shell 脚本获取CPU占用率，作为函数里面的参数。

[baidu_29085129]

#!/bin/bash
num=0
while test $num -le 1
do

/*****
******/
usleep n
done
ehco "over"
想写一个shell脚本，通过获取此时cpu占有率，改变usleep的n，控制此脚本占用cpu的大小，从而达到控制整体cpu使用率的情况，
假如现在cpu是30%,执行脚本，现在输入80%，获取此时cpu使用率，改变休眠时间，从而达到cpu达到80%。难点是不知怎么获取cpu的值作为参数！

[baidu_29085129]

大神们，现身吧！

[奔跑的路]

http://hnsswjjxy-163-com.iteye.com/blog/910322

[baidu_29085129]

#!/bin/ash #echo -n "Enter Your Want Percent" Percent=80 n=100000 while true do CPULOG_1=$(cat /proc/stat | grep 'cpu ' | awk '{print $2" "$3" "$4" "$5" "$6" "$7" "$8}') SYS_IDLE_1=$(echo $CPULOG_1 | awk '{print $4}') Total_1=$(echo $CPULOG_1 | awk '{print $8+$2+$3+$4+$5+$6+$7}') usleep $n CPULOG_2=$(cat /proc/stat | grep 'cpu ' | awk '{print $2" "$3" "$4" "$5" "$6" "$7" "$8}') SYS_IDLE_2=$(echo $CPULOG_2 | awk '{print $4}') Total_2=$(echo $CPULOG_2 | awk '{print $8+$2+$3+$4+$5+$6+$7}') SYS_IDLE=`expr $SYS_IDLE_2 - $SYS_IDLE_1` Total=`expr $Total_2 - $Total_1` let SS=$SYS_IDLE*100/$Total let SYS_USAGE=100-$SS echo $SYS_USAGE if [ $SYS_USAGE -lt $Percent ]; then let n=$n-2500 fi done