Bash - 在perl regex中使用变量和匹配组 [英] Bash - use variable in perl regex together with matching groups

问题描述

这是我在stackoverflow上的第一篇文章，请原谅我，如果我错过了一些重要的东西。
我目前坚持下面的问题。我们的目标是根据我使用 find 准备的文件列表来动态替换端口号。这些文件中的所有端口都以数字4开始，并有5位数字。

现在，棘手的部分，我只替换了数字＃2和＃3例如：

 文件中的旧端口：40380，40381 
新端口in file：41580，40381 
  

我正在使用Sun Solaris 5.10，因此我更喜欢perl for inline replacements

最后一个关键问题是：如何将$ 1（group 1）+ $ PIN_PINNO + $ 3（group 3）组合起来，从而得到如下结果：4 15 80

  NEW_PINNO = 15 
 LOGI = $ HOME / filelist.txt 
 $ b $ $ cat $ LOGI`文件中的b＃端口替换
 
 do 
 perl -pe's / [\：\> \ =] \s *（4） （\d {2}）（\d {2}）\b / $ 1 $ {NEW_PINNO} $ 3 / g'$ file 
 
   
 $ b很多在此先谢谢了解决方案
 
 perl -pse's / [：> =] \s * \K（\ d）\d \d（\d\d）\ b / $ 1 $ pin $ 2 / gx' -  -pin =$ new_pinnofile 
  

• 您的正则表达式将匹配[冒号，大于等号]和空格，但不包括它们在替换中。我正在使用 \K 指令来匹配那些字符，但是忘了它们（ref： http://perldoc.perl.org/perlre.html#Lookaround-Assertions ）


• 使用 -s 选项启用基本开关解析，将shell变量传递到perl而不用引用游戏。 （ref： http://perldoc.perl.org/perlrun.html ）

测试

  new_pinno = 15 
 perl -pse's / [：> =] \s * \K（\ d）\d\d（\d\d）\b / $ 1 $ pin $ 2 / gx ' -  -pin =$ new_pinno<< END 
 var1 = 40380 
 var2 = 40381 
 END 
  

  var1 = 41580 
 var2 = 41581 
  

注意

• 你不应该在shell中使用ALL_CAPS_VARNAMES ，留下那些由shell保留。有一天，你会使用 PATH = something ，然后想知道为什么你的脚本坏了。
  

• 和@ 123的评论已验证。这是从文件读取行的安全方式：
  
  

    while read -r file;做
   perl ...$ file
  完成< $ LOGI
    

  ref： http://mywiki.wooledge.org/DontReadLinesWithFor

this is my first post on stackoverflow, please forgive me if I missed something important. I am currently stuck with the follwing issue. The goal is, to replace port numbers dynamically based on a filelist I prepared with find. All of the ports in those files, start with the number "4" and have 5 digits.

Now the tricky part, I am replacing only digit #2 and #3, and keep positions 1, 4 and 5. Examples:

old port in file: 40380, 40381
new port in file: 41580, 40381

I am working on Sun Solaris 5.10 therefore I prefer perl for inline replacements

Finally the key question: how can I combine $1 (group 1) + $PIN_PINNO + $3 (group 3) so that the result would be: 41580

NEW_PINNO=15
LOGI=$HOME/filelist.txt

# port replacement
for file in `cat $LOGI`
do
    perl -pe 's/[\:\>\=]\s*(4)(\d{2})(\d{2})\b/$1${NEW_PINNO}$3/g' $file
done

many thanks in advance

解决方案

perl -pse 's/ [:>=]\s* \K (\d)\d\d(\d\d) \b/$1$pin$2/gx' -- -pin="$new_pinno" file

• Your regex will match the [colon, greater than, equal sign] and the spaces, but you don't include them in the substitution. I'm using the \K directive to match those characters but then forget about them (ref: http://perldoc.perl.org/perlre.html#Lookaround-Assertions)
• I'm using the -s option to enable "rudimentary switch parsing" to pass the shell variable into perl without playing quoting games. (ref: http://perldoc.perl.org/perlrun.html)

Testing

new_pinno=15
perl -pse 's/ [:>=]\s* \K (\d)\d\d(\d\d) \b/$1$pin$2/gx' -- -pin="$new_pinno" <<END
var1=40380
var2=40381
END

var1=41580
var2=41581

Notes

• you should not use ALL_CAPS_VARNAMES in the shell, leave those to be reserved by the shell. One day, you'll use PATH=something and then wonder why your script is broken.

• and @123's comment is valid. This is the safe way to read lines from a file:

  while read -r file; do
      perl ... "$file"
  done < "$LOGI"
  

  ref: http://mywiki.wooledge.org/DontReadLinesWithFor

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