bash php 中等效的“grep”命令是什么？
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What is the equivalent "grep" command in php?
提问by hillz
Please bear with me since I'm still really new in PHP. So I have a configfile like this:
请耐心等待，因为我对 PHP 还是很陌生。所以我有一个config这样的文件：
profile 'axisssh2'
server '110.251.223.161'
source_update 'http://myweb.com:81/profile'
file_config 'udp.group-1194-exp11nov.ovpn'
use_config 'yes'
ssh_account 'sgdo.ssh'
I want to create a PHP variable named $currentprofilewith the value of axisssh2, the value keeps changing. With grepin bash I can just do
我想创建一个名为PHP变量$currentprofile有值axisssh2，该值不断变化。随着grep在bash我可以做
currentprofile=$(cat config | grep ^profile | awk -F "'" '{print }')
But I have no idea how to do that with PHP. Please kindy help me how to do that, thank you.
但我不知道如何用 PHP 做到这一点。请大神帮我怎么做，谢谢。
UPDATE:
So I tried preg_matchlike this but it only shows the value of 1
更新：所以我尝试preg_match这样，但它只显示值1
$config=file_get_contents('/root/config');
$currentprofile=preg_match('/^profile /', $config);
echo "Current Profile: ".$currentprofile;
Please tell me what's wrong.
请告诉我怎么了。
回答by AbraCadaver
I'm going out on a limb to answer a question that you didn't ask. You'd be better off using parse_ini_string()or fgetcsv(). The .inifile would need the following format profile='axisssh2', so replace the space:
我要出去回答一个你没有问过的问题。最好使用parse_ini_string()或fgetcsv()。该.ini文件需要以下格式profile='axisssh2'，因此请替换空格：
$array = parse_ini_string(str_replace(' ', '=', file_get_contents($file)));
print_r($array);
Yields:
产量：
Array
(
[profile] => axisssh2
[server] => 110.251.223.161
[source_update] => http://myweb.com:81/profile
[file_config] => udp.group-1194-exp11nov.ovpn
[use_config] => yes
[ssh_account] => sgdo.ssh
)
So just:
所以就：
echo $array['profile'];
But the answer to your question would be:
但你的问题的答案是：
preg_matchreturns the number of matches (which is why you get 1) but you can get the actual matches with a capture group which will populate the third argument:
preg_match返回匹配的数量（这就是你得到 1 的原因），但你可以使用捕获组获取实际匹配，该组将填充第三个参数：
$config = file_get_contents('/root/config');
$currentprofile = preg_match("/^profile '(.*)'/", $config, $matches);
echo $matches[1];
