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字符串截取,该怎么解决

时间：2021-07-01 10:21:17 帮助过：7人阅读

字符串截取
bri=bridge0 if1=wan if2=lan if3=vlan mac=00:11:22:1d:2d:3d ip=192.168.100.1 netmask=255.255.255.0 arp=arp state=stop

bri=bridge1 if1=wan mac=00:01:2B:4C:6D:9F ip=192.168.1.125 netmask=255.255.255.0 state=stop

这是我保存的配置文件
但是 if打头的是多选的,个数是不固定的,最多8个,我想取出 if的值
也就是
bridge0 : wan lan vlan
bridge1 : wan
这条数据我转成数组是

PHP code

Array
(
    [0] => Array
        (
            [bri] => bridge0
            [if1] => wan
            [if2] => lan
            [if3] => vlan
            [mac] => 00:11:22:1d:2d:3d
            [ip] => 192.168.100.1
            [netmask] => 255.255.255.0
            [arp] => arp
            [state] => stop
        )

    [1] => Array
        (
            [bri] => bridge1
            [if1] => wan
            [mac] => 00:01:2B:4C:6D:9F
            [ip] => 192.168.1.125
            [netmask] => 255.255.255.0
            [state] => stop
        )

)

请问有什么好的办法么,难道我真要
转成数组之后在根据键值判断

------解决方案--------------------

PHP code

$s = "bri=bridge0 if1=wan if2=lan if3=vlan mac=00:11:22:1d:2d:3d ip=192.168.100.1 netmask=255.255.255.0 arp=arp state=stop";
parse_str(strtr($s, ' ', '&'), $a);
print_r($a);

------解决方案--------------------

不就是找哪个key是if开头的吗.

PHP code[User:root Time:17:19:16 Path:/home/liangdong/php]$ php preg.php 
bridge0: wan lan vlan
bridge1: wan
[User:root Time:17:19:16 Path:/home/liangdong/php]$ cat preg.php

字符串截取,该怎么解决

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