在检查 /etc /group /etc /passwd中检查用户和组是否存在时，运算符和用户输入未找到 /出乎意料

Question

我不明白我的代码怎么了。我正在尝试做到这一点，以便如果在/etc/group和etc/passwd中都找不到ECHO中的用户名和组都没有找到。

如果发现一个但没有发现一个回声，最后发现它们都被发现，那么它们都存在Echo。

这是我的代码：

#/bin/bash

read -p 'Username: ' username
read -p 'Group: ' group

if [ $(grep "^$group:" /etc/group | cut -f 1 -d ':') != $group ] && [ $(grep "^$username:" /etc/passwd | cut -f 1 -d ':') != $username ]
then 
  echo Both not found 

elif $(grep "^$group:" /etc/group | cut -f 1 -d ':') != $group ] || [ $(grep "^$username:" /etc/passwd | cut -f 1 -d ':') != $username ]
then
  echo One exists, one does not.

elif $(grep "^$group:" /etc/group | cut -f 1 -d ':') = $group ] && [ $(grep "^$username:" /etc/passwd | cut -f 1 -d ':') = $username ]
then
  echo both exist
fi

这是输出：

$ ./exercise_10.sh        
Username: user
Group: root
./exercise_10.sh: 10: root: not found
./exercise_10.sh: 14: root: not found
                                                                                                                                                                               
$ ./exercise_10.sh
Username: dfgdfggd
Group: root
./exercise_10.sh: 10: root: not found
./exercise_10.sh: 10: [: !=: unexpected operator
./exercise_10.sh: 14: root: not found
                                                                                                                                                                               
$ ./exercise_10.sh
Username: dfgdg
Group: cgdgdfg
./exercise_10.sh: 6: [: !=: unexpected operator
./exercise_10.sh: 10: !=: not found
./exercise_10.sh: 10: [: !=: unexpected operator
./exercise_10.sh: 14: =: not found

我已经在命令行中测试了GREP和剪切命令，它有效：

$ grep '^user:' /etc/passwd | cut -f 1 -d ':'
user

Answer 1

在Shebang中有一个缺少的！，第一行应该是：

#!/bin/bash

在elif s之后，缺少[。

您可以始终使用shellCheck命令来解析并列出bash脚本中的所有问题：

shellcheck ./exercise_10.sh

最后，您可以依靠grep命令命令返回状态，如果语句像下面的那个：

if grep -q "^${group}:" /etc/group 2>/dev/null && grep -q "^${username}:" /etc/passwd 2>/dev/null ; then
  printf "Both exist.\n"
...

尝试一下，只能解析每个文件一次：

#!/bin/bash

read -rp 'Username: ' lusername
read -rp 'Group: ' lgroup

if grep -q "^${lgroup}:" /etc/group 2>/dev/null ; then
  if grep -q "^${lusername}:" /etc/passwd 2>/dev/null ; then
    printf "Both exist\n"
  else 
    printf "One exists, one does not\n"
  fi
else
  if grep -q "^${lusername}:" /etc/passwd 2>/dev/null ; then
    printf "One exists, one does not\n"
  else
    printf "Both not found\n"
  fi
fi