排序一个网站的访问统计，并按照域名进行倒序排列，如：
http://www.selinux.com/a.html http://bbs.selinux.com/b.html http://nj.selinux.com/a.html http://www.selinux.com/ab.html http://sh.selinux.com/a.html http://bbs.selinux.com/a.html http://www.selinux.com/a.html http://www.selinux.com/a.html http://sh.selinux.com/a.html http://www.selinux.com/123.html http://book.bbs.selinux.com/a.html http://www.selinux.com/b.html
刚稍微写了简单的排序方法：
[root@fc16 ~]# time awk -F selinux '$1 ~/^http:\/\// {print $1}' test_word.txt |awk -F "//" '{print $2}'|sort|uniq -c |sort -rn|awk '{print $1" http://"$2"selinux.com"}'
6 http://www.selinux.com
2 http://sh.selinux.com
2 http://bbs.selinux.com
1 http://nj.selinux.com
1 http://book.bbs.selinux.com
real 0m0.006s
user 0m0.001s
sys 0m0.010s
又感觉这样太麻烦，改了改：
[root@fc16 ~]# time awk -F '/' '$1 ~/^http/ {print $(NF-1)}' test_word.txt |sort |uniq -c|sort -rn
6 www.selinux.com
2 sh.selinux.com
2 bbs.selinux.com
1 nj.selinux.com
1 book.bbs.selinux.com
real 0m0.007s
user 0m0.001s
sys 0m0.010s
