“email ip”正則表達式進入日志文件

Question

我有一個日志文件看起來像：

'User_001','Entered server','email@aol.com','2','','','0','YES','0','0',','0','192.168.1.1','192.168.1.2','0','0','0','0','0','0','0','0','0','1','0','','0','0','0','1'
'User_002','Entered server','email@aol.com','2','','','0','NO','0','0',','0','192.168.1.3','192.168.1.4','0','0','0','0','0','0','0','0','0','1','0','','0','0','0','1'

要么

User_001 Entered server email@aol.com 2 Pool_1 YES 0 0 0 192.168.1.1 192.168.1.2 0 0 0 0 0 0 0 0 0 1 0 0 1
User_002 Entered server email@aol.com 2 Pool_1 NO 0 0 0 192.168.1.3 192.168.1.4 0 0 0 0 0 0 0 0 0 1 0 0 1

而我正試圖以“電子郵件IP”格式制作出口正則表達的內容。

我嘗試使用正則表達式：

([A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,6}(.*)([0-9]{1,3}[\.]){3}[0-9]{1,3})

但是當然不起作用，因為它也獲得了2個匹配字符串之間的全部內容。

如何忽略2個找到的字符串之間的內容？

我試圖否定正則表達式部分沒有成功。

感謝大家提前！

我需要用grep做這個

Answer 1

這是我丑陋的正則表達式解決方案（有效）：

([a-z0-9]+@[a-z0-9.]+).*?([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})

https://www.regex101.com/r/APfJS1/1

 const regex = /([a-z0-9]+@[a-z0-9.]+).*?([0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3})/gi; const str = `User_001','Entered server','email@aol.com','2','','','0','YES','0','0',','0','192.168.1.1','192.168.1.2','0','0','0','0','0','0','0','0','0','1','0','','0','0','0','1'`; let m; while ((m = regex.exec(str)) !== null) { // This is necessary to avoid infinite loops with zero-width matches if (m.index === regex.lastIndex) { regex.lastIndex++; } // The result can be accessed through the `m`-variable. m.forEach((match, groupIndex) => { console.log(`Found match, group ${groupIndex}: ${match}`); }); } 

但正如評論中所提到的：一個好的csv解析器可能會更好！

PHP

$re = '/([a-z0-9]+@[a-z0-9.]+).*?([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})/i';
$str = 'User_001\',\'Entered server\',\'email@aol.com\',\'2\',\'\',\'\',\'0\',\'YES\',\'0\',\'0\',\',\'0\',\'192.168.1.1\',\'192.168.1.2\',\'0\',\'0\',\'0\',\'0\',\'0\',\'0\',\'0\',\'0\',\'0\',\'1\',\'0\',\'\',\'0\',\'0\',\'0\',\'1\'';

preg_match_all($re, $str, $matches);

// Print the entire match result
print_r($matches);