先給大家分享一篇關于shell判斷一個變量是否為空方法總結內容

shell判斷一個變量是否為空方法總結

https://www.jb51.net/article/154835.htm

1.判斷變量

代碼如下:
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

2.判斷輸入參數

代碼如下:
#!/bin/bash
if [ ! -n "$1" ] ;then
    echo "you have not input a word!"
else
    echo "the word you input is $1"
fi

以下未驗證。

3. 直接通過變量判斷

如下所示:得到的結果為: IS NULL
代碼如下:
#!/bin/sh
para1=
if [ ! $para1 ]; then
  echo "IS NULL"
else
  echo "NOT NULL"
fi

4. 使用test判斷

得到的結果就是: dmin is not set!
代碼如下:
#!/bin/sh
dmin=
if test -z "$dmin"
then
  echo "dmin is not set!"
else 
  echo "dmin is set !"
fi

5. 使用""判斷

代碼如下:
#!/bin/sh
dmin=
if [ "$dmin" = "" ]
then
  echo "dmin is not set!"
else 
  echo "dmin is set !"
fi

下面是我在某項目中寫的一點腳本代碼， 用在系統啟動時:
代碼如下:
#! /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