Shell script keeps executing even if was true

I have a simple shell script to check if the user exist, if not, add that user

#!/usr/bin/bash
echo "Enter user"
read -r user
if [ "$(id -u "$user" &>/dev/null)" = 1 ]; then
  echo "user existed"
else
  useradd "$user"
  echo "add password"
  passwd "$user"
  echo "$user" &>>text.txt
  echo "saved"
fi

But it does not display the "echo user exist". instead it displays like so:

Enter user

abcdef

useradd: user 'abcdef' already exists

add password

New password: 

In this case abcdef is an existing user

Please let me know what goes wrong here, thanks a lot.

The following condition in your script:

if [ "$(id -u "$user" &>/dev/null)" = 1 ]; then

will never be true. It would be true only if the output of command id -u "$user" &>/dev/null would be the single digit 1, but as any output from that command is redirected to /dev/null, this will never happen.

Replace that line with

if { id -u "$user" &>/dev/null ; } then

This condition checks the exit status of the command (and not its output like $(...) does) and will be true if the command id -u "$user" &>/dev/null succeeds (that is, user exists) and false if it fails (that is, user doesn't exist).

The answer by @raj is 100% correct for simplifying code.

These are just some alternate ways of using the exit codes and maybe shedding a little more light in how they can be used.

The exit code can be checked by doing an echo $? right after your command of id -u $user &>/dev/null which doesn't need to be encased in $(..) or anything like that.

Just try id -u <username> &>/dev/null and then type in echo $? to check the exit code.

Example:

terrance@terrance-ubuntu:~$ id -u terrance &>/dev/null
terrance@terrance-ubuntu:~$ echo $?
0
terrance@terrance-ubuntu:~$ id -u mark &>/dev/null
terrance@terrance-ubuntu:~$ echo $?
1

In the above example, the user mark does not exist on my computer thus the exit status of 1.

Using the exit code you can put it into a case statement instead of an if..else statement.

#!/bin/bash
echo "Enter username"
read -r user
id -u "$user" &>/dev/null
case $? in
    0) echo "User exists";;
    1) adduser $user
       echo "add password"
       passwd $user
       echo $user >> text.txt
       echo "saved";;
esac

Interestingly enough, using an if [ ]; then can still be used as well by adding an echo $? into your if check, but I did notice that you were checking for 1 to be if exist, but 1 means that the code exited with 1 meaning not found.

Adding it will work like this as well.

#!/bin/bash
echo "Enter username"
read -r user
if [ $(id -u "$user" &>/dev/null; echo $?) == 0 ]; then
    echo "User exists"
else
    adduser $user
    echo "add password"
    passwd $user
    echo $user >> text.txt
    echo "saved"
fi