echo 'Ctrl+C is captured,拷贝已停止'
exit 1
}
isalive(){
out=ps -p $1 2> /dev/null
return $?
}
while 2>1; do
{ SSIZE=/bin/ls -l $SOURCE | gawk "{print \\\$5}"
if [ -f $TARGET ]; then
TSIZE=/bin/ls -l $TARGET | gawk "{print \\\$5}"
else
TSIZE=“0”
fi
PERCENT=echo "scale=2; $TSIZE/$SSIZE*100" | bc -l
RATE=echo "scale=0; 63*$PERCENT/100" | bc -l
BLUE=“\033[3;44m”
NORMAIL=“\033[0;39m”
BAR=$BLUE
i=0
while [ $i -le 62 ]; do
[ $i = $RATE ] && BAR=$BAR"\\033[7;39m"
BAR=$BAR" "
let i=$i+1
done
BAR=$BAR$NORMAIL
echo -en "\r$BAR ${PERCENT}%"
if ! isalive "$CPID"; then echo -en "\n"; exit; fi
sleep 1
}
done
**脚本运行效果:**
[root@centos61 ~]# bash copyfile.sh Demofile Demofile-bak
14.00%^C
Ctrl+C is captured,拷贝已停止

**此时到百分之14的时候按ctrl+c停止了,我们可以核对一下目标文件是否是源文件的百分之14:**
**可以计算出1.5除以9.8确实是百分之14**
[root@centos61 ~]# ls -alh Demofile*
-rw-r–r–. 1 root root 9.8G May 27 17:41 Demofile
-rw-r–r–. 1 root root 1.5G May 27 18:25 Demofile-bak
**shell脚本完美运行!!!!!!**
**完整运行:**
[root@centos61 ~]# time bash copyfile.sh Demofile Demofile-bak
100.00%
real 2m6.804s
user 0m1.769s
sys 0m22.517s
[root@centos61 ~]# ls -al Demofile*
-rw-r–r–. 1 root root 10485760000 May 27 17:41 Demofile
-rw-r–r–. 1 root root 10485760000 May 27 18:36 Demofile-bak

###
### **三,**
### **脚本说明:**
* **$1 代表源文件,本例是Demofile,$2 代表目标文件,本例是Demofile-bak**
* **实际的拷贝命令是在后台运行,因为前台需要显示的是进度条,因此是****$CP "$SOURCE" "$TARGET" &**
>
> **trap "onCtrlC" INT
> function onCtrlC () {
> #捕获CTRL+C,当脚本被ctrl+c的形式终止时同时终止程序的后台进程
> kill -9 ${isalive} ${CPID}
> echo
> echo 'Ctrl+C is captured,拷贝已停止'
> exit 1
> }**
>
>
>
>
**这一块是抓取中断信号,如果不想拷贝了的情况下,ctrl+c停止脚本,这样会使得脚本退出更为优雅。**
>
> **isalive(){
> out=`ps -p $1 2> /dev/null`
> return $?
> }**
>
>
>
>
**这一块是监听拷贝程序的 pid,当返回值为1的时候,表示拷贝完成,程序停止退出。**
**一个循环的debug如下:**
++ ps -p 35785
- out=’ PID TTY TIME CMD
35785 pts/1 00:00:15 cp’ - return 0
- sleep 1
++ /bin/ls -l Demofile
++ gawk ‘{print $5}’ - SSIZE=10485760000
- ‘[’ -f Demofile-bak ‘]’
++ /bin/ls -l Demofile-bak
++ gawk ‘{print $5}’ - TSIZE=10337685504
++ echo ‘scale=2; 10337685504/10485760000*100’
++ bc -l - PERCENT=98.00
++ echo ‘scale=0; 63*98.00/100’
++ bc -l - RATE=61
- BLUE=‘\033[3;44m’
- NORMAIL=‘\033[0;39m’
- BAR=‘\033[3;44m’
- i=0
- ‘[’ 0 -le 62 ‘]’
- ‘[’ 0 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=0+1
- ‘[’ 1 -le 62 ‘]’
- ‘[’ 1 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=1+1
- ‘[’ 2 -le 62 ‘]’
- ‘[’ 2 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=2+1
- ‘[’ 3 -le 62 ‘]’
- ‘[’ 3 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=3+1
- ‘[’ 4 -le 62 ‘]’
- ‘[’ 4 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=4+1
- ‘[’ 5 -le 62 ‘]’
- ‘[’ 5 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=5+1
- ‘[’ 6 -le 62 ‘]’
- ‘[’ 6 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=6+1
- ‘[’ 7 -le 62 ‘]’
- ‘[’ 7 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=7+1
- ‘[’ 8 -le 62 ‘]’
- ‘[’ 8 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=8+1
- ‘[’ 9 -le 62 ‘]’
