如何使用 awk 处理最后一个命令的结果？

Question

I have to count how many users were logged in each day of the week. I can get the data with the last command on the server, but how can I process it with awk? It is a long list of data, the name of the day is in the 4th column. I declared an array like this: t["Sun"]=0; t["Sat"]=0; etc, I can count it from a simple input from the terminal, but how can I use it on the output of last?

If I log in the server and type: last, I get a result like this, each row containing the name of the user, pts/something, ip address, day of the week, time etc:

user pts/0 address Mon etc.

user pts/1 address Wed etc.

I'd like an output like this:

Mon: 10 users were logged in

Tue: 20 user were logged in etc

My code is here, which works with an input from terminal:

awk '
        BEGIN { t["Sun"]=0; t["Sat"]=0; t["Fri"]=0; t["Thu"]=0; t["Wed"]=0; t["Tue"]=0; t["Mon"]=0; }
        { t[substr($4,i,3)]++;} 
        END {for(i in t) print i ": " t[i]}   
'$*

So far to an input in terminal like this:

aa ss dd Fri
a d g Mon
q w e Fri

gives the result:

Wed: 0
Tue: 0
Fri: 2
Thu: 0
Sat: 0
Sun: 0
Mon: 1

How can I use the awk command to process the output of last?

Answer 1

It sounds like you might want something like:

last |
awk '
    BEGIN { split("Sun Mon Tue Wed Thu Fri Sat",days) }
    { numUsers[$4]++ }
    END {
        for (dayNr=1; dayNr in days; dayNr++) {
            dayName = days[dayNr]
            printf "%s: %d users were logged in\n", dayName, numUsers[dayName]
        }
    }
'

so you always get output for every day, including days when no-one logged in, and the days are sorted in the weekday order you prefer.