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[英]how to print stuff in a loop to console, so that unix grep can interact with it?
[英]How to grep a file then print it out? Unix
while true
do
if($update)
then
who | awk {'print$1'} > first_user_list #store original user list
update=false
fi
who | awk {'print$1'} > updated_user_list
(diff first_user_list updated_user_list) | cut -c 3- > in_out_list
inOutVar='cat in_out_list' #<----here's my problem
length_first=$(wc -l < updated_user_list)
length_update=$(wc -l < first_user_list)
if [[ "$length_first" -lt "$length_update" ]]; then
echo -e "$inOutVar" " has logged out"
update=true
elif [ "$length_first" -gt "$length_update" ]; then
echo -e "$inOutVar" " has logged in"
update=true
else
echo No user has logged in/out in the last 3 seconds
fi
sleep 3
done
我将如何打印出已注销然后“已注销”的用户名,例如。
“约翰史密斯已注销”
对于Unix来说还很陌生,任何帮助或建议都很好,在此先感谢:)x
您的问题是您使用引号而不是反引号。 由于将变量设置为等于命令,因此需要使用``或$():
#!/bin/sh
while true; do
if($update)
then
who | awk {'print$1'} > first_user_list #store original user list
update=false
fi
who | awk {'print$1'} > updated_user_list
(diff first_user_list updated_user_list) | cut -c 3- > in_out_list
inOutVar=`cat in_out_list` ## use `` or $(), not ''
length_first=$(wc -l < updated_user_list)
length_update=$(wc -l < first_user_list)
if [[ "$length_first" -lt "$length_update" ]]; then
echo -e "$inOutVar" " has logged out"
update=true
elif [ "$length_first" -gt "$length_update" ]; then
echo -e "$inOutVar" " has logged in"
update=true
else
echo No user has logged in/out in the last 3 seconds
fi
sleep 3
这适用于我的系统。
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