Re: Bash question: get output as a variable?

[Mart Frauenlob <mart.frauenlob@chello.at>]

On 06.02.2010 15:43, Mart Frauenlob wrote:
> On 06.02.2010 14:17, Javier Barroso wrote:
>> On Fri, Feb 5, 2010 at 7:10 PM, Chris Jackson <c.jackson@shadowcat.co.uk> wrote:
>>> Dotan Cohen wrote:
>>>
>>>> I'm scripting a backup solution, the line that does the business looks
>>>> like this:
>>>>
>>>> tar -zcvf - *  --exclude-from $EXCLUDES  | openssl des3 -salt -k $1 |
>>>> dd of=$(hostname)-$(date +%Y%m%d).tbz
>>>>
>>>> Because of the "v" flag tar writes to stdout the name of each file
>>>> copied. How can I get that output redirected to a variable, to use
>>>> later in the script?
>>>>
>>>> Thanks!
>>>>
>>>
>>>
>>> Use $() like you do with the date command. You have to redirect stderr back
>>> to stdout, which means running it in a subshell:
>>>
>>>
>>> FILES=$( ( tar -zcvf - *  --exclude-from $EXCLUDES  | openssl des3 -salt -k
>>> $1 | dd of=$(hostname)-$(date +%Y%m%d).tbz ) 2>&1 )
>>>
>>> It may cause unexpected results if there're spaces in the filenames though.
>>
>> If there are spaces in filenames, you can try:
>>
>> $ n=0; while read l; do files[n]="$l"; ((n++)); done < <((tar -zvcf -
>> * | openssl > $(hostname)-$(date +%Y%m%d)) 2>&1)
>>
>> Regards,
>>
>>
> Warning:
> 

[..]

> read strips of leading and trailing spaces.
> use the $REPLY variable to avoid that.

or by:
IFS=$'\n'

while IFS=$'\n'; read l [...] ; do ...

> 
> rare conditions of newline in filenames is not covered...

> 
>