Test if directory exists or not

Question

Here is my code, and it works but it always says that directory exist, doesnt matter what I write. Moreover it does not print a variable in echo $DIRECTORY. What I need to fix?

#!/bin/sh

if [ -d $DIRECTORY ]; then
echo "Directory exists"
elif [ ! -d $DIRECTORY ]; then
echo "Directory does not exists"
fi
echo $DIRECTORY

Did you assign the DIRECTORY variable beforehand? Like `DIRECTORY=/etc` — Vasily G, Oct 29 '15 at 07:29
I just write ./program(it's a name of program) abc(it's a name of directory, which does not exist) — David Arutiunian, Oct 29 '15 at 07:31
Then substitute `$DIRECTORY` for `$1` (the first argument to the script) — Vasily G, Oct 29 '15 at 07:33
Does this answer your question? [How can I check if a directory exists in a Bash shell script?](https://stackoverflow.com/questions/59838/how-can-i-check-if-a-directory-exists-in-a-bash-shell-script) — Josh Correia, Dec 09 '20 at 18:03

F. Hauri - Give Up GitHub · Accepted Answer · 2021-03-01T07:36:47.687

Passing variables to shell script

You have to instruct you script that DIRECTORY is a variable, passed as first script argument.
You have to enclose your variable into double-quotes in order to ensure spaces and special characters, like empty variables, to be correctly parsed.

Sample:

#!/bin/sh

DIRECTORY="$1"

if [ -d "$DIRECTORY" ]; then
    echo "Directory '$DIRECTORY' exists"
else
    echo "Directory '$DIRECTORY' does not exists"
fi
echo "$DIRECTORY"

Other sample:

#!/bin/bash

DIRECTORY="$1"
FILE="$2"
if [ -d "$DIRECTORY" ]; then
    if [ -e "$DIRECTORY/$FILE" ]; then
        printf 'File "%s" found in "%s":\n  ' "$FILE" "$DIRECTORY"
        /bin/ls -ld "$DIRECTORY/$FILE"
    else
        echo "Directory '$DIRECTORY' exists, but no '$FILE'!"
    fi
else
    echo "Directory '$DIRECTORY' does not exists!"
fi
echo "$DIRECTORY/$FILE"

Yes, but @gniourf_gniourf is right! Usig `elif` there reflect a stupid 3 way boolean logic! — F. Hauri - Give Up GitHub, Oct 29 '15 at 07:40

Test if directory exists or not

1 Answers1

Passing variables to shell script