Bash脚本CD问题

Question

Hi all I have some problems with my script. I've read that changing the current directory from within a script is a bit of an issue. Basically I am looking for a single php file with a project folder and any sub-folders in it. And I want to change the directory to where that folder is and perform a command for it. So far no luck.

function findPHP(){
declare -a FILES
FILES=$(find ./ -name \*.php)
for file in "${FILES[@]}"
do

DIR=`dirname file`
( cd $DIR && doSomethingInThisDir &(...))

done

Any help would be greatly appreciated.

Answer 1

You are trying to iterate over FILES as an array, but it only has one element. In order to make the result of your subshell into an array, you can:

FILES=($(find ./ -name \*.php))

Note that it splits file names on spaces, so even though you properly quote below, it won't help. Alternatively, you could just let it split below (ie using your existing FILES ) and use instead:

for file in $FILES

If you are using bash 4 , you may want to have a look at recursive globbing... this would make it a bit easier:

for file in **/*.php

Note that you have to have the globstar shell option set, which you could enable with shopt -s globstar . This way is simpler and won't break on whitespace.

Also, you probably want $file here:

DIR=`dirname $file`

Or just use parameter expansion:

DIR=${file%/*}

Answer 2

There is no reason to use an array, or store the file list in anyway. If your find supports -execdir (eg gnufind 4.2.27), then use it. Otherwise, cd in a subshell as you have done:

#!/bin/bash
doSomethingInThisDir() ( cd $(dirname $1); ... )
export -f doSomethingInThisDir
find . -type f -exec bash -c 'doSomethingInThisDir {}' \;

I have defined the function using () instead of {} , but that is not necessary in this case. Normally, using () causes the function to run in a subshell, but that happens here anyway because find runs a separate process for each file.