帮助使用命令行来提取stdout上的数据片段 [英] help using command line to extract snippets of data on stdout

问题描述

我想要提取以下字符串/数据的选项：

  / work / foo / processed / 25 
 / work / foo / processed / myproxy 
 / work / foo / processed / sample 
  

=或=

  25 
 myproxy 
示例
  
 
 
 
但是如果我看到这两个会有帮助。
 
 
 
 c> cut 或perl或任何其他工作方式：
 找到3项
 drwxr-xr-x  -  foo_hd foo_users 0 2011-03-16 18:46 / work / foo / processed / 25 
 drwxr-xr-x  -  foo_hd foo_users 0 2011-04-05 07:10 / work / foo / processed / myproxy 
 drwxr-x ---  -  foo_hd testcont 0 2011-04-08 07:19 / work / foo / processed / sample 
  
执行 cut -d-f6 会得到 foo_users ， testcont 。我尝试将字段增加到更高的值，我只是不能得到我想要的。
 
 
 
我不知道 / code>是有益于这个或像perl的东西？ 
基本目录将保持静态 / work / foo / processed 。
 
 
 
第一行找到Xn项已删除。 
 
          code><您的命令> | grep -P -o'[\ / \.\w] + $'
 
如果'/ work / foo / processed'目录总是静态的，那么：
 
< Your Command> | grep -P -o'\ / work\ / foo\ / processed \ /.+$'
 
 -o：仅显示与PATTERN匹配的匹配线的部分。 
 -P：将PATTERN解释为Perl正则表达式。 
  

在此示例中，输入中的最后一个字符将匹配。
（字也可以包含点），因此文件名，如'text_file1.txt'，可以匹配）。
当然，你可以根据你的要求改变模式。

I would like the option of extracting the following string/data:

/work/foo/processed/25
/work/foo/processed/myproxy
/work/foo/processed/sample

=or=

25
myproxy
sample

But it would help if I see both.

From this output using cut or perl or anything else that would work:

Found 3 items
drwxr-xr-x   - foo_hd foo_users          0 2011-03-16 18:46 /work/foo/processed/25
drwxr-xr-x   - foo_hd foo_users          0 2011-04-05 07:10 /work/foo/processed/myproxy
drwxr-x---   - foo_hd testcont           0 2011-04-08 07:19 /work/foo/processed/sample

Doing a cut -d" " -f6 will get me foo_users, testcont. I tried increasing the field to higher values and I'm just not able to get what I want.

I'm not sure if cut is good for this or something like perl? The base directories will remain static /work/foo/processed.

Also, I need the first line Found Xn items removed. Thanks.

解决方案

Try this out :

<Your Command> | grep -P -o '[\/\.\w]+$' 

OR if the directory '/work/foo/processed' is always static then:

<Your Command>| grep -P -o '\/work\/foo\/processed\/.+$' 

-o : Show only the part of a matching line that matches PATTERN.
-P : Interpret PATTERN as a Perl regular expression.

In this example, the last word in the input will be matched . (The word can also contain dot(s)),so file names like 'text_file1.txt', can be matched). Ofcourse, you can change the pattern, as per your requirement.

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