shell脚本输出/etc/passwd中shell类型对应的用户名及其个数

#!/bin/bash
users=`wc -l /etc/passwd  | cut -d‘ ‘ -f1`                     
declare -i bashcount=0
declare -i nologincount=0
for i in `seq 1 $users`;do
        usershell=`head -$i /etc/passwd | tail -1 | cut -d: -f7 | cut -d‘/‘ -f3`
        if [ "$usershell" == "bash" ] ;then
                let bashcount+=1
                bash_array[$(($bashcount-1))]=`head -$i /etc/passwd | tail -1 | cut -d: -f1`
        elif [ "$usershell" == "nologin" ];then
                let nologincount+=1
                nologin_array[$(($nologincount-1))]=`head -$i /etc/passwd | tail -1 | cut -d: -f1`    
        else
                echo "other shell" &>/dev/null
        fi  
done
for x in ${bash_array[@]};do
        bashusers=$bashusers,$x
done  
echo "bash have $bashcount "users",they are:"$bashusers
for y in ${nologin_array[@]};do
        nologinusers=$nologinusers,$y
 done
echo "/sbin/nologin "have "$nologincount "users",they are:"$nologinusers

shell脚本输出/etc/passwd中shell类型对应的用户名及其个数

原文：https://blog.51cto.com/11342825/2425128