[SOLVED] shell script extension changer

[king vash]

I am working on a shell script that takes an old extension, a new extension and a directory and renames all the old extension to new extension in the directory (and subdirectories)

so far I have this.


#!/bin/sh
if [ $# = 3 ]
then
if [ ! $2:~0,1 = "." ]
then
oldext="."$2
else
oldext=$2
fi

if [ ! $3:~0,1 = "." ]
then
newext="."$3
else
newext=$3
fi

filelist=$(find "$1" -name *$2)
echo $filelist
for name in "$filelist"
do
# echo "$name"
name2=`echo "$name" | sed -n -e "s!$oldext!$newext!gp"`
# mv "$name" "$name2"
# echo "$name2"
done
else
echo "takes three parameters : location old extention and new extention"
fi

The problem is because I am ignoring spaces in $filelist in the for loop it only sends one long filename (all the file names seperated by a space)

[yabbadabbadont]

man rename
;)

[ghostdog74]

filelist=$(find "$1" -name *$2)
echo $filelist
for name in "$filelist"
do
# echo "$name"
name2=`echo "$name" | sed -n -e "s!$oldext!$newext!gp"`
# mv "$name" "$name2"
# echo "$name2"
done
else


use a while read loop instead of for loop


....
find "$1" -name *$2 | while read name
do
#mv file...
done
....

also , in bash, don't have to use external commands like sed to change extension. for eg


# file="test.txt"
# echo ${file/.txt/.jpg}
test.jpg


see my sig for bash link

[king vash]

The while read loop fixed the problem thanks!

but now I have a different problem. I wanted it to count the files changes and if it changed more than one file to return a successful message.


j=0;
find "$1" -name *$oldext | while read name
do
oldname=$name
newname=`echo $name | sed -e "s!$oldext!$newext!"`
j=`expr $j + 1`
echo $j
done
echo $j
if [ $j -gt 2 ]
then
echo "Yeah! "$j" files extensions changed"
fi

how ever this returns

1
2
3
4
.
.
.
47
0

any variable gets reset after the script. how ever if I change the initialization to "10" then 10 will be returned at the end.

[ghostdog74]

i don't understand.

[king vash]

if i use the line


find "$1" -name *$oldext | while read name

then any variables set or changed in my loop get reset to the values they had before the loop.

If j=0 coming into the loop, and then I add 1 to it a couple of times and exit the loop, j will be reset to 0. If j=10 coming into the loop I can change it and do anything I want but the second I exit the loop j will be reset to 10.

I tried a couple of variations to see it they had the same problem.
this still has the problem


find "$1" -name *$oldext | while [ $j -lt 20 ]

this didn't


while [ $j -lt 20 ]

this did


find "$1" -name *$oldext | for name in "a as asd asdf"


so you can see anything that has the

find "$1" -name *$oldext
exhibits this problem

[dwhitney67]

The value does not change when you complete the while-loop because when you pipe the results from 'find' to the 'while', you are in essence creating a sub-shell which will have a copy of your variable 'j'. Upon returning to the top-level shell, 'j' is still at the same value it originally held.

If there is a way around this, I am not aware of it.

I did some Googling and found a document that provides a better explanation from what I provided above.
http://www.cs.cornell.edu/Courses/cs114/2002fa/lect9.pdf