1.判断变量 read -p "input a Word :" word if [ ! -n "$word" ] ;then echo "you have not input a word!
1.判断变量
read -p "input a Word :" word
if [ ! -n "$word" ] ;then
echo "you have not input a word!"
else
echo "the word you input is $word"
fi
2.判断输入参数
#!/bin/bash
if [ ! -n "$1" ] ;then
echo "you have not input a word!"
else
echo "the word you input is $1"
fi
以下未验证。
3. 直接通过变量判断
如下所示:得到的结果为: IS NULL
#!/bin/sh
para1=
if [ ! $para1 ]; then
echo "IS NULL"
else
echo "NOT NULL"
fi
4. 使用test判断
得到的结果就是: dmin is not set!
#!/bin/sh
dmin=
if test -z "$dmin"
then
echo "dmin is not set!"
else
echo "dmin is set !"
fi
5. 使用""判断
#!/bin/sh
dmin=
if [ "$dmin" = "" ]
then
echo "dmin is not set!"
else
echo "dmin is set !"
fi
下面是我在某项目中写的一点脚本代码, 用在系统启动时:
#! /bin/bash
echo "Input Param Is [$1]"
if [ ! -n "$1" ] ;then
echo "you have not input a null word!"
./app1;./app12;./app123
elif [ $1 -eq 2 ];then
./app12;./app123
elif [ $1 -eq 90 ];then
echo "yy";
fi
--结束END--
本文标题: Shell脚本中判断输入变量或者参数是否为空的方法
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