2014-09-09shell练习

0

本帖最后由 wyatt88 于 2014-9-9 10:54 编辑

#!/bin/bash
count=`awk -F':' '$3>=500' /etc/passwd|wc -l`
if [ $count -eq 0 ]
then
echo "none" > /dev/null
else
echo "have" >/dev/null
echo $count
fi

0

Stary-liu

USER=awk -F':' '$3>=500 {print $1}' /etc/passwd |wc -l

if [$USER -gt 0]; then
echo "$USER common users"
eles
echo "no common users"
fi

0

泡沫。

不会写，求分享更好的思路

cat /etc/passwd|awk -F ":" '{print $NF }'|sort -n|uniq -c|awk '/bash$/ {print $1}'

0

wyatt88

wyatt88 发表于 2014-9-9 10:52
#!/bin/bash
count=`awk -F':' '$3>=500' /etc/passwd|wc -l`
if [ $count -eq 0 ]

对哈思维定式啊谢啦铭哥{:5_121:}

0

小雨

#!/bin/bash

awk -F ':' '$3>500' /etc/passwd > num.txt
if [ -e num.txt ];then
echo `wc -l num.txt` | awk -F '' '{print $1}'
fi

0

陈沛

本帖最后由陈沛于 2014-9-9 11:54 编辑

#!/bin/bash
# 2014-09-09
# chenpei
# More than 500 user query

num=`awk -F':' '$3>=500 {print $1}' /etc/passwd | wc -l`
if [ $num -gt 0 ]
  then
      echo "users: $num"
else
      echo "No ordinary users."
fi

0

So Long

#!/bin/bash

/bin/cp /etc/passwd  /chengcheng/haha.txt
a=0
for i in `awk -F ':' '$3>=500 {print $3}'  /chengcheng/haha.txt`;do

      if [ $i -ge 500 ];then
            a=$[$a+1]
      else

            echo "the user's uid greater 500 is not exist"
      fi
done
echo "the user's uid greater 500 is exist,it has $a users"

0

Louis

#!/bin/bash
## This script is for count the sum of general users.
## Writed by Louis on 2014/09/09 13:05
num=`awk -F: '$3>=500 {print $1}' /etc/passwd|wc -l`
if [ $num -eq 0 ]; then
echo "There is no general user on this system."
else
echo "The sum of general users numbers is: $num"
fi

0

乌托邦

oh

0

ocean

cat /etc/passwd |awk -F: '$3>=500'|wc -l{:5_121:}

0

coffee

#!/bin/bash

user=`awk -F':' '$3>=500 {print $1}' /etc/passwd |wc -l`

if [ $user -ne 0 ] ;then
echo $user
else
echo "No ordinary users"
fi

0

chenzudao1234

kankan

0

Wagskun

看看

0

So Long

程城发表于 2014-9-9 13:04
#!/bin/bash

/bin/cp /etc/passwd /chengcheng/haha.txt

铭哥说得对，我把问题复杂化了。继续学习学习。

0

butterfly梧桐雨

[root@myCentOS practice]# cat homework1.sh
#!/bin/bash
# This is a shell practice for count the number of normal users
# 20140909 by zhuzhu
num=`awk -F ':' '$3 >=500 {print $1}' /etc/passwd | wc -l`
if [ -z $num ]; then
echo "None normal user"
else
echo "Total $num normal users"
fi

[root@myCentOS practice]# bash homework1.sh
Total 1 normal users

0

hongyufeng2014

很好很好很好很好

0

soar

#!/bin/bash
cat /etc/passwd | awk -F ":" '{print $3 }'> temp.user
declare -i n=0;
for i in `cat temp.user`
do
      if [ $i -ge 500 ]
      then
            n=$n+1
      fi
done
echo "the count of user is:$n"

0

齐天大圣

#!/bin/bash
n=`awk -F':' '$3>=500{print $1}' /root/passwd|wc -l`
if [ $n -eq 0 ];then
echo "没有自定义用户"
else
echo $n
fi

0

寻@轨迹

#!/bin/bash
#Writen by yuhui!
count=`awk -F ':' '$3 >= 500' /etc/passwd | wc -l`
person=`awk -F ':' '$3 >= 500' /etc/passwd | awk -F ':' '{print $1}'`
if [ $count -eq 0 ]
then
      echo " 没有普通用户！"
else
      echo " 有 $count 个普通用户"
      echo $person
fi

0

第六感

本帖最后由第六感于 2014-9-10 11:57 编辑

#!/bin/bash
#
USERID=`cut -d: -f3 /etc/passwd`
for i in `echo $USERID`;do
      if [ $i -gt 500 ];then
            let j++
      fi
done
echo  $j

0

huifeidexiaxia

基础还是不扎实，有思路，不知怎么动笔，要多写写

0

寒一

#! /bin/bash

awk -F ':' '$3>=500 {print $3}' /etc/passwd |wc -l

0

木字当头

#!/bin/bash

a=`awk -F ':' '$3>=500' /etc/passwd|wc -l `

if [ $a -ge 1 ]

then

echo "the system have $a个user-difined"

else

echo "the system have no user-difined"
fi

0

zyfeifie

#!/bin/bash
n=`awk -F ':' '$3>499' /etc/passwd | wc -l `
if [ $n -gt 0 ]
echo "this server has $n normal user"
else
echo ‘have no normal users’
fi

