多行模式的php preg_match_all

Question

我被困在这个问题上，希望任何一个人都能帮我。

我有一个配置文件，其中包含以下行：

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来匹配接口的名称、vdom、vlanid、ip和备用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);

对于第一个.+，一切都是匹配的，而不是其他的！

如有任何建议

Answer 1

我喜欢regex

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

)

编辑回答后续

(?<=\vedit ") --这是在构造后面的一个正面的外观，必须放在括号中。这个位?<=指定了\v后面的正向匹配垂直空格和edit "实际上是匹配编辑“，后面是(\w+)，意思是尽可能多次匹配单词字符，把它放在括号中会创建一个捕获组，以便稍后引用匹配。

可以将组名称添加到捕获组中，以便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/';