Check Whether a User Exists

Question

I want to create a script to check whether a user exists. I am using the logic below:

# getent passwd test > /dev/null 2&>1
# echo $?
0
# getent passwd test1 > /dev/null 2&>1
# echo $?
2

So if the user exists, then we have success, else the user does not exist. I have put above command in the bash script as below:

#!/bin/bash

getent passwd $1 > /dev/null 2&>1

if [ $? -eq 0 ]; then
    echo "yes the user exists"
else
    echo "No, the user does not exist"
fi

Now, my script always says that the user exists no matter what:

# sh passwd.sh test
yes the user exists
# sh passwd.sh test1
yes the user exists
# sh passwd.sh test2
yes the user exists

Why does the above condition always evaluate to be TRUE and say that the user exists?

Where am I going wrong?

UPDATE:

After reading all the responses, I found the problem in my script. The problem was the way I was redirecting getent output. So I removed all the redirection stuff and made the getent line look like this:

getent passwd $user  > /dev/null

Now my script is working fine.

I think that when you use `$?` you get the return code of the `test` command. See answer below. — user000001, Feb 11 '13 at 12:53
Is that the *exact* code you are running? No "echo" debug statements between the call to `getent` and your `if` statement? That would cause `$?` to check the exit status of `echo`, not your call to `getent`. — chepner, Feb 11 '13 at 14:09
Code works on non-empty input, but if `$1` is not set, the if statement will return true. From my testing, solution would be to wrap `$1` in double quotes: `getent passwd "$1" > /dev/null 2&>1`. — Vince, Jul 24 '17 at 14:28
@slayedbylucifer The reason your `getent` command does not work is that you have a typo: `2&>1` __is not a valid redirection__. — jimis, Feb 11 '21 at 16:09

score 384 · Answer 1 · edited Jun 12 '23 at 12:16

384

You can also check user by id command.

id -u name gives you the id of that user. if the user doesn't exist, you got command return value ($?)1

And as other answers pointed out: if all you want is just to check if the user exists, use if with id directly, as if already checks for the exit code. There's no need to fiddle with strings, [, $? or $():

if id "$1" >/dev/null 2>&1; then
    echo 'user found'
else
    echo 'user not found'
fi

(no need to use -u as you're discarding the output anyway)

Also, if you turn this snippet into a function or script, I suggest you also set your exit code appropriately:

#!/bin/bash
user_exists(){ id "$1" &>/dev/null; } # silent, it just sets the exit code
if user_exists "$1"; code=$?; then  # use the function, save the code
    echo 'user found'
else
    echo 'user not found' >&2  # error messages should go to stderr
fi
exit $code  # set the exit code, ultimately the same set by `id`

edited Jun 12 '23 at 12:16

Agostino

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answered Feb 11 '13 at 12:29

Kent

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3

That's what I use. See http://superuser.com/questions/336275/find-out-if-user-name-exists – guettli Oct 11 '13 at 11:09
24

great answer - just a minor tweak that works nicely... user_exists=$(id -u user > /dev/null 2>&1; echo $?) user_exists=0 if "user" exists, or 1 if not – Pancho Jun 02 '15 at 22:37
Thats great, id command is available even in yocto images, while getent is not. – HimanshuR Apr 04 '16 at 06:46
5

@Pancho I think your variable should be named user_doesnt_exist – Will Sheppard Aug 31 '16 at 09:22
@WillSheppard - the variable as named is correct. Run the command from the bash shell and you will see that it returns 0 if the user exists and 1 if the user doesn't exist. This is as distinct from boolean logic which I assume you are referring to. If you want to get fancy and make the response align that's great, but it's not necessary in order to achieve the distinction. (look at Kent's answer above and you will see my comment simply expands on his answer) – Pancho Aug 31 '16 at 14:44
@Pancho Can you show any example code which uses the variable, where your name makes sense? – Will Sheppard Aug 31 '16 at 15:20
@WillSheppard : echo $(id -u RUBBISHUSER > /dev/null 2>&1; echo $?) ....the output is 1 (because RUBBISHUSER doesn't exist) then: echo $(id -u root > /dev/null 2>&1; echo $?) ...the output is 0 (because the root user exists). It's really not rocket science, not sure why you don't just open a bash shell and test it yourself. Again, I am not attempting to apply boolean logic here simply extending the original answer. – Pancho Aug 31 '16 at 18:19
1

