shell判断输入变量或者参数是否为空

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

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，转自http://blog.csdn.net/lllxy/article/details/3255554

2. 直接通过变量判断

如下所示:得到的结果为: IS NULL

1. #!/bin/sh
2. para1=
3. if

   [

   !

   $para1

   ];

   then

4. echo

   "IS NULL"

5. else
6. echo

   "NOT NULL"

7. fi

3. 使用test判断

得到的结果就是: dmin is not set!

1. #!/bin/sh
2. dmin=
3. if

   test

   -z

   "$dmin"

4. then
5. echo

   "dmin is not set!"

6. else
7. echo

   "dmin is set !"

8. fi

4. 使用""判断

1. #!/bin/sh
2. dmin=
3. if

   [

   "$dmin"

   =

   ""

   ]

4. then
5. echo

   "dmin is not set!"

6. else
7. echo

   "dmin is set !"

8. fi