我如何在每一条匹配的线周围显示前面和后面的5条线?
当前回答
让我们用一个例子来理解。我们可以使用带有选项的grep:
-A 5 # this will give you 5 lines after searched string.
-B 5 # this will give you 5 lines before searched string.
-C 5 # this will give you 5 lines before & after searched string
实例File.txt包含6行,以下是操作。
[abc@xyz]~/% cat file.txt # print all file data
this is first line
this is 2nd line
this is 3rd line
this is 4th line
this is 5th line
this is 6th line
[abc@xyz]~% grep "3rd" file.txt # we are searching for keyword '3rd' in the file
this is 3rd line
[abc@xyz]~% grep -A 2 "3rd" file.txt # print 2 lines after finding the searched string
this is 3rd line
this is 4th line
this is 5th line
[abc@xyz]~% grep -B 2 "3rd" file.txt # Print 2 lines before the search string.
this is first line
this is 2nd line
this is 3rd line
[abc@xyz]~% grep -C 2 "3rd" file.txt # print 2 line before and 2 line after the searched string
this is first line
this is 2nd line
this is 3rd line
this is 4th line
this is 5th line
记住选项的技巧:选项:
-一个 -> 之后的平均值-B级 -> B表示之前。-c类 -> 介于两者之间
其他回答
-A和-B将起作用,-C n(对于n行上下文)或-n(对于n个上下文……只要n是1到9)也会起作用。
Grep有一个名为Context Line Control的选项,您可以使用--Context,
| grep -C 5
or
| grep -5
应该会成功的
这是awk中的@Ygor解决方案
awk 'c-->0;$0~s{if(b)for(c=b+1;c>1;c--)print r[(NR-c+1)%b];print;c=a}b{r[NR%b]=$0}' b=3 a=3 s="pattern" myfile
注意:用前后的行数替换a和b变量。
它对于不支持grep的-A、-B和-C参数的系统特别有用。
使用grep
$ grep --help | grep -i context
Context control:
-B, --before-context=NUM print NUM lines of leading context
-A, --after-context=NUM print NUM lines of trailing context
-C, --context=NUM print NUM lines of output context
-NUM same as --context=NUM
对于BSD或GNUgrep,可以使用-B num设置匹配前的行数,使用-A num设置匹配后的行数。
grep -B 3 -A 2 foo README.txt
如果希望前后行数相同,可以使用-C num。
grep -C 3 foo README.txt
这将显示前3行和后3行。
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