测试开发进阶(四十一)

while 条件do    执行语句done

#!/bin/bash#Author:zhongxina=1while [ ${a} -lt 5 ]do     echo ${a}    a=$((${a}+1))done

case 变量名 in    值1)        指令1    值2)        指令2    值3)        指令3    *)        # 条件都不满足        指令4esac

#!/bin/bash#Author:zhongxincase "$2" in    +)        echo "$1 + $3 = $(($1+$3))"        ;;    -)        echo "$1 - $3 = $(($1-$3))"        ;;    \*)        echo "$1 * $3 = $(($1*$3))"        ;;    /)        echo "$1 / $3 = $(($1/$3))"        ;;    *)        echo "$2 不是合法的运算符"        ;;esac

$ grep mail passwd

$ grep New passwd$ grep New passwd -i

$ grep New passwd -i -c1

$ grep New passwd -i -n

$ grep -r "bash" ./

$ grep -rl "bash" ./

$ grep "s{2}" passwd -E

$ grep "oo" passwd -w$ grep "root" passwd -w

$ grep "$USER" passwd  > /dev/null && echo "$USER是存在的"|| echo "$USER是不存在的"

# 查看ps中包含docker的内容$ ps -aux | grep dockerroot     24521  0.0  0.0  14224  1088 pts/0    S+   23:21   0:00 grep --color=auto docker# 排除grep$ ps -aux | grep docker | grep -v grep# 查看符合的个数$ ps -aux | grep docker | grep -v grep | wc -l0

$ sed "p" passwd

$ sed -n "p" passwd

$ sed "/mail/ p" passwd -n

$ sed -n "/^s/ p" passwd

$ sed -n "/^mail/,/^s/ p" passwd

$ sed '3,5 i Add content before the current line' passwd

$ sed -i '3,5 i Add content before the current line' passwd

$ echo "testfile" >> a.txt$ sed -i '/man/,/mail/ r a.txt' passwd

$ sed -i '/man/,/mail/ w b.txt' passwd

$ sed -i '/test/ d' passwd

$ sed '3,6 s/nologin/666/' passwd |head