Bash 脚本 - 检查用户是否登录
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Bash Script - Check if user is logged in or not
提问by LinuxN00b
I'm trying to write a script that checks if the user (that is sent as an argument) is logged in or not. I want the answer to return 0 to the shell if the user is logged in and 1 to the shell if the user is not logged in.
我正在尝试编写一个脚本来检查用户（作为参数发送）是否已登录。如果用户已登录，我希望答案将 0 返回到 shell，如果用户未登录，则将 1 返回到 shell。
But I have run into some problem. I get the message "You need to enter a user" everytime i try to run the script even if I send a user as an argument.
但我遇到了一些问题。每次尝试运行脚本时，我都会收到消息“您需要输入用户”，即使我将用户作为参数发送。
#!/bin/bash
function checkUser
{
status=0
for u in $(who | awk '{print }' | sort | uniq)
do
if [ "$u" = "" ]; then
status=1
fi
done
if [ "$status" = "1" ]; then
echo "$user is logged in."
exit 0
else
echo "$user is not logged in."
exit 1
fi
}
if [[ -eq 0 ]] ; then
echo 'You need to enter a user'
read u
else
user=
fi
采纳答案by mf_starboi_8041
You have to provide following to check whether user has provide argument or not.
您必须提供以下内容来检查用户是否提供了参数。
if [ $# -eq 0 ] ; then
$#used for number of argument provided in command line
$#用于命令行中提供的参数数量
Your full code should look like this,
你的完整代码应该是这样的，
#!/bin/bash
function checkUser {
status=0
for u in $(who | awk '{print }' | sort | uniq)
do
if [ "$u" == "" ]; then
return 0
fi
done
return 1
}
if [ $# -eq 0 ] ; then
echo 'You need to enter a user'
read user
checkUser $user
ret_val=$?
else
user=
checkUser $user
ret_val=$?
fi
if [ $ret_val -eq 0 ]; then
echo "User Logged In"
exit 0
else
echo "User Not Logged In"
exit 1
fi
回答by Ville Oikarinen
The if [[ $1 -eq 0 ]]check is outside the function body. That's why it doesn't have access to the parameter $1of the function invocation. Also, you are not even calling the function anywhere.
该if [[ $1 -eq 0 ]]检查是在函数体外部。这就是为什么它无权访问$1函数调用的参数。此外，您甚至没有在任何地方调用该函数。
Try this:
尝试这个：
#!/bin/bash
set -eu
[ $# == 1 ] || {
echo "Usage: ##代码## USERNAME"
exit 1
}
USER=
who | awk '{print }' | sort | uniq | grep -q "^$USER$" && {
echo "$USER is logged in"
exit 0
} || {
echo "$USER is not logged in"
exit 1
}
UPDATE: exit for correct exit code
更新：退出以获得正确的退出代码
