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Linux三剑客企业级经典面试题解答实战

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标签:linux三剑客

Linux三剑客企业级经典面试题解答实战

 

说明:来自早晨老男孩教育在线班一期同学的面试题和学生给出的答案整理

面试题:请过滤oldboy.log中在device: {}里面出现了多少次oldboy,过滤并统计出来。

oldboy is a linuxer.

device: {

oo

oldboy

no sql

this is log

niu niu

}

oldboy

device: {

oldboy

no sql

this is log

niu niu

}

oldboy

device: {

oldboy

no sql

this is log

niu niu

}

device: {

oldboy

no sql

this is log

niu niu

}

 

解答:本题可利用sedawk取区间的用法:

1)利用数字行数取区间

[oldboy@oldboy ~]$ seq 10 >test.log      

[oldboy@oldboy ~]$ sed -n ‘2,5p‘ test.log

2

3

4

5

[oldboy@oldboy ~]$ awk ‘NR>1&&NR<6‘test.log              

2

3

4

5

2)利用字符串匹配取区间

本题可以取以"device"开头以 "}"结尾,然后将里面的符合要求的字符串过滤计数就可以了。

取区间的方法:

sed -n ‘/^device/,/\}$/p‘ oldboy.log

awk ‘/device: {/, /}/‘ oldboy.log

提示:

sed -n ‘/区间开始标识/,/区间结束标识/p‘ oldboy.log

 

3)本题完整答案:这里给3个方法

sed -n ‘/^device/,/\}$/p‘ oldboy.log|grep -w"oldboy"|wc -l

awk ‘/device: {/, /}/‘ oldboy.log|grep -w oldboy|wc -l

awk ‘/device: {/,/}/ {if($0=="oldboy") count++}END{print count}‘ oldboy.log

演示:

[oldboy@oldboy ~]$ sed -n ‘/^device/,/\}$/p‘oldboy.log|grep -w "oldboy"|wc -l                      

4

[oldboy@oldboy ~]$ awk ‘/device: {/,/}/ {if($0=="oldboy")count++} END{print count}‘ oldboy.log

4

[oldboy@oldboy ~]$ awk ‘/device: {/, /}/‘ oldboy.log|grep-w oldboy|wc -l                      

4

4)考虑到面试题的企业里的通用性,即一行还可能有多个oldboy,即原题改为如下:

[oldboy@oldboy ~]$ cat oldboy.log

oldboy is a linuxer.

device: {

oo

oldboy oldboy oldboy

no sql

this is log

niu niu

}

oldboy

device: {

oldboy oldboy

no sql

this is log

niu niu

}

oldboy oldboy

device: {

oldboy oldboy

no sql

this is log

niu niu

}

device: {

oldboy oldboy

no sql

this is log

niu niu

}

oldboy

因此本题最佳答案,老男孩老师从学生的解答中选举优秀答案如下:

sed -n ‘/^device/,/\}$/p‘ oldboy.log|xargs -n 1|sort|grep-w oldboy|wc -l

awk ‘/device: {/, /}/‘ oldboy.log|grep -w oldboy|tr" " "\n"|wc -l

awk ‘/{/,/}/{a+=gsub("oldboy","")}END{print a}‘ oldboy.log

sed -n ‘/{/,/}/{/oldboy/p}‘  oldboy.log | tr ‘ ‘ ‘\n‘ | wc -l

演示:

[oldboy@oldboy ~]$ sed -n ‘/^device/,/\}$/p‘oldboy.log|xargs -n 1|sort|grep -w oldboy|wc -l

9

[oldboy@oldboy ~]$ awk ‘/device: {/, /}/‘ oldboy.log|grep-w oldboy|tr " " "\n"|wc -l

9

[root@littleboy ~]# awk ‘/{/, /}/{a+=gsub("oldboy","")}END{printa}‘ oldboy.log

9

[root@littleboy ~]# sed -n ‘/{/,/}/{/oldboy/p}‘oldboy.log|tr ‘ ‘ ‘\n‘| wc -l

9

普通人员可以就题论题,高手的解答多考虑通用性,高手和低手的差别就在于此!


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本文出自 “老男孩linux培训” 博客,请务必保留此出处http://oldboy.blog.51cto.com/2561410/1908510

Linux三剑客企业级经典面试题解答实战

标签:linux三剑客

原文地址:http://oldboy.blog.51cto.com/2561410/1908510

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