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Top Forums Shell Programming and Scripting Problem scripting a copy and renaming shell executable Post 302932171 by sea on Monday 19th of January 2015 03:56:45 PM
Old 01-19-2015
Quote:
Originally Posted by ilovedoritos
i ran this
Code:
cd Desktop/sourcepictures
index="0000000" 
for image in $(cat list.txt) ; do 
cp ${sourcepictures}/$image orderedpictures/${index}.jpeg
index = $(printf %06d $(( $index + 1 )) )
done

i have the code in a file named "test.command" on the desktop

i have the folder named "sourcepictures" on the desktop and the jpeg files are in that folder and inside the "sourcepictures" folder i have the folder named "orderedpictures". also, list.txt is in "sourcepictures" folder. it found the list.txt

what do i have to put in the script to get it to find the files and directories
You've used php/lua/java settings to assign a value, but its bash.
index = $(printf %06d $(( $index + 1 )) ) --> index=$(printf %06d $(( $index + 1 )) )

And regarding the folders, i now assume it lookes like (yes the renaming of the folders is on purpose, just for the example):
  • $HOME/Desktop/test.sh
  • $HOME/Desktop/pictures-source
  • $HOME/Desktop/pictures-ordered


Now according to your posted code snippet, i would change that to:
test.sh
Code:
#!/bin/bash
SOURCES="$HOME/Desktop/sourcepictures"
TARGET="$HOME/Desktop/orderedpictures"
LIST="$HOME/Desktop/list.txt"
index="0000000" 

for image in $(cat $LIST) ; do 
    ln -s $SOURCES/$image  $TARGET/$index.jpg
    index=$(printf %06d $(( $index + 1 )) )
done

If list.txt contains only the chars (a,b,c,d...) you need to add ".jpg" after "$image".
If list.txt contains the proper image name, you wont need to change anything.

hth & gn8
 

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