@WillSheppard : At a general level boolean handling in bash is in my opinion less than clear and I typically avoid implicit boolean inferences in my code. The following make interesting reading and interestingly both favour 0 = true and = false for bash purposes, and while not de facto, align with the variable name I have used from a boolean perspective also: http://stackoverflow.com/a/5431932/3051627 . http://stackoverflow.com/questions/2933843/why-0-is-true-but-false-is-1-in-the-shell . – Pancho Aug 31 '16 at 18:53
Thanks for highlighting the difference between boolean and return/error code. Your variable name $user_exists implies boolean. Perhaps it should be called $user_check_error :-) #namingishard – Will Sheppard Sep 01 '16 at 09:28
1

Is it a vanilla linux command? Or does it come with an optional package? I think it's in "coreutils" so it's not universal. – Sandburg Jul 12 '19 at 06:59
2

Thanks to @Pancho's comment I could achieve that, just by sending the error to the void and doing ```validuser(){ [[ -n $(id -u "$1" 2>/dev/null) ]] && echo 1 || echo 0; }```. That, pretty much returns 1 if user exists, or 0 if it does not. – lu1s Feb 06 '20 at 19:11
1

The status code of the last executed command inside the function is what function returns in bash. So simply put the command inside the function: `user_exists() { id -u $1 > /dev/null 2>&1 }` (this is not a one liner) and use it `if ! user_exists username; then....` – muratgozel Jul 09 '20 at 08:54
@muratgozel: your function can perfectly be a one-liner if you just add a `; ` before the closing `}` – MestreLion Aug 16 '20 at 04:00
This will be error "permission denied" on create /dev/null. – vee Aug 14 '21 at 14:31

score 40 · Answer 2 · answered Feb 11 '13 at 14:08

There's no need to check the exit code explicitly. Try

if getent passwd $1 > /dev/null 2>&1; then
    echo "yes the user exists"
else
    echo "No, the user does not exist"
fi

If that doesn't work, there is something wrong with your getent, or you have more users defined than you think.

score 37 · Answer 3 · answered Feb 11 '13 at 10:52

37

Why don't you simply use

grep -c '^username:' /etc/passwd

It will return 1 (since a user has max. 1 entry) if the user exists and 0 if it doesn't.

answered Feb 11 '13 at 10:52

poitroae

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23

This doesn't work if the server is using another system to manage uses such as NIS or LDAP. In this case, you can use `getent passwd | grep -c '^username:'` – Toby Jackson Nov 12 '13 at 16:54
1

this will return 1 if you have a username, let's say "test1" and want to create a new username called "test". – Marin N. May 22 '18 at 20:08
2

@MarinNedea the colon `:` after the username protects against substring matching by matching with the field separator, so the effect you describe won't happen. – Stéphane Gourichon May 27 '18 at 19:01
1

`/etc/passwd` depends on `nsswitch.conf` settings – Gert van den Berg Oct 06 '20 at 10:05

Daniel Sokolowski · Answer 4 · 2018-08-14T19:52:11.930

This is what I ended up doing in a Freeswitch bash startup script:

# Check if user exists
if ! id -u $FS_USER > /dev/null 2>&1; then
    echo "The user does not exist; execute below commands to crate and try again:"
    echo "  root@sh1:~# adduser --home /usr/local/freeswitch/ --shell /bin/false --no-create-home --ingroup daemon --disabled-password --disabled-login $FS_USER"
    echo "  ..."
    echo "  root@sh1:~# chown freeswitch:daemon /usr/local/freeswitch/ -R"
    exit 1
fi

Fleshgrinder · Answer 5 · 2022-02-20T09:56:52.967

24

By far the simplest solution:

if id -u "$user" >/dev/null 2>&1; then
    echo 'user exists'
else
    echo 'user missing'
fi

