幫助監(jiān)視/etc/passwd文件是否正常(P90 練習(xí)6.7)

1）找出有UID0的所有項(xiàng)

2）找出有重復(fù)UID的所有項(xiàng)

3）找出有重復(fù)登錄名的所有項(xiàng)

4）找出沒有口令的所有項(xiàng)

5）找出沒有作廢日期的所有項(xiàng)

以下是實(shí)現(xiàn)的shell腳本：

#!/bin/bash#監(jiān)視/etc/passwd正常

#找出有UID 0的用戶echo "----------------------------------"user1=`less /etc/passwd | awk -F: '$3==0 {PRint $1}'`if [ -z $user1 ];then echo "1.there is no user's UID equals 0"elseecho -n "1.the user of UID equals 0 have: "$user1echo ""fi

#找出有重復(fù)UID的用戶echo "----------------------------------"user2=`awk -F: 'BEGIN{ORS=","}NR==FNR {a[$3]++} NR>FNR&&a[$3]>1 {print $1,$3}' /etc/passwd /etc/passwd`if [ -z $user2 ];then echo "2.there is no user's UID repeat"elseecho -n "2.the user of repeat UID have: "$user2echo ""fi

#找出有重復(fù)登錄名的用戶echo "---------------------------------"user3=`awk -F: 'BEGIN{ORS=","}NR==FNR {a[$1]++} NR>FNR&&a[$1]>1 {print $1,$3}' /etc/passwd /etc/passwd`if [ -z $user3 ];then echo "3.there is no user's login name repeat"elseecho -n "3.the user of repeat login name have: "$user3echo ""fi

#找出沒有口令的所有用戶echo "---------------------------------"user4=`sudo awk -F: 'BEGIN{ORS=","} length($2)<30 {print $1}' /etc/shadow`if [ -z $user4 ];then echo "4.there is no user have no passWord"elseecho -n "4.the user of no password have: "$user4echo ""fi

#找出沒有作廢日期的所有用戶echo "---------------------------------"user5=`sudo awk -F: 'BEGIN{ORS=","} length($7)<1 {print $1}' /etc/shadow`if [ -z $user5 ];then echo "5.there is no user have no outdate time"elseecho -n "5.the user of no outdate time have: "$user5echo ""fiecho "---------------------------------"