bash - 使用 vanilla Ubuntu，一个用于搜索和替换/取消注释代码块的脚本？

Question

我正在尝试在开发环境的 Ubuntu 14.04 中自动安装，其中一部分要求我取消注释 Nginx vHost 配置的代码块:

    #location ~ \.php$ {
    #       fastcgi_split_path_info ^(.+\.php)(/.+)$;
    #       # NOTE: You should have "cgi.fix_pathinfo = 0;" in php.ini
    #
    #       # With php5-cgi alone:
    #       fastcgi_pass 127.0.0.1:9000;
    #       # With php5-fpm:
    #       fastcgi_pass unix:/var/run/php5-fpm.sock;
    #       fastcgi_index index.php;
    #       include fastcgi_params;
    #}

我试过sed:

sed -i "s/#location ~ \.php$ {
        #       fastcgi_split_path_info ^(.+\.php)(/.+)$;
        #       # NOTE: You should have \"cgi.fix_pathinfo = 0;\" in php.ini
        #
        #       # With php5-cgi alone:
        #       fastcgi_pass 127.0.0.1:9000;
        #       # With php5-fpm:
        #       fastcgi_pass unix:/var/run/php5-fpm.sock;
        #       fastcgi_index index.php;
        #       include fastcgi_params;
        #}/location ~ \.php$ {
          fastcgi_split_path_info ^(.+\.php)(\/.+)$;
          # NOTE: You should have \"cgi.fix_pathinfo = 0;\" in php.ini

          # With php5-cgi alone:
          fastcgi_pass 127.0.0.1:9000;
          # With php5-fpm:
          fastcgi_pass unix:\/var\/run\/php5-fpm.sock;
          fastcgi_index index.php;
          include fastcgi_params;
        }/g" /etc/nginx/sites-available/nginx.dev;

但这会返回:

sed: -e expression #1, char 22: unterminated `s' command

我认为这与语法错误有关，我试图逃避 /和 "字符，但我认为这还不够/完全正确。

我发现了这个:https://unix.stackexchange.com/a/26289/138115

这表明 perl 在这里可能是一个很好的解决方案，因为它是随 Ubuntu 一起安装的，所以我尝试了它:

perl -0777 -i.original -pe 's/#location ~ \.php$ {\n        #       fastcgi_split_path_info ^(.+\.php)(\/.+)$;\n        #       # NOTE: You should have "cgi.fix_pathinfo = 0;" in php.ini\n        #\n        #       # With php5-cgi alone:\n        #       fastcgi_pass 127.0.0.1:9000;\n        #       # With php5-fpm:\n        #       fastcgi_pass unix:\/var\/run\/php5-fpm.sock;\n        #       fastcgi_index index.php;\n        #       include fastcgi_params;\n        #}/slocation ~ \.php$ {\n  fastcgi_split_path_info ^(.+\.php)(\/.+)$;\n  NOTE: You should have "cgi.fix_pathinfo = 0;" in php.ini\n  With php5-cgi alone:\n  fastcgi_pass 127.0.0.1:9000;\n  With php5-fpm:\n  fastcgi_pass unix:\/var\/run\/php5-fpm.sock;\n  fastcgi_index index.php;\n  include fastcgi_params;\n}/igs' /etc/nginx/sites-available/nginx.dev;

但这会产生一大堆语法错误:

syntax error at -e line 1, near "(."
Unknown regexp modifier "/v" at -e line 1, within string
Unknown regexp modifier "/r" at -e line 1, within string
Unknown regexp modifier "/h" at -e line 1, within string
Unknown regexp modifier "/5" at -e line 1, within string
Not enough arguments for index at -e line 1, near "index."
syntax error at -e line 1, near "n}"
Execution of -e aborted due to compilation errors.

我以前为各种环境编写过很多这样的脚本，但我总是尽量避免替换多行文本，因为我从来没有能够做到正确。今天，我已经花了 3 个小时，但我仍然没有真正了解如何完成这项工作。如果有人可以分享一些对此的输入/见解以及如何实现它，那么将不胜感激，谢谢!

编辑 1:

使用方括号进行简单转义:

#!/bin/bash
#/etc/nginx/sites-available/test.sh
file=$(<default.file);
search=$(<nginx_search.txt);
replace=$(<nginx_replace.txt);
$file =~ s[$search][$replace]g;
echo "$file" > "/etc/nginx/sites-available/test.file";
# Outputs notice: ./test.sh: line 6: #: command not found

test.file已创建，但它包含 default.file 的原始值没有修改。

在 perl 中测试后，我收到:

syntax error at ./perl.perl line 6, near "(."
Unknown regexp modifier "/v" at ./perl.perl line 6, within string
Unknown regexp modifier "/r" at ./perl.perl line 6, within string
Unknown regexp modifier "/h" at ./perl.perl line 6, within string
Unknown regexp modifier "/5" at ./perl.perl line 6, within string
syntax error at ./perl.perl line 9, near ";
}"
Execution of ./perl.perl aborted due to compilation errors.

第 6 行内容如下:

#       # NOTE: You should have "cgi.fix_pathinfo = 0;" in php.ini

Answer 1

注意 - 你提到(和标签)perl .以下来自 perl看法。其中一些可能适用于传统的外壳，但我不能确定到底是什么。 perl确实支持一些超出基础 POSIX 的正则表达式内容规范。

像这样的模式的问题在于你的模式中有定界符。您的初始失败，因为它将处理这个斜线:

    #       fastcgi_pass unix:/var/run/php5-fpm.sock;

作为模式中的“分割点”。您可以通过转义分隔符来处理此问题，但实际上更好的技巧是 - 使用其他地方不存在的分隔符。在上面，我建议你可以使用方括号:

my $str = "some fish"; 
$str =~ s[some][more]g;

print $str;

虽然作为替代方案 - 您可以使用 range operator如果它在两个指定的分隔符内，则测试为“真”:

while ( <> ) {
   if ( m|\#location.*php\$ \{| .. m|^\s*#\}| ) {
       s/#//;
   }
   print ;
}

例如。:

#!/usr/bin/perl
use strict;
use warnings;


while ( <DATA> ) {
   if ( m|\#location.*php\$ \{| .. m|^\s*#\}| ) {
       #note - no g modifier, so we only do the first per line
       s/#//;
   }
   print ;
}

__DATA__
# Some stuff
we don't care about this line
#and this shouldn't be changed

#but after this point, it should be!

   #location ~ \.php$ {
    #       fastcgi_split_path_info ^(.+\.php)(/.+)$;
    #       # NOTE: You should have "cgi.fix_pathinfo = 0;" in php.ini
    #
    #       # With php5-cgi alone:
    #       fastcgi_pass 127.0.0.1:9000;
    #       # With php5-fpm:
    #       fastcgi_pass unix:/var/run/php5-fpm.sock;
    #       fastcgi_index index.php;
    #       include fastcgi_params;
    #}


# and now we stop, and leave this bit alone. 

    more stuff; 
    here; 
    # and another comment line

如果您在两个分隔符之间(上面的位置 php 和“关闭波浪括号”)，这有条件地应用转换。

你可以单线化这个:

perl -ne 'if ( m|\#location.*php\$ \{| .. m|^\s*#\}| ) { s/#//g; } print' myfile

(如果您想就地编辑，请添加 -i)。