php preg_match_all 多行模式

Question

我被这个问题困住了，希望有人能帮助我。

我有一个包含以下行的配置文件：

config system interface
edit "internal1"
    set vdom "root"
    set ip 192.168.1.1 255.255.255.0
    set allowaccess ping https ssh http fgfm capwap
    set type physical
    set snmp-index 1
next
edit "internal2"
    set vdom "root"
    set ip 192.168.20.2 255.255.255.0
    set allowaccess ping https ssh http fgfm capwap
    set type physical
    set snmp-index 2
    Set secondary-IP enable
      config secondaryip
        edit 1
          set ip 192.168.21.2 255.255.255.0
        next
        edit 2
          set ip 192.168.22.2 255.255.255.0
        next
      end
next
edit "internal3"
    set vdom "root"
    set ip 192.168.30.3 255.255.255.0
    set allowaccess ping https ssh http fgfm capwap
    set type physical
    set snmp-index 3
    Set secondary-IP enable
      config secondaryip
        edit 1
          set ip 192.168.31.3 255.255.255.0
        next
      end
next
end
....

并希望使用以下正则表达式匹配接口的名称、vdom、vlanid、ip 和辅助 ip：

preg_match_all("/edit .+(\s+config secondaryip\r?\n(\s+edit \d+\r?\n.+\s+next\r?\n){1,}\s+end\r?\n)?.+next\r?\n/s", $configFile, $matched_interfaces);

第一个 .+ 是所有匹配的，而不是其他的！

感谢任何建议

Answer 1

我喜欢正则表达式

$regex = '/(?<=\vedit ")(\w+)|(?<=vdom ")(\w+)|(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})(?=\s+set)|(\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})(?=\s+next)/';
print_r(preg_match_all($regex, $configFile, $matched_interfaces));

输出是

Array
(
[0] => Array
(
[0] => internal1
[1] => root
[2] => 192.168.1.1 255.255.255.0
[3] => internal2
[4] => root
[5] => 192.168.20.2 255.255.255.0
[6] => 192.168.21.2 255.255.255.0
[7] => 192.168.22.2 255.255.255.0
[8] => internal3
[9] => root
[10] => 192.168.30.3 255.255.255.0
[11] => 192.168.31.3 255.255.255.0
)

[1] => Array
(
[0] => internal1
[1] => 
[2] => 
[3] => internal2
[4] => 
[5] => 
[6] => 
[7] => 
[8] => internal3
[9] => 
[10] => 
[11] => 
)

[2] => Array
(
[0] => 
[1] => root
[2] => 
[3] => 
[4] => root
[5] => 
[6] => 
[7] => 
[8] => 
[9] => root
[10] => 
[11] => 
)

[3] => Array
(
[0] => 
[1] => 
[2] => 192.168.1.1 255.255.255.0
[3] => 
[4] => 
[5] => 192.168.20.2 255.255.255.0
[6] => 
[7] => 
[8] => 
[9] => 
[10] => 192.168.30.3 255.255.255.0
[11] => 
)

[4] => Array
(
[0] => 
[1] => 
[2] => 
[3] => 
[4] => 
[5] => 
[6] => 192.168.21.2 255.255.255.0
[7] => 192.168.22.2 255.255.255.0
[8] => 
[9] => 
[10] => 
[11] => 192.168.31.3 255.255.255.0
)

)

编辑以回答跟进

(?<=\vedit ") 这是构造背后的积极观点，必须放在括号中。此位 ?<= 指定它在 \v 后面的正面外观匹配垂直空格，并且 edit " 字面匹配编辑 "。紧随其后的是 (\w+)，这意味着匹配单词字符的次数与你可以，把它放在括号中创建一个捕获组，这样你可以稍后引用匹配。正向后看意味着 (\w+) 模式只有在它之前的后向序列也匹配时才会匹配

您可以将组名称添加到您的捕获组中，以将它们作为命名数组返回

$regex = '/(?<=edit ")(?<name>\w+)\K|(?<=vdom ")(?<vdom>\w+)|(?<ip>\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})(?=\s+set)|(?<secondary>\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3}\s\d{1,3}\.\d{1,3}\.\d{1,3}\.\d{1,3})(?=\s+next)\K/';

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