某日在网上溜达，看见一网友博客（http://rainbird.blog.51cto.com/211214/165779）上的两道题，作者说没有两年linux道行,很难做出其中三道。这里我们看看这些题究竟有多难（仅针对脚本编程）。
第一道：
- 取出/etc/passwd中shell出现的次数?
顺便贴一下我的/etc/passwd文件
- root:x:0:0:root:/root:/bin/bash?
- bin:x:1:1:bin:/bin:/sbin/nologin?
- daemon:x:2:2:daemon:/sbin:/sbin/nologin?
- adm:x:3:4:adm:/var/adm:/sbin/nologin?
- lp:x:4:7:lp:/var/spool/lpd:/sbin/nologin?
- sync:x:5:0:sync:/sbin:/bin/sync?
- shutdown:x:6:0:shutdown:/sbin:/sbin/shutdown?
- halt:x:7:0:halt:/sbin:/sbin/halt?
- mail:x:8:12:mail:/var/spool/mail:/sbin/nologin?
- news:x:9:13:news:/etc/news:?
- uucp:x:10:14:uucp:/var/spool/uucp:/sbin/nologin?
- operator:x:11:0:operator:/root:/sbin/nologin?
- games:x:12:100:games:/usr/games:/sbin/nologin?
- gopher:x:13:30:gopher:/var/gopher:/sbin/nologin?
- ftp:x:14:50:FTP?User:/var/ftp:/sbin/nologin?
- nobody:x:99:99:Nobody:/:/sbin/nologin?
- nscd:x:28:28:NSCD?Daemon:/:/sbin/nologin?
- vcsa:x:69:69:virtual?console?memory?owner:/dev:/sbin/nologin?
- rpc:x:32:32:Portmapper?RPC?user:/:/sbin/nologin?
- mailnull:x:47:47::/var/spool/mqueue:/sbin/nologin?
- smmsp:x:51:51::/var/spool/mqueue:/sbin/nologin?
- oprofile:x:16:16:Special?user?account?to?be?used?by?OProfile:/home/oprofile:/sbin/nologin?
- pcap:x:77:77::/var/arpwatch:/sbin/nologin?
- dbus:x:81:81:System?message?bus:/:/sbin/nologin?
- avahi:x:70:70:Avahi?daemon:/:/sbin/nologin?
- sshd:x:74:74:Privilege-separated?SSH:/var/empty/sshd:/sbin/nologin?
- rpcuser:x:29:29:RPC?Service?User:/var/lib/nfs:/sbin/nologin?
- nfsnobody:x:4294967294:4294967294:Anonymous?NFS?User:/var/lib/nfs:/sbin/nologin?
- haldaemon:x:68:68:HAL?daemon:/:/sbin/nologin?
- avahi-autoipd:x:100:156:avahi-autoipd:/var/lib/avahi-autoipd:/sbin/nologin?
- xfs:x:43:43:X?Font?Server:/etc/X11/fs:/sbin/nologin?
- ntp:x:38:38::/etc/ntp:/sbin/nologin?
这道题真的很简单，要知道处理文本那可是Perl的强项。熟悉perl的童鞋，一个hash就能搞定。
- #!/usr/bin/perl?-w?
- ?
- use?strict;?
- ?
- my?%hash;?
- ?
- open?my?$file,'<','/etc/passwd'?or?die?"$!n";?
- ?
- while?(?<$file>?)?{?
- ???????chomp;?
- ???????my?@shell?=?split?/:/;?
- ?
- ???????if?(?$shell[6]?)?{?
- ??????????if?(?exists?$hash{$shell[6]}?)?{?
- ??????????????$hash{$shell[6]}++;?
- ??????????}?else?{?
- ??????????????$hash{$shell[6]}?=?1;?
- ??????????}?
- ??????}?
- ?
- }?
- ?
- close?$file;?
- ?
- while?(?my?(?$key,$val?)?=?each?%hash?)?{?
- ???????print?"$keyt$valn";?
- }?
output:
- [root@Xen?~]#?perl?c.pl?
- /sbin/halt??????1?
- /bin/sync???????1?
- /sbin/nologin???27?
- /sbin/shutdown??1?
- /bin/bash???????1?
第二道：
employee文件中记录了工号和姓名
- 100?Jason?Smith?
- 200?John?Doe?
- 300?Sanjay?Gupta?
- 400?Ashok?Sharma??
bonus文件中记录工号和工资
- 100?$5,000?
- 200?$500?
- 300?$3,000?
- 400?$1,250??
要求把两个文件合并并输出如下
处理结果:
- 400?ashok?sharma?$1,250?
- 100?jason?smith??$5,000?
- 200?john?doe??$500?
- 300?sanjay?gupta??$3,000?
这道题也不是特别的难，两个hash就可以搞定的
- #!/usr/bin/perl?-w?
- ?
- use?strict;?
- ?
- my?(%A,%B);?
- ?
- open?my?$file,'employee.txt'?or?die?"$!n";?
- ?
- while?(?<$file>?)?{?
- ???????chomp;?
- ???????/^(d+)s+w+s+w+/;?
- ???????$A{$1}?=?$_?if?$1?;?
- }?
- ?
- close?$file;?
- ?
- open?$file,'bonus.txt'?or?die?"$!n";?
- ?
- while?(?<$file>?)?{?
- ???????chomp;?
- ???????/^(d+)s+($S+)/;?
- ???????$B{$1}?=?$2?if?(?$1?||?$2?);?
- }?
- ?
- close?$file;?
- ?
- for?my?$num?( keys?%A?)?{?
- ?
- ????if?(?exists?$B{$num}?)?{?
- ????????$A{$num}?.=?"t$B{$num}";?
- ????}?
- ?
- }?
- ?
- print?join?"n",sort values?%A;?
- print qq/n/;?
output:
- [root@Xen?~]#?perl?hebin.pl?
- 100?Jason?Smith?$5,000?
- 200?John?Doe????$500?
- 300?Sanjay?Gupta????????$3,000?
- 400?Ashok?Sharma????????$1,250?
怎么样，看完上面的代码，您有学习perl的冲动了没？
