Actually in the bash script there is a command that passes "--options='-t 0
-v 0'" as argument to an executable. I just found if I double quote
--options='-t 0 -v 0' as argument to the bash script, then the '-t 0 -v 0'
as a whole can be preserved until reaching the command invoking the
executable. However the '-t 0 -v 0' is not passed as a whole to the
executable, but splitted by spaces again.
This may sounds like the problem of calling the executable, however when
directly invoking the executable from terminal, either --options='-t 0 -v 0'
or --options="-t 0 -v 0" without quoting could be passed successfully as
argument.
Hope that I could explain my question clearly enough. Really appreciate your
advice!
lehe wrote:
>
> Thanks Mike.
> Could you explain it a little? I don't quite get it. How to apply this to
> argument parsing?
>
>
> Mike Frysinger wrote:
>>
>> On Thursday 09 April 2009 16:46:27 lehe wrote:
>>> I was wondering how to pass arguments with space inside. For example, my
>>> bash script looks like:
>>>
>>> #!/bin/bash
>>> ARG_OPTS=""
>>> while [[ -n "$1" ]];
>>> ARG_OPTS="${ARG_OPTS} $1"
>>> shift
>>> done
>>>
>>> If I pass an argument like "--options='-t 0 -v 0'", then it would be
>>> splitted by the spaces inside, ie "--options='-t", "0", "-v" and "0".
>>>
>>> How can I achieve what I wish?
>>
>> use arrays
>>
>> $ f=( a "b c" d)
>> $ printf '%s\n' "$...@]}"
>> a
>> b c
>> d
>> -mike
>>
>>
>>
>
>
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