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
我采用紧凑的方式:
grep -5 string file
这相当于:
grep -A 5 -B 5 string file
其他回答
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
如果您经常搜索代码,AG银搜索器比grep更高效(即更快)。
使用-C选项显示上下文行。
Eg:
ag -C 3 "foo" myFile
line 1
line 2
line 3
line that has "foo"
line 5
line 6
line 7
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
在/some/file.txt中搜索“17655”,显示前后10行上下文(使用Awk),输出前有行号,后有冒号。当grep不支持-[ACB]选项时,在Solaris上使用此选项。
awk '
/17655/ {
for (i = (b + 1) % 10; i != b; i = (i + 1) % 10) {
print before[i]
}
print (NR ":" ($0))
a = 10
}
a-- > 0 {
print (NR ":" ($0))
}
{
before[b] = (NR ":" ($0))
b = (b + 1) % 10
}' /some/file.txt;
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