shell判断一个变量是否为空方法总结
https://www.jb51.net/article/154835.htm
1.判断变量
复制代码代码如下:
read -p "input a word :" word
if [ ! -n "$word" ] ;then
echo "you have not input a word!"
else
echo "the word you input is $word"
fi
if [ ! -n "$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
复制代码代码如下:
#!/bin/bash
if [ ! -n "$1" ] ;then
echo "you have not input a word!"
else
echo "the word you input is $1"
fi
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
复制代码代码如下:
#!/bin/sh
para1=
if [ ! $para1 ]; then
echo "IS NULL"
else
echo "NOT NULL"
fi
para1=
if [ ! $para1 ]; then
echo "IS NULL"
else
echo "NOT NULL"
fi
4. 使用test判断
得到的结果就是: dmin is not set!
复制代码代码如下:
#!/bin/sh
dmin=
if test -z "$dmin"
then
echo "dmin is not set!"
else
echo "dmin is set !"
fi
dmin=
if test -z "$dmin"
then
echo "dmin is not set!"
else
echo "dmin is set !"
fi
5. 使用""判断
复制代码代码如下:
#!/bin/sh
dmin=
if [ "$dmin" = "" ]
then
echo "dmin is not set!"
else
echo "dmin is set !"
fi
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:
复制代码代码如下:
#! /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