if f:
           date = f.group(1)
   if g:
           filename = g.group(1)
   if a and b and c and d and e and f and g:
           coldata.append(permissions)
           coldata.append(links)
           coldata.append(owner)
           coldata.append(group)
           coldata.append(size)
           coldata.append(date)
           coldata.append(filename)
           Final_list.append(coldata)

但这段代码不适用于我得到的日期

Print Final_list

 [['-rw-r-----','1','root','groupA','0','2014-08-04','a.log'],['-rw-------','1','root','groupA','48132720','Aug','core'],['-rwxr-xr-x','1','root','groupA','208','Mar','restart.sh'],['lrwxrwxrwx','1','root','groupA','49','2014-06-06','../combined_ABC/scripts'],['drwxrwxrwx','3','root','groupA','1.0K','Apr','logs']]

预期产出：-

[['-rw-r-----','1','root','groupA','0','2014-08-04 10:42 ','a.log'],['-rw-------','1','root','groupA','48132720','Aug  1 23:45','core'],['-rwxr-xr-x','1','root','groupA','208','Mar 13 17:18','restart.sh'],['lrwxrwxrwx','1','root','groupA','49','2014-06-06 09:32','MerSem -> ../combined_ABC/scripts'],['drwxrwxrwx','3','root','groupA','1.0K','Apr 22 16:12','logs']]

---

os.stat不会给你这些信息吗？我会尝试使用split来分割空格，然后你应该将所有字段准确地分开。最好使用stat。ls只是stat系统调用的一个报告生成器，供人使用。使用stat来检索每个文件所需的信息，而不是试图解析ls的输出，在大多数情况下，ls的输出是这样的。带空格的文件名将破坏此功能，符号链接将破坏此功能。各种日期格式将打破这一点，设备文件将打破这一点。

s="""
-rw-r-----   1 root groupA     0 2014-08-04 10:42 a.log
-rw-------   1 root  groupA  48132720 Aug  1 23:45 core
-rwxr-xr-x   1 root  groupA      208 Mar 13 17:18 restart.sh
lrwxrwxrwx 1 root groupA   49 2014-06-06 09:32 MerSem -> ../combined_ABC/scripts
drwxrwxrwx   3 root groupA     1.0K Apr 22 16:12 logs
"""

line = s.lstrip().split("\n")[0].split() # split on newline and split the first line 

permission = line[0]
links = line[1]
owner = line[2]
group = line[3]
size = line[4]
date = line[5]
Name = line[-1]

print permission,links,owner,group,size,date,Name
-rw-r----- 1 root groupA 0 2014-08-04 a.log