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检查另一个用户的文件访问权限的命令

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检查另一个用户的文件访问权限的命令

/我需要知道如何检查给定用户的有效文件访问权限，但从目标文件或目录开始、检查组等手动执行此操作需要很长时间。

答案1

据我所知，唯一的方法是要么执行您所描述的操作，然后根据有效用户/组检查每个权限集。或者您可以尝试设置 sudo 以便能够执行test(1).

sudo -u luser test -x ~juser/bin/myprogram

就像你说的，检查有效的用户/组权限：

:

# called as $0 usertocheck pathname {r|w|x}
# for example, permcheck luser ~juser/bin/myprogrm x
# displays either "root", "user", "groups", "other" or "none"
user=$1
file=$2
smode=$3
# if user has no access from state, an empty string is returned, fuid,
# fgid and fmode would become empty strings as well; the end result is
# always showing 'none' even if $user has access (except $user == 'root')
set -- $(stat -L -c '%u %g %a' $file 2>&-)
awk -f $tmpawk \
    -veuid="$(id -u $user)" \
    -vgrp="$(id -G $user)" \
    -vfuid="$1" \
    -vfgid="$2" \
    -vfmode="$(echo ibase=8\;$3 | bc)" \
    -vsmode="$smode" \
'BEGIN {
  if (euid == 0) { print "root"; exit; }
  split(grp,Groups);
  omode = fmode % 8; gmode = int(fmode / 8 % 8);
  umode = int(fmode / 64 % 8);
  # set up tests
  # these could be function, but not all version of awk has a function
  # statement
  if (smode == "r") {
    utest = int(umode / 4);
    gtest = int(gmode / 4);
    otest = int(omode / 4);
  }
  if (smode == "w") {
    utest = int(umode / 2 % 2);
    gtest = int(gmode / 2 % 2);
    otest = int(omode / 2 % 2);
  }
  if (smode == "x") {
    utest = (int(umode >= 4) && umode % 2);
    gtest = (int(gmode >= 4) && gmode % 2);
    otest = (int(omode >= 4) && omode % 2);
  }
  if (utest && fuid == euid) { print "user"; exit; }
  for (idx in Groups) {
    if (gtest && Groups[idx] == fgid) { print "group"; exit; }
  }
  if (otest) { print "other"; exit; }
  print "none";
}
'

在我的 Ubuntu 11.04 系统上，运行此脚本平均需要大约 16 毫秒。此外，stat 不需要读取/执行

答案2

该stat命令会打勾吗？

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