Test if file exists with a function in shell script

Question

I have the following script, basically a function and an IF

file_exists() {

    if [ -f "$1" ]; then
        return 0
    else
        return 1
    fi
}

if [[ $(file_exists "LICENSE") ]]; then
    echo "YES"
else
    echo "NO"
fi

But this code always returns NO. I know the IF statement expects to get a 0 to be true, but I don't understand why it doesn't work

Also consider `-s` which will test `"True if file exists and has a size greater than zero."` — David C. Rankin, Dec 30 '19 at 05:11
Your function does not write anything to stdout. Hence, `$(file_exists.....) is always empty. `[[ '' ]]` (i.e. on an empty string) takes by design the `else` branch. — user1934428, Dec 30 '19 at 07:30

Michael Jarrett · Accepted Answer · 2019-12-30T00:32:24.810

3

When using the return value of a function in an if-statement, you do not need to wrap it in [[]]. You can replace

if [[ $(file_exists "LICENSE") ]]; then

with

if file_exists "LICENSE"; then

As for the convention of 0=true and 1=false, it's not preferred to write them out explicitly in return statements. The body of your file_exists function can be reduced to

file_exists() {
    [ -f "$1" ]
}

edited Dec 30 '19 at 00:32

answered Dec 30 '19 at 00:28

Michael Jarrett

425
3
10

score -1 · Answer 2 · answered Dec 30 '19 at 07:40

-1

IMHO if..else redundant here. Use your function like this:

file_exists $FILENAME && echo ok || echo notok

answered Dec 30 '19 at 07:40

Ivan

6,188
1
16
23

Test if file exists with a function in shell script

2 Answers2