通过脚本判断输入是否是数字执行后续命令。
#!/bin/bash
## 方法1,有点问题
#if [[ "$1" =~ "^[[:digit:]]*$" ]];then
#if [[ "$1" =~ "^[0-9]+$" ]];then
# echo "$1 is number."
#else
# echo 'no.'
#fi
## 方法2, 可以，不过不是bash实现的，是使用了grep的正则
#if grep '^[[:digit:]]*$' <<< "$1";then
# echo "$1 is number."
#else
# echo 'no.'
#fi
## 方法3
#if [ "$1" -gt 0 ] 2>/dev/null ;then
# echo "$1 is number."
#else
# echo 'no.'
#fi
## 方法4,case
#case "$1" in
# [1-9][0-9]*)
# echo "$1 is number."
# ;;
# *)
# ;;
#esac
## 方法5,awk
#echo $1| awk '{print($0~/^[-]?([0-9])+[.]?([0-9])+$/)?"number":"string"}'
## 方法5,awk
#if [ -n "$(echo $1| sed -n "/^[0-9]\+$/p")" ];then
# echo "$1 is number."
#else
# echo 'no.'
#fi
## 方法6,expr
expr $1 "+" 10 &> /dev/null
if [ $? -eq 0 ];then
echo "$1 is number"
else
echo "$1 not number"
fi 