sh: Test for existence of files - bash

Question

How does one test for the existence of files in a directory using bash?
if ... ; then
echo 'Found some!'
fi
To be clear, I don't want to test for the existence of a specific file. I would like to test if a specific directory contains any files.
I went with:
(
shopt -s dotglob nullglob
existing_files=( ./* )
if [[ ${#existing_files[#]} -gt 0 ]] ; then
some_command "${existing_files[#]}"
fi
)
Using the array avoids race conditions from reading the file list twice.

Answer 1

From the man page:
-f file
True if file exists and is a regular file.
So:
if [ -f someFileName ]; then echo 'Found some!'; fi
Edit: I see you already got the answer, but for completeness, you can use the info in Checking from shell script if a directory contains files - and lose the dotglob option if you want hidden files ignored.

Answer 2

I typically just use a cheap ls -A to see if there's a response.
Pseudo-maybe-correct-syntax-example-ahoy:
if [[ $(ls -A my_directory_path_variable ) ]] then....
edit, this will work:
myDir=(./*) if [ ${#myDir[#]} -gt 1 ]; then echo "there's something down here"; fi

Answer 3

You can use ls in an if statement thus:
if [[ "$(ls -a1 | egrep -v '^\.$|^\.\.$')" = "" ]] ; then echo empty ; fi
or, thanks to ikegami,
if [[ "$(ls -A)" = "" ]] ; then echo empty ; fi
or, even shorter:
if [[ -z "$(ls -A)" ]] ; then echo empty ; fi
These basically list all files in the current directory (including hidden ones) that are neither . nor ...
If that list is empty, then the directory is empty.
If you want to discount hidden files, you can simplify it to:
if [[ "$(ls)" = "" ]] ; then echo empty ; fi
A bash-only solution (no invoking external programs like ls or egrep) can be done as follows:
emp=Y; for i in *; do if [[ $i != "*" ]]; then emp=N; break; fi; done; echo $emp
It's not the prettiest code in the world, it simply sets emp to Y and then, for every real file, sets it to N and breaks from the for loop for efficiency. If there were zero files, it stays as Y.

Answer 4

Try this
if [ -f /tmp/foo.txt ]
then
echo the file exists
fi
ref: http://tldp.org/LDP/abs/html/fto.html
you may also want to check this out: http://tldp.org/LDP/abs/html/fto.html
How about this for whether directory is empty or not
$ find "/tmp" -type f -exec echo Found file {} \;

Answer 5

#!/bin/bash
if [ -e $1 ]; then
echo "File exists"
else
echo "Files does not exist"
fi

Answer 6

I don't have a good pure sh/bash solution, but it's easy to do in Perl:
#!/usr/bin/perl
use strict;
use warnings;
die "Usage: $0 dir\n" if scalar #ARGV != 1 or not -d $ARGV[0];
opendir my $DIR, $ARGV[0] or die "$ARGV[0]: $!\n";
my #files = readdir $DIR;
closedir $DIR;
if (scalar #files == 2) { # . and ..
exit 0;
}
else {
exit 1;
}
Call it something like emptydir and put it somewhere in your $PATH, then:
if emptydir dir ; then
echo "dir is empty"
else
echo "dir is not empty"
fi
It dies with an error message if you give it no arguments, two or more arguments, or an argument that isn't a directory; it's easy enough to change if you prefer different behavior.

Answer 7

# tested on Linux BASH
directory=$1
if test $(stat -c %h $directory) -gt 2;
then
echo "not empty"
else
echo "empty"
fi

Answer 8

For fun:
if ( shopt -s nullglob ; perl -e'exit !#ARGV' ./* ) ; then
echo 'Found some!'
fi
(Doesn't check for hidden files)