将grep的-v选项与-A选项结合使用 [英] Combine -v option for grep with -A option

问题描述

我想问是否可以结合-v和-A？

我有示例文件：

  abc 
 1 
 2 
 3 
 ACB 
 def 
 abc 
 1 
 2 
 3 
 ABC 
 xyz 
  

与-AI可以看到我想要剪切的部分：

  $ grep abc -A 4 grep_v_test.txt 
 abc 
 1 
 2 
 3 
 ACB 
  -  
 abc 
 1 
 2 
 3 
 ABC 
  

它有一些选项可以指定某些内容仅用于查看

  def 
 xyz 
  

？

我发现这个答案 - 在grep中结合-v标志和-A标志但它不适用于我，我试过了

  $ sed -e/ abc / {2; 2; d}grep_v_test.txt 
 sed：-e表达式＃1，字符8：未知命令：`;'
  

也



  $ sed/ abc / 2dgrep_v_test.txt 
 sed：-e表达式＃1，字符6：未知命令：`2'
  

或

  $ sed/ abc / + 2dgrep_v_test.txt 
 sed：-e表达式＃1，字符6：未知命令：`+'
  



Sed版本是：



  $ sed --version 
 GNU sed版本4.2.1 
  

edit1：
根据评论我对这两种解决方案都进行了一些尝试，但是它不起作用，因为我想为 grep -v -A 1 abc


c $ c>我希望行abc和1被删除，但其余的将被打印 awk'c&&！ - c; / abc / {c = 2}'grep_v_test.txt 只打印包含2的行，这不是我想要的。



非常相似它与sed

  $ sed -n'/ abc / {n; n; p}'grep_v_test.txt 
 2 
 2 
  

edit2：
它看来，我无法正确描述它，让我再试一次。

$ b $ <什么 grep -AN abc文件确实是在abc后面打印N行。我想删除 grep -A 会显示的内容，所以在一个文件中

  abc 
 1 
 2 
 3 
 ACB 
 def 
 DEF 
 abc 
 1 
 2 
 3 
 ABC 
 xyz 
 XYZ 
  

我会只需将abc部分删除到ABC，然后打印剩下的部分：

abc

1

2

3

ACB

def

DEF

abc

1

2

3

ABC

xyz

XYZ

... awk解决方案只打印def和xyz并跳过DEF和XYZ ......

解决方案

跳过5行上下文从最初的匹配行开始：

  $ awk'/ abc / {c = 5} c&& c-- {next} 1'文件
 def 
 xyz 
  

请参阅为匹配模式提取第N行为其他相关脚本。

wrt下面的评论，这里是这个答案和 @ fedorqui的答案之间的区别：

  $ cat file 
现在是我们不满的冬季
 
 abc 
 1 
 2 
 bar 
 
 $ awk'/ abc / {c = 3} c&& c  -  {next} 1'文件
现在是我们不满$ b的冬季
 $ b bar 
 
 $ awk'/ abc / {c = 0} c ++> 2'文件
 bar 
  

看看@ fedorqui的脚本如何无条件地跳过文件的前两行？

I'd like to ask is it possible to combine somehow -v with -A?

I have example file:

abc
1
2
3
ACB
def
abc
1
2
3
ABC
xyz

with -A I can see the parts I want to "cut":

$ grep abc -A 4 grep_v_test.txt
abc
1
2
3
ACB
--
abc
1
2
3
ABC

it there some option to specify something to see only

def
xyz

?

I found this answer - Combining -v flag and -A flag in grep but it is not working for me, I tried

$ sed -e "/abc/{2;2;d}" grep_v_test.txt
sed: -e expression #1, char 8: unknown command: `;'

also

$ sed "/abc/2d" grep_v_test.txt
sed: -e expression #1, char 6: unknown command: `2'

or

$ sed "/abc/+2d" grep_v_test.txt
sed: -e expression #1, char 6: unknown command: `+'

Sed version is:

$ sed --version
GNU sed version 4.2.1

edit1: Based on comment I experimented a little bit with both solution, but it is not working as I want to

for grep -v -A 1 abc I would expect line abc and 1 to be removed, but the rest will be printed awk 'c&&!--c; /abc/ {c=2}' grep_v_test.txt prints just the line containing 2, which is not what I wanted.

Very similar it is with sed

$ sed -n '/abc/{n;n;p}' grep_v_test.txt
2
2

edit2: It seems, I'm not able to describe it properly, let me try again.

What grep -A N abc file does is to print N lines after abc. I want to remove what grep -A will show, so in a file

abc
1
2
3
ACB
def
DEF
abc
1
2
3
ABC
xyz
XYZ

I'll just remove the part abc to ABC and I'll print the rest:

abc
1
2
3
ACB
def
DEF
abc
1
2
3
ABC
xyz
XYZ

so 4 lines will remain... The awk solution prints just def and xyz and skips DEF and XYZ...

解决方案

To skip 5 lines of context starting with the initial matching line is:

$ awk '/abc/{c=5} c&&c--{next} 1' file
def
xyz

See Extract Nth line after matching pattern for other related scripts.

wrt the comments below, here's the difference between this answer and @fedorqui's answer:

$ cat file
now is the Winter
of our discontent
abc
1
2
bar

$ awk '/abc/{c=3} c&&c--{next} 1' file
now is the Winter
of our discontent
bar

$ awk '/abc/ {c=0} c++>2' file
bar

See how the @fedorqui's script unconditionally skips the first 2 lines of the file?

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