如何执行shell变量中设置的sh命令？

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 中运行它工作正常。

我想用这种方法将它添加到我的脚本中：

#!/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

当我执行脚本时出现此错误：

SIOCGIFINDEX：没有这样的设备

我测试了这些脚本并且有效：

#!/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

我看不到错误，你能解释一下错误吗？

Answer 1

Better (and safer) to use a shell function rather that a variable:

#!/bin/sh

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

# call it
myfunc

如何执行shell变量中设置的sh命令？

How to execute sh command set in shell variable?

linux

shell

sh