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Old 12-20-2004, 07:28 AM   #1
drisay
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bash programming


hi all,

i'm currently trying to teach myself bash. in the script i'm trying to rename all files from IMG_00xx.JPG to PIC_00xx.JPG. Can anyone spot the error.

Note: I was reading up on sed on a few web sites and this seems to be the way they assign the result to a variable... but obviously it's not working. The output of tmpvar is simply "$i | sed "s/IMG/PIC/"".

script:
Code:
#!/bin/bash

tmpvar=""

for i in $(ls *.JPG); do
  tmpvar='echo $i | sed "s/IMG/PIC/"'
  mv $i $tmpvar
done;

Result:
$ . pics.sh
mv: when moving multiple files, last argument must be a directory
Try `mv --help' for more information.
mv: when moving multiple files, last argument must be a directory
Try `mv --help' for more information.
mv: when moving multiple files, last argument must be a directory
Try `mv --help' for more information.
mv: when moving multiple files, last argument must be a directory
Try `mv --help' for more information.
mv: when moving multiple files, last argument must be a directory
Try `mv --help' for more information.

Thanks in advance,
drisay.
 
Old 12-20-2004, 07:32 AM   #2
acid_kewpie
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you're using quotes and not backticks, so rather than tmpvar being the output of that command, it IS that command....

so it's
Code:
tmpvar=`echo.....`
rather than
Code:
tmpvar='echo...'
 
Old 12-20-2004, 07:40 AM   #3
drisay
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wow, ok thanks. i never would have found that.
 
Old 12-20-2004, 08:19 AM   #4
bigearsbilly
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or, the new way is


Code:
tmpvar=$(   echo $i | sed "s/IMG/PIC/"     )
 
Old 12-20-2004, 08:25 AM   #5
acid_kewpie
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that echo and pipe always bugs me... would be nicer to be able to give the variable as a parameter to sed in the first place really.
 
Old 12-20-2004, 08:52 AM   #6
bigearsbilly
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true, true....

could do:
Code:
ls -1 *.jpg|sed -n '
s/\.jpg/\.IMG/g
s/.*/& &/
s/\.IMG/\.jpg/
p' |  xargs -n2 cp
or better:

Code:
ls -1 *.jpg|sed -n '
s/\.jpg/\.IMG/g
s/.*/& &/
s/\.IMG/\.jpg/
p'     > ~/1
give the output an eyeball
to check it's ok then do

. ~/1

so you've got a temp record too.

billy
 
Old 12-20-2004, 09:09 AM   #7
bigearsbilly
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OR, best of all:

Code:
for file in *.jpg
do  
    echo  $file ${file%.jpg}.IMG
done
 
  


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