Loop with variable in bash script - bash

Question

I deploy my website using a post-receive git hook. Within that hook I use the YUI compressor to minify js and css files:
export TEMP=/var/www/example.com
GIT_WORK_TREE=/var/www/example.com git checkout master -f
#Minify mit YUI
(cd $TEMP/css && min style.css && rm style.css && mv style.min.css style.css)
(cd $TEMP/addons/css && min bootstrap.css && rm bootstrap.css && mv bootstrap.min.css
(cd $TEMP/js && min script.js && rm script.js && mv script.min.js script.js)
(cd $TEMP/addons/js && min startup.js && rm startup.js && mv startup.min.js startup.js)
Now I would like to not specify the exact files, but to search for all js and css files through all folders in $TEMP and repeat the minify procedure each time.
Could any one help me with the right loop and search-syntax for this case?
Thanks!

Answer 1

Just a guess here, but what about constructs like this?
find $TEMP -name \*.css -exec sh -c 'f="{}"; min "$f" && mv "${f%.css}.min.css" "$f"' \;
The idea is that the find command finds all your CSS files, then executes the min and mv commands. No need to rm, as the mv will overwrite.
You can probably figure out the equivalent line for your JavaScript files. :-)
Note that I haven't tested this, as I don't use min, but this isn't a question about minifying or YUI, it's a question about how to execute a command on multiple files in a directory tree.
UPDATE:
You can skip some files by putting extra logic into the find conditions:
find $TEMP -name \*.js -and -not -name \*.min.js -exec ...

Answer 2

In a single command line using bash terminal:
TEMP="/var/www/example.com"; pwd="$PWD"; IFS=$'\n'; for f in `find $TEMP -name "*.js" -or -name "*.css"`; do dir=`dirname $f`; cd "$dir"; min "$f"; ext="${f##*.}"; f2=`echo $f | sed "s/$ext$/min\.$ext/g"`; rm -f "$f"; mv "$f2" "$f"; done; cd "$pwd"
or if you place it in a sh file, then:
#!/bin/bash
TEMP="$1";
IFS=$'\n';
for f in `find $TEMP -name "*.js" -or -name "*.css"`; do
dir=`dirname $f`
cd "$dir"
min "$f"
ext="${f##*.}"
f2=`echo $f | sed "s/$ext$/min\.$ext/g"`
rm -f "$f"
mv "$f2" "$f"
done
and call the script with the path as an argument or hardcode it directly into the file.
The above algorithm takes care of both js and css files. It moves into the directory where the file is located and create the temporary file in that very same directory, as your original command does. It also should work even if you have spaces in file or directory names or if files contain the substring .js or .css within their names and not only in the extension.