查找并替换文件中的配置行

Question

我正在尝试编写一个 bash 脚本来在配置文件中进行一些替换。

conf 测试文件是这样的：

pm.max_children = 50
bla bla bla pm.max_children
pm.start_servers = 10
pm.min_spare_servers = 10
pm.max_spare_servers = 30
;pm.max_requests = 20000

我需要做的替换是：

pm.max_children = 5
pm.start_servers = 1
pm.min_spare_servers = 1
pm.max_spare_servers = 3
pm.max_requests = 2000

我需要从 bash 脚本运行这个程序。

Answer 1

这是使用 sed 的更简单的版本：

#!/usr/bin/sed -f
s/^\(pm.max_children = \).*/\15/
s/^\(pm.start_servers = \).*/\11/
s/^\(pm.min_spare_servers = \).*/\11/
s/^\(pm.max_spare_servers = \).*/\13/
s/^\(pm.max_requests = \).*/\12000/

您可以将其放入单独的文件中并从脚本中调用它吗？

我同意 Kerrek SB 的观点——如果你能解释一下你想要实现的目标，这将会有所帮助！

Answer 2

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

my %vals = (
    'pm.max_children'      => 5,
    'pm.start_servers'     => 1,
    'pm.min_spare_servers' => 1,
    'pm.max_spare_servers' => 3,
    'pm.max_requests'      => 2000,
);

while (<DATA>) {
    if ( my($param) = /^(\S+)\s*=/ and exists $vals{$1} ) {
        s/\d+$/$vals{$param}/;
    }
    print;
}

__DATA__
pm.max_children= 50
bla bla bla pm.max_children
pm.start_servers = 10
pm.min_spare_servers = 10
pm.max_spare_servers = 30
pm.max_requests = 20000