第三题:有个文件,就是passwd.back:root:!:0:0::/:/usr/bin/kshroot:1:3:4:5:m:/proot:2:9:3:8:n:qdaemon:!:1:1::/etc:bin:!:2:2::/bin:sys:!:3:3::/usr/sys:adm:!:4:4::/var/adm:uucp:!:5:5::/usr/lib/uucp:guest:!:100:100::/home0/guest:nobody:!:4294967294:4294967294:...
显示全部第三题:
有个文件,就是passwd.back:
root:!:0:0::/:/usr/bin/ksh
root:1:3:4:5:m:/p
root:2:9:3:8:n:q
daemon:!:1:1::/etc:
bin:!:2:2::/bin:
sys:!:3:3::/usr/sys:
adm:!:4:4::/var/adm:
uucp:!:5:5::/usr/lib/uucp:
guest:!:100:100::/home0/guest:
nobody:!:4294967294:4294967294::/:
lpd:!:9:4294967294::/:
lp:*:11:11::/var/spool/lp:/bin/false
invscout:*:6:12::/var/adm/invscout:/usr/bin/ksh
snapp:*:200:13:snapp login user:/usr/sbin/snapp:/usr/sbin/snappd
ipsec:*:201:1::/etc/ipsec:/usr/bin/ksh
nuucp:*:7:5:uucp login user:/var/spool/uucppublic:/usr/sbin/uucp/uucico
esaadmin:*:811:0::/home0/esaadmin:/usr/bin/ksh
jiang:!:202:0::/home0/jiang:/bin/bash
sshd:*:203:201::/var/empty:/usr/bin/ksh
ldap:*:204:1::/home0/ldap:/usr/bin/ksh
用grep命令列出uid在200-299之间的所有用户。
Answer:
$ cat passwd.back |grep "2[0-9][0-9]:"
(没有测试,自己试一下。不能写成$ cat passwd.back |grep "2[0-99]:" //会把uid为2位(20~29)的也会抽取出来。)
To omit ":" is done in "2[0-9][0-9]:" , the ":" is used for matching accurately.
":" is a separation in the file named passwd.back.
"2[0-9][0-9]:" also can be done.
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