如何在shell变量中执行sh命令集？

Question

我有这个linux命令:

candump -l -e -x -s 0 -n 10 any,0~0,#FFFFFFFF 2>/dev/null> /tmp/can.log&

当我直接在shell中运行它时,它可以正常工作.

我想用这个方法在我的脚本中添加它:

#!/bin/sh

# I tried this 2 syntax    
# MYVAR="candump -l -e -x -s 0 -n 10 any,0~0,#FFFFFFFF 2> /dev/null > /tmp/can.log &"
MYVAR='candump -l -e -x -s 0 -n 10 any,0~0,#FFFFFFFF 2> /dev/null > /tmp/can.log &'
$MYVAR

当我执行我的脚本时,我收到此错误:

SIOCGIFINDEX:没有这样的设备

我测试了这些脚本并且它有效:

#!/bin/sh

MYVAR='ls -l'
$MYVAR

结果:

total 8
drwxr-xr-x  2 root root     0 Nov  5  2015 bin
drwxr-xr-x  2 root root     0 Oct 22  2015 boot
drwxr-xr-x  5 root root 13460 Jan  1 00:00 dev
drwxr-xr-x  8 root root     0 Nov  5  2015 etc
drwxr-xr-x  3 root root     0 Nov  5  2015 home
-rwsr-xr-x  1 root root   258 Nov  5  2015 init
drwxr-xr-x  3 root root     0 Nov  5  2015 lib
drwxr-xr-x  3 root root     0 Nov  5  2015 media
drwxr-xr-x  3 root root     0 Nov  5  2015 mnt
dr-xr-xr-x 67 root root     0 Jan  1 00:00 proc
drwx------  2 root root     0 Oct 22  2015 root
drwxr-xr-x  4 root root     0 Jan  1 00:00 run
drwxr-xr-x  2 root root     0 Nov  5  2015 sbin
dr-xr-xr-x 14 root root     0 Jan  1 00:00 sys
-rwxr-xr-x  1 root root    33 Jan  1 00:04 test
drwxrwxrwt  2 root root     0 Jan  1 00:00 tmp
drwxr-xr-x  9 root root     0 Oct 22  2015 usr
drwxr-xr-x  7 root root     0 Oct 22  2015 var

我看不出错误,你能解释一下我的错误吗？

Answer 1

更好(和更安全)使用shell函数而不是变量:

#!/bin/sh

myfunc() {
   candump -l -e -x -s 0 -n 10 any,0~0,#FFFFFFFF 2>/dev/null >/tmp/can.log &
}

# call it
myfunc