Shell编程之变量的高级用法详解_linux shell_

 var1="hello world hello world" # 所谓最长和最短其实就是正则表达式中的贪婪模式和非贪婪模式 # 删除匹配规则为*lo的数据（开头最短删除） echo ${var1#*lo} -> world hello world # 删除匹配规则为*lo的数据（开头最长删除） echo ${var1##*lo} -> world # 删除匹配规则为hel*的数据（尾部最短删除） echo ${var1%hel*} ->hello world # 删除匹配规则为hel*的数据（尾部最长删除） echo ${var1%%hel*} -> # 替换第一个旧字符串world为Beijing echo ${var1/world/Beijing} ->hello Beijing hello world # 替换第一个旧字符串world为Beijing echo ${var1/world/Beijing} ->hello Beijing hello Beijing 

 # x=${y-新值} echo ${y=$var1} ->hello world hello world y= echo ${y-$var1} -> y=hello Beijing echo ${y-$var1} ->hello Beijing # x=${y:-新值} echo ${y:-$var1} ->hello world hello world y= echo ${y:-$var1} ->hello world hello world y=hello Beijing echo ${y:-$var1} ->hello Beijing # x=${y+新值} echo ${y+$var1} -> y= echo ${y+$var1} ->hello world hello world y=hello Beijing echo ${y+var1} ->hello world hello world # x=${y:+新值} echo ${y:+$var1} -> y= echo ${y:+$var1} -> y=hello Beijing echo ${y:+var1} ->hello world hello world # x=${y=新值} x=${y=$var1} echo $x ->hello world hello world echo $y ->hello world hello world y= x=${y=$var1} echo $x -> echo $y -> y=hello Beijing x=${y=$var1} echo $x ->hello world hello world echo $y ->hello Beijing # x=${y:=新值} x=${y:=$var1} echo $x ->hello world hello world echo $y ->hello world hello world y= x=${y:=$var1} echo $x ->hello world hello world echo $y ->hello world hello world y=hello Beijing x=${y:=$var1} echo $x ->hello Beijing echo $y ->hello Beijing # x=${y?新值} echo ${y?$var1} ->-bash: y: hello world hello world y= echo ${y?$var1} -> y=hello Beijing echo ${y?$var1} ->hello Beijing # x=${y:?新值} echo ${y?$var1} ->-bash: y: hello world hello world y= echo ${y:?$var1} ->-bash: y: hello world hello world y=hello Beijing echo ${y:?$var1} ->hello Beijing 

 # 方法一 var1="hello world" echo ${#var1} ->11 # 方法二 len=`expr length "$var1"` echo $len ->11 ## var1有空格，必须加双引号，否则报错 expr: syntax error ## var2没有空格，可以不需要加双引号 var2=hello expr length $var2 5 

 expr index "$var1" llo ->3 expr index "$var1" lole ->2 

 expr match "$var1" hel ->3 expr match "$var1" llo ->0 

 # 方法一 echo ${var1:2} ->llo world # 方法二 echo ${var1:2:5} ->llo w # 方法三 echo ${var1: -2} ->ld echo ${var1: -2:1} ->l # 方法四 echo ${var1:(3)} ->lo world echo ${var1:(3):2} ->lo echo ${var1:(-2):1} ->l # 方法五 expr substr "$var1" 2 5 ->ello

 var2=hello echo ${var2:2:5} ->llo expr substr $var2 2 5 ->ello 

 #!/bin/bash string="Bigdata process framework is Hadoop,Hadoop is an open source project." function print_tips { echo "==============================" echo "1->打印string长度" echo "2->删除字符串中所有的Hadoop" echo "3->替换第一个Hadoop为Mapreduce" echo "4->替换所有Hadoop为Mapreduce" echo "q|Q->退出" echo "==============================" } function len_of_string { echo "${#string}" } function del_hadoop { echo "${string//Hadoop/}" } function rep_first_hadoop_mapreduce { echo "${string/Hadoop/Mapreduce}" } function rep_all_hadoop_mapreduce { echo "${string//Hadoop/Mapreduce}" } while true do echo "" echo "" echo "" echo "" echo "【string=$string】" print_tips read -p "Please input your choice (1|2|3|4|q|Q):" choice case $choice in 1)len_of_string;; 2)del_hadoop;; 3)rep_first_hadoop_mapreduce;; 4)rep_all_hadoop_mapreduce;; q|Q)exit;; *)echo "Error input,only in (1|2|3|4|q|Q)";; esac done

 # 获取系统的所有用户并输出 #!/bin/bash index=1 for user in `cat /etc/passwd | cut -d ":" -f 1` do echo "这是第 $index 个用户，名称为：$user" index=$(($index + 1)) # $(())主要用来进行整数运算，引用变量前面可以加$也可以不加 done # 根据系统时间计算今年或明年 #!/bin/bash year=`date +%Y` echo "今年是$year年，明年是$(($year+1))年" # 根据系统时间获取今年还剩下多少星期，已经过了多少星期 #!/bin/bash # 计算现在是今年第几天 #!/bin/bash days=$(date +%j) weeks=$((days/7)) remain_weeks=$((52-weeks)) echo "今年还剩$remain_weeks个星期，已经过了$weeks个星期" # 判定nginx进程是否存在，若不存在则自动拉起进程 #!/bin/bash nginx_process_num=$(ps -ef | grep nginx | grep -v grep | wc -l) if [ $nginx_process_num -eq 0 ];then echo "nginx未启动，拉起中..." systemctl start nginx else echo "nginx已启动..." fi 

 # 设置为只读 var1="hello wordl" var1="hello shell" declare -r var1 var1="hello java" ->-bash: var1: readonly variable # 设置为整形 num1=20 num2=$num1+30 echo $num2 ->20+30 #linux默认设置为字符串 declare -i num3 num3=$num1+30 echo $num
                
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