The >/dev/null 2>&1 can be shortened to &>/dev/null in Bash, and if you only want to know if a user does not exist:

if ! id -u "$user" >/dev/null 2>&1; then
    echo 'user missing'
fi

edited Feb 20 '22 at 09:56

answered Oct 30 '19 at 14:32

Fleshgrinder

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The best answer – MestreLion Aug 16 '20 at 03:36

score 14 · Answer 6 · answered May 09 '16 at 12:10

14

I suggest to use id command as it tests valid user existence wrt passwd file entry which is not necessary means the same:

if [ `id -u $USER_TO_CHECK 2>/dev/null || echo -1` -ge 0 ]; then 
echo FOUND
fi

Note: 0 is root uid.

answered May 09 '16 at 12:10

Valdas Rapševičius

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4
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score 13 · Answer 7 · edited Jun 15 '18 at 03:13

13

I was using it in that way:

if [ $(getent passwd $user) ] ; then
        echo user $user exists
else
        echo user $user doesn\'t exists
fi

edited Jun 15 '18 at 03:13

Craig S. Anderson

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answered Feb 24 '16 at 08:18

Takeda

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A bad approach. You can get rid of both `[ ]` and `$()` and simply check for `getent`'s _exit code_ directly instead of using its _output_ – MestreLion Aug 16 '20 at 03:31

score 7 · Answer 8 · answered Aug 17 '18 at 11:18

7

Script to Check whether Linux user exists or not

Script To check whether the user exists or not

#! /bin/bash
USER_NAME=bakul
cat /etc/passwd | grep ${USER_NAME} >/dev/null 2>&1
if [ $? -eq 0 ] ; then
    echo "User Exists"
else
    echo "User Not Found"
fi

answered Aug 17 '18 at 11:18

Bakul Gupta

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Why use `cat` ? `grep $USER_NAME /etc/passwd >/dev/null 2>&1` – Alexx Roche Dec 04 '18 at 13:31
Yes Alexx Roche,there is no need to use cat,I have use in order to have better way to visualise the concept for beginner. – Bakul Gupta Dec 05 '18 at 05:52
Unneccary use of cat, dependent on nsswitch only being configured to use files. – Gert van den Berg Oct 06 '20 at 10:07
This won't work in case the input username matches some other parts of each line in passwd. – Anutrix Dec 14 '22 at 11:37

score 6 · Answer 9 · answered Oct 01 '17 at 00:32

6

Late answer but finger also shows more information on user

  sudo apt-get finger 
  finger "$username"

answered Oct 01 '17 at 00:32

jackotonye

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OP might not be using Debian. – narendra-choudhary Mar 13 '20 at 08:45

score 5 · Answer 10 · answered Jul 11 '19 at 05:51

5

Using sed:

username="alice"
if [ `sed -n "/^$username/p" /etc/passwd` ]
then
    echo "User [$username] already exists"
else
    echo "User [$username] doesn't exist"
fi

answered Jul 11 '19 at 05:51

Stephane

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score 4 · Answer 11 · answered Feb 09 '15 at 05:15

Actually I cannot reproduce the problem. The script as written in the question works fine, except for the case where $1 is empty.

However, there is a problem in the script related to redirection of stderr. Although the two forms &> and >& exist, in your case you want to use >&. You already redirected stdout, that's why the form &> does not work. You can easily verify it this way:

getent /etc/passwd username >/dev/null 2&>1
ls

You will see a file named 1 in the current directory. You want to use 2>&1 instead, or use this:

getent /etc/passwd username &>/dev/null

This also redirects stdout and stderr to /dev/null.

Warning Redirecting stderr to /dev/null might not be such a good idea. When things go wrong, you will have no clue why.

score 4 · Answer 12 · answered Feb 19 '19 at 02:02

4

user infomation is stored in /etc/passwd, so you can use "grep 'usename' /etc/passwd" to check if the username exist. meanwhile you can use "id" shell command, it will print the user id and group id, if the user does not exist, it will print "no such user" message.

answered Feb 19 '19 at 02:02

Yu Chai

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4

IT is stored in /etc/passwd IF nsswitch.conf's passwd setting is set to files... This approach breaks if network-based logins are involved... – Gert van den Berg Oct 06 '20 at 10:09

score 3 · Answer 13 · answered Feb 11 '13 at 11:00

3

Depending on your shell implementation (e.g. Busybox vs. grown-up) the [ operator might start a process, changing $?.