0

小海

编写思路：普通用户的uid号数都大于等于500，/etc/passwd中第三个字段是uid,一次提取出uid然后判断是否大于等于500就可以判断是否存在普通用户，num计数，存在一次递加1，最终就是总的个数。

#! /bin/bash
#查看Linux系统中是否有自定义用户（普通用户），若是有，一共有几个

num=0
for i in `cat /etc/passwd | awk -F ':' '{print $3}'`
do
      if [ $i -ge 500 ]
      then
            num=$[$num+1]
      fi
done
if [ $num -gt 0 ]
then
      echo "存在普通用户，一共有："$num
else
      echo "不存在普通用户"
fi

0

游夜

#!/bin/bash
user=`cat /etc/passwd|awk -F ':' '$3>499 {print $3}'|wc -l`
echo "There have $user users."

0

楓瀛夢鞢

#! /bin/bash
a=`awk -F ':' '$3 >= 500 ' /etc/passwd|awk -F ':' '{print NR}'|xargs|awk '{print NF}' `
if [ $a -eq 0 ] ;then
echo "No ordinary users.";
else
echo "The ordinary users is total $a";
fi

0

阳光

   1 #!/bin/bash
   2 num=0
   3
   4
   5 for i in `cat /etc/passwd |awk -F ':' '{print $3}'`; do
   6       if [ $i -ge 500 ] ; then
   7          num=$[$num+1]
   8       fi
   9 done
   10 echo $num

0

黑色的梦

1

0

larson

#!/bin/bash
#this script is statistical the average user in the system

awk -F ':' '{print $3}' /etc/passwd > /tmp/user_id.txt
echo -n "">/tmp/1.txt
for n in `cat /tmp/user_id.txt`;do
if [ $n -ge 500 ]; then
echo $n >> /tmp/1.txt
fi
done
m=`cat /tmp/1.txt | wc -l`
echo "The number of prdinary users:" $m

0

潘有成

oh,yeah,ohyea

0

丁张龙

#!/bin/bash
num=`cat /etc/passwd|awk -F ':' '$3>=500'|grep "bash"|wc -l`
echo "you have $num users"

0

我想做个好人

{:4_91:}

0

cmzsteven

{:4_91:}

0

zzfeng2012

...

0

yaabb163

dddddddddd

0

zhangw

1

0

qq20847697

#!/bin/bash
#
a=`awk -F: '$3>=500' /etc/passwd|wc -l`
if [ $a -eq 0 ] ;
then echo "No ordinary users."
else
echo "$a ordinary users."
fi

0

pykihwfn

看看

0

llzdwyp

cat /etc/passwd |awk -F: '$3>=500 {print $1,$3}'

0

iwannachange

#!/bin/bash
a=0
for I in `cut -d: -f3 /etc/passwd`;do
if [ $I -ge 500 ] ;then
let a=$a+1
fi
done
echo "your system created $a users"

0

qiaoxin360

#!/bin/bash
user=`awk -F ':' '$3>=1000' /etc/passwd`
echo "$user

0

巧杜鹃

学习

0

鑫柏

不知道行不行 awk -F ':' '$3>=500' /etc/passwd |wc -l

0

Armani

aa=`cat /etc/passwd | awk -F:  '$3>=500 {print $3}' | wc -l`
if  [ $aa  -eq 0 ];then
echo "You machine  don't have  common  user!"
else
echo "You machine  have  "$aa" users!"
fi

0

307141950

cat /etc/passwd |grep /bin/bash |sort -n

0

shajoe

awk -F ":" '$3>500' /etc/passwd |wc -l

0

ifconfig2

1

0

chenqi

awk -F':' '$3>500' passwd |wc -l

0

Shawn

awk -F':' '$3>=500 {print $1}' /etc/passwd |wc -l

0

不怕不怕

急急急

0

664778717@qq.co

#!/bin/bash
#date 2015/11/10
#wite by cong
usernum=`nl /etc/passwd|awk -F: '$3>=500'|wc -l`
if [ $usernum -gt 0 ];
then
echo "$usernum 个普通用户"
else
echo "没有普通用户"
fi

0

ldp840611

看目的

0

Rohero

学习

0

J!_yuan

awk -F: '$3>"0"' /etc/passwd |wc -l

0

翟厚翔

cat /etc/passwd|awk -F':' '{ if($3 >= 500) print $3}'|wc -l

0

石头

{:4_91:}

0

蓝色风暴

awk -F ':' '$3>=500{print $0}' /etc/passwd | wc -l

0

pbw19950507

学习

0

我是学渣

kk

0

loveyouhyf

#! /bin/bash
##看看你的Linux系统中是否有自定义用户（普通用户），若是有，一共有几个
##write by 2015-12-26
a=`awk -F ':' '$3>499 {print $0}' /etc/passwd|wc -l`
if [ $a -gt 0 ]; then
echo "the system have $a user-define user(s),is:"
awk -F ':' '$3>499 {print $0}' /etc/passwd
else
echo "the system have not user-define user(s)"
fi

