How to find a file on the $PATH in Bash?

Question

The utility which will find a program on the path but what about an arbitrary data file?

I must be searching for the wrong stuff but I can't find any way to do this with a standard utility so I thought of writing a little Bash script to do it.

The $PATH environment variable is separated by colons, but trying to set IFS=':' and then iterate over the results is not working for me. Here's what I've got so far:

IFS=':' DIRS=($PATH)
for d in $DIRS; do echo $d; done

At this point it only outputs the first entry in my path rather than all of them.

Thoughts? If there's already a standard command that does this then there's no reason to write a script ...

Arbitrary data files probably shouldn't be in directories lists in `PATH`. — chepner, Mar 27 '19 at 17:01

Gilles Quénot · Answer 1 · 2020-05-22T13:43:39.170

10

Using this, using bash parameter expansions and find:

find ${PATH//:/\/ } -name 'file'

edited May 22 '20 at 13:43

answered Mar 27 '19 at 16:58

Gilles Quénot

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6

Suppose there's a directory in the PATH that has a space in it. Or a file in a directory in the PATH with a space in the filename. – glenn jackman Mar 27 '19 at 17:06
3

You're also not expanding the files in the *last* PATH entry. – glenn jackman Mar 27 '19 at 17:11
1

This solution would be great if I could use `find` in a way that searches the `$PATH` but replacing colons with spaces isn't going to work – Neil C. Obremski Mar 27 '19 at 17:56
2

@GillesQuenot I've upvoted your `find` solution because I like it _but_ it won't work when there's spaces in directories that are in `$PATH` – Neil C. Obremski Mar 28 '19 at 15:42
I never use spaces especially for a PATH directory, and I like my solution too ^^ – Gilles Quénot Mar 28 '19 at 16:01
If you do have spaces in your `$PATH` but have managed to split the entries into an array, simply use `find "${array[@]}" -name "$file"` – tripleee Jun 10 '22 at 05:50

tripleee · Answer 2 · 2019-03-27T17:13:09.697

The error is not with the assignment, but with not looping over the array.

IFS=: dirs=($PATH)
for dir in "${dirs[@]}"; do
    echo "$dir"
done

works for me; or more succinctly

printf '%s\n' "${dirs[@]}"

Like you discovered, just $dirs only returns the first item from the array.

To actually traverse these directories looking for a particular file, maybe try

desired_file_name=$1  # or whatever
for dir in "${dirs[@]}"; do
    test -e "$dir/$desired_file_name" || continue
    echo "$0: found $dir/$desired_file_name" >&2
done

Another approach is

find "${dirs[@]}" -name "$desired_file_name"

(You should prefer lower case for your private variables, so I changed DIRS to dirs. Less shouting is good for the eyes, too.)

Huh, I guess I didn't test what I wrote. – glenn jackman Mar 27 '19 at 17:16 — glenn jackman, Mar 27 '19 at 17:16

score 2 · Answer 3 · answered Mar 27 '19 at 17:04

To find a file named $name on your PATH using a bash loop:

oldIFS=$IFS
IFS=:
for d in $PATH
do
    [ -f "$d/$name" ] && echo "Found $d/$name"
done
IFS=$oldIFS

On many systems, such as debian, which is just a shell script and it uses a very similar loop. Have a look at less /usr/bin/which.

glenn jackman · Accepted Answer · 2019-03-27T17:20:09.987

A few of things going on here:

~~IFS=':' DIRS=($PATH)~~

bash will expand the variable before setting IFS and DIRS, so it's too late by then. You'll need something crazy like
```
dirs=()
while IFS= read -r -d: dir || [ "$dir" ]; do dirs+=("$dir"); done <<<"$PATH"
```

-- nope, that was wrong. First the $PATH variable is expanded, then IFS is set, then the DIRS element is split using IFS because it is unquoted.

for d in $DIRS; do echo $d; done

To iterate over all the elements of an array, you need
```
for d in "${dirs[@]}" ...
```
With an array, $DIRS is equal to ${DIRS[0]}, i.e. the first entry.
Don't use ALLCAPS varnames. It's too easy to overwrite a crucial system variable.
Quote your variables unless you know exactly what will happen if you don't.

How to find a file on the $PATH in Bash?

4 Answers4