Shell脚本中判断输入变量或者参数是否为空的方法

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