0

DSH20150916

学习

0

weifeng1463

pk

0

Linuxpp

继续学习

0

Landon

asdas

0

maria

#!/bin/bash
count=`awk -F ':' '$3>=500' /etc/passwd|wc -l `
if (($count==0))
then echo '系统中没有普通用户！'
else
echo '系统中有'$count'个普通用户！'
fi

0

重庆-刘鹏

自定义用户

0

yanggang

ff

0

369666951

本帖最后由 369666951 于 2015-12-30 14:57 编辑

#!/bin/bash
a=`awk -F ':' '$3>=500' /etc/passwd|wc -l`
if [ $a -ne 0 ]
then
echo $a
else
echo "no uid>=500"
fi

0

一笑而过

啊实打实的

0

rolay8

#!/bin/bash
cat /etc/passwd|while read line
do
if [ `echo -n $line|awk -F ":" '{print $3}'` -ge 500 ]; then
echo `echo -n $line|awk -F ":" '{print $1}'`
fi
done

0

rolay8

awk 里面判断{:4_118:} 一行就搞定。。。学习了

0

沧海一叶

#!/bin/bash
#
count=`awk -F':' '$3>499 {print $0}' /etc/passwd|wc -l`
echo $count

0

hlymlv

#!/bin/bash
m=`cat /etc/passwd|wc -l`
for i in `seq 1 $m`
do
n=`sed -n "$i"p /etc/passwd|awk -F ':' '{print $3}'`
if [ "$n" -gt 500 ];then
sum=`expr $sum + 1`
sed -n "$i"p /etc/passwd|awk -F ':' '{print $1}'
fi
done
echo $sum

0

xteplinux

{:4_91:}

0

prospect

#!/bin/sh
num=0
for i in `cat /etc/passwd |awk -F':' '{print $3}'`
do
if [ $i -ge 500 ];then
num=$[$num + 1]
fi
done
echo $num

0

出VU时代

学习了

0

licengceng

学习

0

qiqige

学习

0

t0ny1988

#!/bin/bash
user=`cat /etc/passwd | awk -F ':' '$3>=500' |wc -l`
if [ $user -eq 0 ]
then
echo "没有普通用户"
else
echo "有 $user 个普通用户"
fi

0

lin19890913

看看

0

落涧飞鹰

看看

0

jokerhuman

学习

0

北辰星

学习

0

北辰星

本帖最后由北辰星于 2016-2-4 15:12 编辑

#!/bin/bash
u=$[`awk -F ':' '{print $6}' /etc/passwd |grep '/home' |wc -l`]
if [ $u -gt 0 ]
then
echo "该系统中含有 $u 个自定义用户"
else
echo "该系统中没有自定义用户"
fi
为什么都喜欢用 uid大于500 这个条件呢，其实直接匹配出有/home这个字符串不就说明了是自定义的用户吗？而且uid大于500就一定是自定义的用户吗？

0

lyhabc

#!/bin/bash
>/tmp/1.txt
for i in $(seq 1 $(sed -n ""p /etc/passwd|wc -l))
do
n=$(sed -n "$i"p /etc/passwd|awk -F ":" '{print $3}')
if [ $n -ge 500 ]
then
echo $n >>/tmp/1.txt
fi
done

echo "user is $(wc -l /tmp/1.txt|awk '{print $1}')"

0

thedawn

1

0

lin13750529011

谢谢

0

zmh0415

本帖最后由 zmh0415 于 2016-2-27 20:35 编辑

a=`awk -F':' '$3>=500 {print $3}' /etc/passwd|wc -l`
if [ $a -gt 1 ]
then
echo "had created normal user"
num=$[$a-1]
echo "user numbers is : $num "
else
echo "no user had been created"
fi

0

HMOM

#!/bin/bash
sum=`awk -F':' '$3 >= 500' /etc/passwd | wc -l`
uname=`awk -F: '$3 >= 500 {print $1," Uid=",$3}' /etc/passwd`
if [ $sum -eq 0 ];then
      echo "The system exist $sum users."
else
      echo "The system exist $sum users:"
      echo "$uname"
fi

0

蔡炳森

#!/bin/bash

m=`cat /etc/passwd |awk -F':' '$3>= 500 {print $3}' |wc -l`
echo $m

0

wangzai

学习

0

ayuan124

本帖最后由 ayuan124 于 2016-3-23 18:08 编辑

#!/bin/bash
sum=`awk -F ':' '$3>=500' /etc/passwd|wc -l`
if [ $sum -eq 0 ];then
echo "no ordinary user"
else
echo "ordinary user have $sum"
fi

0

向南看是晴空

学习了

0

gxp2008

cat /etc/passwd |awk -F ':' '$3>=500 {print $1}'|wc -l

0

liang

awk -F':' '$3>=500' /etc/passwd|wc -l

0

西瓜.M

#!/bin/bash #display the amount of common users count=`awk -F ':' '$3>=500' /etc/passwd|wc -l` if [ $count>0 ];then echo $count else echo "common users is not exist" fi