bash中带有空格的回调脚本

Question

I wrote a little script that check something, and i want a command to be executed from within the script if the test was successfull. And i don't want to hardcode the command, but give it as argument to it like a callback script.

The command I testing with is /usr/bin/xmessage -buttons "button a","button b" some text to test . Running it within a terminal standalone works fine, no quotation marks needed for the last text.

The script looks like this:

#!/bin/bash
echo "$1"
$1

But when running /path/to/script.bash '/usr/bin/xmessage -buttons "button a","button b" some text to test' it looks like this , though the echo looks right.

When using "$1" instead of $1 it complains it couldn't find the file. Anyone got ideas how to fix the behavior with the space?

Answer 1

You'll need to use an array.

See the following link on how to fix your problem: I'm trying to put a command in a variable, but the complex cases always fail!

Answer 2

I think you should use eval :

#!/bin/bash
echo "$1"
eval $1

Answer 3

I suggest you write the script like this:

#!/bin/bash
echo ${1+"$@"}
${1+"$@"}

And call it like this:

/path/to/script.bash /usr/bin/xmessage -buttons "button a","button b" some text to test

Answer 4

尝试使用eval "$1"或重新设计，以便在所有选项之后指定回调-常见的安排将类似于script.bash -options arguments etc -- /usr/bin/xmessage -buttons "button a","button b" some text to test