- ‘[’ 9 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=9+1
- ‘[’ 10 -le 62 ‘]’
- ‘[’ 10 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=10+1
- ‘[’ 11 -le 62 ‘]’
- ‘[’ 11 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=11+1
- ‘[’ 12 -le 62 ‘]’
- ‘[’ 12 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=12+1
- ‘[’ 13 -le 62 ‘]’
- ‘[’ 13 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=13+1
- ‘[’ 14 -le 62 ‘]’
- ‘[’ 14 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=14+1
- ‘[’ 15 -le 62 ‘]’
- ‘[’ 15 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=15+1
- ‘[’ 16 -le 62 ‘]’
- ‘[’ 16 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=16+1
- ‘[’ 17 -le 62 ‘]’
- ‘[’ 17 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=17+1
- ‘[’ 18 -le 62 ‘]’
- ‘[’ 18 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=18+1
- ‘[’ 19 -le 62 ‘]’
- ‘[’ 19 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=19+1
- ‘[’ 20 -le 62 ‘]’
- ‘[’ 20 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=20+1
- ‘[’ 21 -le 62 ‘]’
- ‘[’ 21 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=21+1
- ‘[’ 22 -le 62 ‘]’
- ‘[’ 22 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=22+1
- ‘[’ 23 -le 62 ‘]’
- ‘[’ 23 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=23+1
- ‘[’ 24 -le 62 ‘]’
- ‘[’ 24 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=24+1
- ‘[’ 25 -le 62 ‘]’
- ‘[’ 25 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=25+1
- ‘[’ 26 -le 62 ‘]’
- ‘[’ 26 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=26+1
- ‘[’ 27 -le 62 ‘]’
- ‘[’ 27 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=27+1
- ‘[’ 28 -le 62 ‘]’
- ‘[’ 28 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=28+1
- ‘[’ 29 -le 62 ‘]’
- ‘[’ 29 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=29+1
- ‘[’ 30 -le 62 ‘]’
- ‘[’ 30 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=30+1
- ‘[’ 31 -le 62 ‘]’
- ‘[’ 31 = 61 ‘]’
- BAR='\033[3;44m ’
- let i=31+1
- ‘[’ 32 -le 62 ‘]’
为了做好运维面试路上的助攻手,特整理了上百道 【运维技术栈面试题集锦】 ,让你面试不慌心不跳,高薪offer怀里抱!
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- 174道运维工程师面试题
1、什么是运维?
2、在工作中,运维人员经常需要跟运营人员打交道,请问运营人员是做什么工作的?
3、现在给你三百台服务器,你怎么对他们进行管理?
4、简述raid0 raid1raid5二种工作模式的工作原理及特点
5、LVS、Nginx、HAproxy有什么区别?工作中你怎么选择?
6、Squid、Varinsh和Nginx有什么区别,工作中你怎么选择?
7、Tomcat和Resin有什么区别,工作中你怎么选择?
8、什么是中间件?什么是jdk?
9、讲述一下Tomcat8005、8009、8080三个端口的含义?
10、什么叫CDN?
11、什么叫网站灰度发布?
12、简述DNS进行域名解析的过程?
13、RabbitMQ是什么东西?
14、讲一下Keepalived的工作原理?
15、讲述一下LVS三种模式的工作过程?
16、mysql的innodb如何定位锁问题,mysql如何减少主从复制延迟?
17、如何重置mysql root密码?
加入社区》https://bbs.youkuaiyun.com/forums/4304bb5a486d4c3ab8389e65ecb71ac0
in有什么区别,工作中你怎么选择?
8、什么是中间件?什么是jdk?
9、讲述一下Tomcat8005、8009、8080三个端口的含义?
10、什么叫CDN?
11、什么叫网站灰度发布?
12、简述DNS进行域名解析的过程?
13、RabbitMQ是什么东西?
14、讲一下Keepalived的工作原理?
15、讲述一下LVS三种模式的工作过程?
16、mysql的innodb如何定位锁问题,mysql如何减少主从复制延迟?
17、如何重置mysql root密码?
加入社区》https://bbs.youkuaiyun.com/forums/4304bb5a486d4c3ab8389e65ecb71ac0