Try

getent passwd $1 > /dev/null 2&>1
RES=$?

if [ $RES -eq 0 ]; then
    echo "yes the user exists"
else
    echo "No, the user does not exist"
fi

answered Feb 11 '13 at 11:00

Eugen Rieck

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changing `$?` gives me the same output. This is the bash version: `GNU bash, version 4.1.5(1)-release (i486-pc-linux-gnu)` – slayedbylucifer Feb 11 '13 at 11:17
Try `if test "$RES" == "0"; then ...` – Eugen Rieck Feb 11 '13 at 12:00
1

`$?` would be expanded before `[` runs, so that's not a problem. – chepner Feb 11 '13 at 14:10
2

Instead of testing for the result of the command, you can test the command directly: `if getent passwd $1 > /dev/null 2&>1; then echo "yes the user exists" else echo "No, the user does not exist" fi` – GMartinez Sep 20 '18 at 09:05

score 2 · Answer 14 · answered May 02 '17 at 10:13

2

Login to the server. grep "username" /etc/passwd This will display the user details if present.

answered May 02 '17 at 10:13

syam

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This will also match any username which contains "username". E.g. "some_username". You would want to do `if grep -q "^username:" /etc/passwd; then ...` – maedox Jun 12 '18 at 10:58

score 1 · Answer 15 · answered Oct 15 '19 at 07:40

Below is the script to check the OS distribution and create User if not exists and do nothing if user exists.

#!/bin/bash

# Detecting OS Ditribution
if [ -f /etc/os-release ]; then
    . /etc/os-release
    OS=$NAME
elif type lsb_release >/dev/null 2>&1; then
OS=$(lsb_release -si)
elif [ -f /etc/lsb-release ]; then
    . /etc/lsb-release
    OS=$DISTRIB_ID
else
    OS=$(uname -s)
fi

 echo "$OS"

 user=$(cat /etc/passwd | egrep -e ansible | awk -F ":" '{ print $1}')

 #Adding User based on The OS Distribution
 if [[ $OS = *"Red Hat"* ]] || [[ $OS = *"Amazon Linux"* ]] || [[ $OS = *"CentOS"*  
]] && [[ "$user" != "ansible" ]];then
 sudo useradd ansible

elif [ "$OS" =  Ubuntu ] && [ "$user" != "ansible" ]; then
sudo adduser --disabled-password --gecos "" ansible
else
  echo "$user is already exist on $OS"
 exit
fi

score 1 · Answer 16 · answered Jan 27 '20 at 16:33

1

Create system user some_user if it doesn't exist

if [[ $(getent passwd some_user) = "" ]]; then
    sudo adduser --no-create-home --force-badname --disabled-login --disabled-password --system some_user
fi

answered Jan 27 '20 at 16:33

luchaninov

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score 1 · Answer 17 · answered Jun 10 '20 at 07:09

I like this nice one line solution

getent passwd username > /dev/null 2&>1 && echo yes || echo no

and in script:

#!/bin/bash

if [ "$1" != "" ]; then
        getent passwd $1 > /dev/null 2&>1 && (echo yes; exit 0) || (echo no; exit 2)
else
        echo "missing username"
        exit -1
fi

use:

[mrfish@yoda ~]$ ./u_exists.sh root
yes
[mrfish@yoda ~]$ echo $?
0

[mrfish@yoda ~]$ ./u_exists.sh
missing username
[mrfish@yoda ~]$ echo $?
255

[mrfish@yoda ~]$ ./u_exists.sh aaa
no
[mrfish@indegy ~]$ echo $?
2

score 0 · Answer 18 · edited Sep 18 '21 at 17:55

0

echo "$PASSWORD" | su -c "cd /" "$USER"
if [ "$?" = "0" ];then
 echo "OK"
else
 echo "Error"
fi

edited Sep 18 '21 at 17:55

NuLo

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answered Sep 18 '21 at 10:33

wiktorm

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score 0 · Answer 19 · answered Feb 01 '22 at 11:39

0

#!/bin/bash
read -p "Enter your Login Name: " loginname
home=`grep -w $loginname /etc/passwd | cut -ef:6 -d:`
if [ $home ]
    echo "Exists"
else
    echo "Not Exist"
fi

answered Feb 01 '22 at 11:39

محمد صبرى فضه

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1
5

Check Whether a User Exists

19 Answers19

Script To check whether the user exists or not

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