123

I want to obtain home dir of any user with echo

echo ~puchuu
>> /home/puchuu

But I cant use variable

echo ~$USER
>> ~puchuu
echo `echo ~$USER`
>> ~puchuu
1
  • Please do not use eval or bash -c with a variable. I added an answer that works safely for an Linux/Unix/macOS system with bash (even if you are not using bash as your shell, it likely has bash available because bashisms are everywhere). superuser.com/a/1613980/3376 Dec 31, 2020 at 6:23

7 Answers 7

124

You can use eval:

eval echo ~$USER

But see Andrew's comment and glenn's reply below.

10
  • 11
    In bash eval isn't needed it with just echo ~$username it's okay, but in sh eval is needed if is a variable Nov 25, 2015 at 14:23
  • This sometimes gives the wrong value, maybe the home folder of a previous account with the same username?
    – Andrew
    Dec 12, 2015 at 18:28
  • @AndrewMacFie: What do you mean by "previous"?
    – choroba
    Dec 12, 2015 at 18:49
  • 1
    @choroba Add a user, delete the user, then add a user with the same username. If the user's home folder is different the second time, this command gives the original home folder rather than the current one. glenn jackman's answer gives the current one.
    – Andrew
    Dec 12, 2015 at 18:59
  • 1
    Can confirm that this won't work at all if you have anything other than letters and numbers in a username. The method provided by Evan Carroll & Glen Jackmans answer below appears to work at least on Ubuntu 18. E.g: $( getent passwd "john-smith" | cut -d: -f6 )
    – Sk446
    Nov 25, 2019 at 11:33
100

This might work for you:

homedir=$( getent passwd "$USER" | cut -d: -f6 )

This will also work on users that are not you. For instance,

homedir=$( getent passwd "someotheruser" | cut -d: -f6 )
1
  • 1
    This is legit, using getenv rather than assuming the location of passwd is even a step further than assuming the location of home is /home/ Dec 1, 2016 at 22:19
18

It seems you are that user -- why not

echo $HOME

?

2
  • 7
    This won't work if you are in a sudo'ed environment and did not pass sudo the -H or -i flags - $HOME will still be the previous (sudo'ing) user's home directory. Jun 12, 2015 at 11:32
  • Thanks, worked. Apr 2, 2021 at 18:13
7

There is a safe way to do this!

on Linux/BSD/macOS/OSX without sudo or root

user=pi
user_home=$(bash -c "cd ~$(printf %q "$user") && pwd")

NOTE: The reason this is safe is because bash (even versions prior to 4.4) has its own printf function that includes:

%q quote the argument in a way that can be reused as shell input

See: help printf

Compare to how other answers respond to code injection

# "ls /" is not dangerous so you can try this on your machine
# But, it could just as easily be "sudo rm -rf /*"
$ user="root; ls /"
$ printf "%q" "$user"
root\;\ ls\ /

# This is what you get when you are PROTECTED from code injection
$ user_home=$(bash -c "cd ~$(printf "%q" "$user") && pwd"); echo $user_home
bash: line 0: cd: ~root; ls /: No such file or directory

# This is what you get when you ARE NOT PROTECTED from code injection
$ user_home=$(bash -c "cd ~$user && pwd"); echo $user_home
bin boot dev etc home lib lib64 media mnt ono opt proc root run sbin srv sys tmp usr var /root

$ user_home=$(eval "echo ~$user"); echo $user_home
/root bin boot dev etc home lib lib64 media mnt ono opt proc root run sbin srv sys tmp usr var

on Linux/BSD/macOS/OSX as root

If you are doing this because you are running something as root then you can use the power of sudo:

user=pi
user_home=$(sudo -u "$user" sh -c 'echo $HOME')

on Linux/BSD (but not modern macOS/OSX) without sudo or root

If not, the you can get it from /etc/passwd. There are already lots of examples of using eval and getent, so I'll give another option:

user=pi
user_home=$(awk -v u="$user" -v FS=':' '$1==u {print $6}' /etc/passwd)

I would really only use that one if I had a bash script with lots of other awk oneliners and no uses of cut. While many people like to "code golf" to use the fewest characters to accomplish a task, I favor "tool golf" because using fewer tools gives your script a smaller "compatibility footprint". Also, it's less man pages for your coworker or future-self to have to read to make sense of it.

2
  • 2
    But "$user" instead of $USER
    – pihentagy
    Apr 7, 2021 at 11:02
  • cd ... && pwd is not needed, a simple echo will suffice. Also, I suggest to improve quoting: ~$(printf '%q' "$user")
    – MestreLion
    Nov 22, 2021 at 6:00
2

ZSH users can place the tilde (~) outside the expression. This does not work on Bash:

echo ~`echo $USER`
1
  • Maybe works, but overly complicated (and archaic with the backticks). Just use echo ~$user or echo ~"$user" if you are afraid of usernames containing "funny" characters.
    – pihentagy
    Jan 22, 2021 at 13:11
1

Once you login, run cd to go to your home directory, then run pwd to print the working directory.

2
  • That's not going to work; the idea here is to do this in shell script, and presumably without requiring the credentials of the user in question...
    – SamB
    Aug 27, 2016 at 4:57
  • 1
    -1 for the same reason SamB mentioned. I edited your answer because I want you to see how to use Markdown formatting, and to be more concise. You also missed the much easier method, which is just echo $HOME.
    – wjandrea
    Sep 4, 2016 at 22:35
-1

how about using realpath instead of eval:

realpath ~$USER

because eval can execute anything, where as realpath will not.

3
  • This doesn't work for me. It just treats ~$USER as a filename in the current directory. For example if USER="bob" and PWD=/home/bob/Desktop, this expands to "/home/bob/Desktop/~bob"
    – pavon
    Feb 15 at 23:31
  • ~$USER syntax only work switch the user home, problem is it doesn't expand properly when used in variable and user must be valid.
    – rho
    Feb 16 at 11:06
  • From what I can tell realpath doesn't evaluate the ~bob syntax at all. Rather, in some cases the bash shell expands it and passes the expanded result into realpath.
    – pavon
    Feb 16 at 18:15

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