How to check if running as root in a bash script

Question

I\'m writing a script that requires root level permissions, and I want to make it so that if the script is not run as root, it simply echoes \"Please run as root.\" and exit

Answer 1

In this answer, let it be clear, I presume the reader is able to read bash and POSIX shell scripts like dash.

I believe there is not much to explain here since the highly voted answers do a good job of explaining much of it.

Yet, if there is anything to explain further, don't hesitate to comment, I will do my best filling the gaps.

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Optimized all-round solution for performance and reliability; all shells compatible

New solution:

# bool function to test if the user is root or not
is_user_root () { [ ${EUID:-$(id -u)} -eq 0 ]; }

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Benchmark (save to file is_user_root__benchmark)

#+------------------------------------------------------------------------------+
#|                           is_user_root() benchmark                           |
#|                  "Bash is fast while Dash is slow in this"                   |
#|                          Language: POSIX shell script                        |
#|                        Copyright: 2020 Vlastimil Burian                      |
#|                      M@il: info[..]vlastimilburian[..]cz                     |
#|                               License: GPL 3.0                               |
#|                                 Version: 1.1                                 |
#+------------------------------------------------------------------------------+

readonly iterations=10000

# intentionally, the file does not have executable bit, nor it has no shebang
# to use it, just call the file directly with your shell interpreter like:

# bash is_user_root__benchmark
# dash is_user_root__benchmark

is_user_root () { [ ${EUID:-$(id -u)} -eq 0 ]; }

print_time   () { date +"%T.%2N"; }
print_start  () { printf '%s' 'Start  : '; print_time; }
print_finish () { printf '%s' 'Finish : '; print_time; }

printf '%s\n' '___is_user_root()___'; print_start
                   
i=1; while [ $i -lt $iterations ]; do
    is_user_root
    i=$((i + 1))
done; print_finish

Examples of use and duration:

$ dash is_user_root__benchmark 
___is_user_root()___
Start  : 03:14:04.81
Finish : 03:14:13.29

$ bash is_user_root__benchmark 
___is_user_root()___
Start  : 03:16:22.90
Finish : 03:16:23.08

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Explanation

Since it is multitude times faster to read the $EUID standard bash variable, the effective user ID number, than executing id -u command to POSIX-ly find the user ID, this solution combines both into a nicely packed function. If, and only if, the $EUID is for any reason not available, the id -u command will get executed, ensuring we get the proper return value no matter the circumstances.

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Why I post this solution after so many years the OP has asked

Well, if I see correctly, there does seem to be a missing piece of code above.

You see, there are many variables which have to be taken into account, and one of them is combining performance and reliability.

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Portable POSIX solution + Example of usage of the above function

#!/bin/sh

# bool function to test if the user is root or not (POSIX only)
is_user_root() { [ "$(id -u)" -eq 0 ]; }

if is_user_root; then
    echo 'You are the almighty root!'
    exit 0 # implicit, here it serves the purpose to be explicit for the reader
else
    echo 'You are just an ordinary user.' >&2
    exit 1
fi

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Conclusion

As much as you possibly don't like it, the Unix / Linux environment has diversified a lot. Meaning there are people who like bash so much, they don't even think of portability (POSIX shells). Others like me prefer the POSIX shells. It is nowadays a matter of personal choice and needs.