使用 sed 打印每个匹配模式的第一行出现

Question

I would like to filter the output of the utility last based on a variable set of usernames.

This is sample output from last unfiltered,

reboot   system boot  server   Wed Apr  6 13:15 - 14:24  (01:09)    
user1    pts/0        server   Wed Apr  6 13:08 - 13:15  (00:06)    
reboot   system boot  system   Wed Apr  6 13:08 - 13:15  (00:06)    
user1    pts/0        server   Wed Apr  6 13:06 - down   (00:01)    
reboot   system boot  system   Wed Apr  6 13:06 - 13:07  (00:01)    
user1    pts/0        server   Wed Apr  6 12:59 - down   (00:06)  

What I would like to do is pipe the output of last to sed. Then, using sed I would print the first occurrence of each specified user name ie their last log entry in wtmp. The output should appear as so,

 reboot   system boot  server   Wed Apr  6 13:15 - 14:24  (01:09)    
 user1    pts/0        server   Wed Apr  6 13:08 - 13:15  (00:06)

The sed expression that I particularly like is,

last|sed '/user1/{p;q;}'

Unfortunately this only gives me the ability to match the first occurrence of one username. Using this syntax is there a way I could specify a multiple of usernames? Thanks in advance!

Answer 1

awk is better fit here than sed due to awk's ability to use associative arrays:

last | awk '!seen[$1]++'

reboot   system boot  server   Wed Apr  6 13:15 - 14:24  (01:09)
user1    pts/0        server   Wed Apr  6 13:08 - 13:15  (00:06)