Bash脚本：如何检查是否只有一个root id？

Question

如何编辑我的脚本，以便检查是否只有一个根ID？

期望输出

Audit criteria: There is only one root id 

Vulnerability: Yes 

Details: See below 


root:!:0:0::/:/usr/bin/bash 

jdoe:*:0:1:John Doe:/home/jdoe:/usr/bin/bash 

脚本

#!/bin/bash 

isVulnerable="No" 
isVulnerable="Yes"  

cat /etc/passwd | cut -f3 -d":" | sort -n | /usr/bin/uniq -c | while read x ;    
do 
    [ -z "${x}" ] && break 
    set - $x 

if [ "$1" -gt 1 ]; then 
    users=`/bin/gawk -F: '($3 == n) { print $1 }' n=$2 /etc/passwd | /usr/bin/xargs` 

echo "Audit Criteria: Duplicate UID ($2): ${users}" 
echo "Vulnerability: ${isVulnerable}" 
echo "Details: see below" 
echo 
grep "x:0:" /etc/passwd 

else 

echo "All user id are unique" 
fi 

done 

Answer 1

这是很方便的使用，以收集与重复字段行AWK：

get_dups() { 
    awk -F':' '$3 == 0 { if (dup++) print } END { exit(dup > 1) }' /etc/passwd 
} 

如果在/etc/passwd文件多个零的用户ID，功能与非零退出状态，并将具有重复的root用户标识的行打印到标准输出。否则，退出状态为零。

用法：

dups="$(get_dups)" 
if [ $? -eq 0 ]; then 
    vulnerability='No' 
    msg='There is only one root ID' 
else 
    vulnerability='Yes' 
    msg='There are multiple root IDs' 
fi 
printf '%15s: %s\n' 'Audit criteria' "$msg" 
printf '%15s: %s\n' 'Vulnerability' "$vulnerability" 

[ -z "$dups" ] && dups='All user IDs are unique' 
printf '\n%s\n' "$dups" 

Answer 2

你可以这样做：

ROOT_COUNT=$(cut -f3 -d":" </etc/passwd | grep -c ^0$) 

然后，如果ROOT_COUNT含有一种大于1，你^ h AVE多个用户使用UID 0