正则表达式在两个字符串之间得到一个字符串

Question

我有一根绳子：

# userid name uniqueid connected ping loss state rate
# 529 2 "arioch83" STEAM_1:0:86796179 55:58 99 0 active 80000
# 619 3 "Snake" STEAM_1:0:27678629 06:42 61 0 active 80000
# 622 4 "Captain_Selleri" STEAM_1:1:47927314 03:25 44 0 active 80000
# 583 5 "krN[786]" STEAM_1:1:14638235 28:53 53 0 active 128000
# 621 6 "Giack" STEAM_1:1:67468100 04:44 129 0 active 80000
# 326 7 "Urkrass" STEAM_1:0:55150382  3:02:31 51 0 active 80000
#613 "Vinny" BOT active
# 584 9 "Tkappa" STEAM_1:0:32266787 27:55 360 0 active 80000
# 605 10 "Narpok19" STEAM_1:0:44838130 14:36 67 0 active 80000
# 551 11 "robbetto83" STEAM_1:1:63675894 50:10 86 0 active 80000
# 530 12 "XxKazuyaxX" STEAM_1:0:18676379 55:57 68 0 active 80000
# 623 13 "beut3d - Keyser Söze" STEAM_1:0:14500718 00:29 70 0 active 80000
# 602 15 "Homroy" STEAM_1:1:7870901 16:34 169 0 active 80000
# 607 16 "[Bloody]Eagle" STEAM_1:1:59567346 09:14 77 0 active 80000
#615 "Jeff" BOT active
# 587 18 "Heisenberg" STEAM_1:1:61427218 25:15 81 0 active 80000
#end

我想得到字符串之间的"#用户I名称，唯一的连接平丢失率“和”#结束“。我尝试了几个regex，基于我在堆栈溢出中发现的内容，但是没有一个有效:(这是我最后的尝试。)

var matches = text.match(/# userid name uniqueid connected ping loss state rate(.+)\&#end/);
                if (matches.length > 1) {
                    alert(matches[1]);
                }

你看我对regex一无所知..。关于这些有什么好的指导吗？

谢谢,

Answer 1

你需要能够跨越多条线进行匹配。您正在寻找s (多特)修饰符。不幸的是，这个修饰符在JavaScript中不存在。一个常见的解决方法是用以下内容替换点.：

[\S\s]    

因此，您可以使用以下正则表达式：

/# userid name uniqueid connected ping loss state rate([\S\s]*?)#end/

解释

(               # group and capture to \1:
  [\S\s]*?      #   any character of: 
                #    non-whitespace (all but \n, \r, \t, \f, and " "), 
                #    whitespace (\n, \r, \t, \f, and " ") 
                #    (0 or more times matching the least amount possible)
)               # end of \1

JavaScript演示

Answer 2

使用正则表达式的另一种方法是通过\n拆分字符串并提取所需的部分(从第二行到第二行到最后一行的所有内容)：

var contents = "your contents here...";

var contentArray = contents.split( "\n" );

// extract the array elements from the second to the second from last and re-join with `\n`
console.log( contentArray.slice( 1, contentArray.length - 2 ).join( "\n" ) );

Answer 3

获取#userid和#end未启动的所有行

/^(?!#\s*(userid|end)).*/gmi

在线演示