Re: 7z command works fine on command line but not in script
On 19/02/14 19:47, berenger.morel@neutralite.org wrote:
> Hello.
>
> I made a script to extract music from a jamendo archive, but for a
> reason I do not know, 7z does not accept the command line. I also echoed
> it, to be able to know what it tries to run, and it works fine when ran
> on command line.
>
> Here is the script:
>
> #!/bin/sh
>
> for i in *.zip;
> do
> artiste=`echo $i|cut -f1 -d-|sed -e 's/^ *//g' -e 's/ *$//g'`
> album=`echo $i|cut -f2 -d-|sed -e 's/^ *//g' -e 's/ *$//g'`
> mkdir "$artiste/$album" -p
> 7z x \"$i\" -o\"$artiste/$album\" || rmdir "$artiste/$album" -p
> echo 7z x \"$i\" -o\"$artiste/$album\"
> done
>
> With, for example, this archive:
> http://www.jamendo.com/fr/list/a69778/monument, the folders are
> correctly created, but it prints:
> Error: Incorrect command line
> 7z x "Shearer - Monument - a69778 --- Jamendo - MP3 VBR 192k.zip"
> -o"Shearer/Monument"
>
> Do someone knows what I am doing wrong?
>
>
Maybe passing "-" without a recognised qualifier (after the
$artiste/$album\) ??
Kind regards
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