从右到左匹配正则表达式？

Question

有没有办法从右到左匹配正则表达式？我要找的是一个正则表达式，它可以

MODULE WAS INSERTED              EVENT
LOST SIGNAL ON E1/T1 LINK        OFF
CRC ERROR                        EVENT
CLK IS DIFF FROM MASTER CLK SRC  OF

从下面的输入

CLI MUX trap received: (022) CL-B  MCL-2ETH             MODULE WAS INSERTED              EVENT   07-05-2010 12:08:40
CLI MUX trap received: (090) IO-2  ML-1E1        EX1    LOST SIGNAL ON E1/T1 LINK        OFF     04-06-2010 09:58:58
CLI MUX trap received: (094) IO-2  ML-1E1        EX1    CRC ERROR                        EVENT   04-06-2010 09:58:59
CLI MUX trap received: (009)                            CLK IS DIFF FROM MASTER CLK SRC  OFF     07-05-2010 12:07:32

如果我可以完成从右到左的匹配，我可以写下(EVENT|OFF)右边的所有内容，直到第二次出现一个以上的空格+

我今天所做的最好的事情就是用正则表达式获取从(022)到事件的所有内容

CLI MUX trap received: \([0-9]+\)[ ]+(.*[  ]+(EVENT|OFF))

但这并不是我真正想要的:)

编辑:它的语言是什么？它实际上是一个过滤器的配置字符串，但我猜它使用的是标准的GNU C Regex库。

edit2:我喜欢关于按长度剪切的答案，但Amarghosh可能更符合我的要求。我真的不知道为什么我没有想过只削减长度，比如：

^.{56}(.{39}).*$

非常感谢你的快速回答...

Answer 1

如果保证标记由多个空格分隔，并且保证EVENT|OFF之前的字符串中的单词只由一个空格分隔，则可以查找以单个空格分隔的单词，后跟空格，然后是EVENT或OFF

var s = "CLI MUX trap received: (022) CL-B  MCL-2ETH             MODULE WAS INSERTED              EVENT   07-05-2010 12:08:40"
        + "\nCLI MUX trap received: (090) IO-2  ML-1E1        EX1    LOST SIGNAL ON E1/T1 LINK        OFF     04-06-2010 09:58:58"
        + "\nCLI MUX trap received: (094) IO-2  ML-1E1        EX1    CRC ERROR                        EVENT   04-06-2010 09:58:59"
        + "\nCLI MUX trap received: (009)                            CLK IS DIFF FROM MASTER CLK SRC  OFF     07-05-2010 12:07:32"

var r = /\([0-9]+\).+?((?:[^ ]+ )* +(?:EVENT|OFF))/g;
var m;
while((m = r.exec(s)) != null)
  console.log(m[1]);

输出：

MODULE WAS INSERTED              EVENT
LOST SIGNAL ON E1/T1 LINK        OFF
CRC ERROR                        EVENT
CLK IS DIFF FROM MASTER CLK SRC  OFF

正则表达式：/\([0-9]+\).+?((?:[^ ]+ )* +(?:EVENT|OFF))/g

\([0-9]+\)       #digits in parentheses followed by  
.+?              #some characters - minimum required (non-greedy)  
(                #start capturing 
(?:[^ ]+ )*      #non-space characters separated by a space  
` +`             #more spaces (separating string and event/off - 
                 #backticks added for emphasis), followed by
(?:EVENT|OFF)    #EVENT or OFF
)                #stop capturing

Answer 2

在.NET中，您可以使用RightToLeft option：

Regex RE = new Regex(Pattern, RegexOptions.RightToLeft);
Match theMatch = RE.Match(Source);

Answer 3

使用regex，您可以简单地替换以下代码：

^.{56}|.{19}$

使用空字符串。

但实际上，您只需要使用substring函数将字符串从"position 56“剪切到" string -length - 19”。这比正则表达式更容易、更快。

下面是一个用JavaScript编写的例子，其他语言的工作原理大同小异：

var lines = [
  'CLI MUX trap received: (022) CL-B  MCL-2ETH             MODULE WAS INSERTED              EVENT   07-05-2010 12:08:40',
  'CLI MUX trap received: (090) IO-2  ML-1E1        EX1    LOST SIGNAL ON E1/T1 LINK        OFF     04-06-2010 09:58:58',
  'CLI MUX trap received: (094) IO-2  ML-1E1        EX1    CRC ERROR                        EVENT   04-06-2010 09:58:59',
  'CLI MUX trap received: (009)                            CLK IS DIFF FROM MASTER CLK SRC  OFF     07-05-2010 12:07:32'
];
for (var i=0; i<lines.length; i++) {
  alert( lines[i].substring(56, lines[i].length-19) );
}