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Old 05-18-2010, 06:31 PM   #1
cade
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Registered: May 2010
Posts: 3

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BASH: Listing contents of *.zip file into an array


Hi, I am fairly new to programming and I need some help with my script.
I want to list the contents of a zip file amd put each entry into an array.
I've been doing
Code:
array=($(unzip*-qq*-l*"/path/to/file.zip"*|*awk*'{print*$4}'))
which works alright, and when I do
Code:
   for str in ${array[@]}; do
       echo "$str"
   done
It gives
Code:
   asdf/something
   base/asdf/something/the_quick_brown_fox_jumped_over_the_lazy_dog
.. until an item has a space in it's name
Code:
   for str in ${array[@]}; do
      echo "$str"
   done
Gives
Code:
   asdf/something
   base/asdf/something/the
   quick
   brown
   fox
   jumped
   over
   the
   lazy
   dog
Thanks

Last edited by cade; 05-18-2010 at 06:37 PM.
 
Old 05-18-2010, 08:57 PM   #2
grail
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Registered: Sep 2009
Location: Perth
Distribution: Manjaro
Posts: 7,192

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Try a while/read loop instead:
Code:
while read line
do
<your stuff here>
done< <(<your unzip stuff here>)
Be aware of the space between the two signs <, after done
 
Old 05-18-2010, 09:06 PM   #3
ta0kira
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Registered: Sep 2004
Distribution: FreeBSD 9.1, Kubuntu 12.10
Posts: 3,078

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You need to use a different field separator ($IFS) that excludes spaces:
Code:
OLD_IFS="$IFS"
IFS=$'\n'
array=($(zip stuff...))
IFS="$OLD_IFS"

#...

for I in `seq 0 1 $((${#array[*]}-1))`; do
  #stuff with ${array[$I]}
done
Kevin Barry
 
Old 05-18-2010, 09:35 PM   #4
ntubski
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Distribution: Debian
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Code:
   for str in "${array[@]}"; do
      echo "$str"
   done
EDIT: this assumes array has been set correctly

Last edited by ntubski; 05-20-2010 at 09:25 AM. Reason: confirm ta0kira's assumption
 
Old 05-19-2010, 04:13 PM   #5
ta0kira
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Registered: Sep 2004
Distribution: FreeBSD 9.1, Kubuntu 12.10
Posts: 3,078

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Quote:
Originally Posted by ntubski View Post
Code:
   for str in "${array[@]}"; do
      echo "$str"
   done
Thanks. I'm assuming this is just a correction to my loop and not to the IFS part.
Kevin Barry
 
  


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