Question

鉴于此文：

1/12/2011
I did something.

10/5/2013
I did something else.

Here is another line.

And another.

5/17/2014
Lalala.
More text on another line.

我想使用正则表达式（或者其他一些方法？）来获得这个：

["1/12/2011", "I did something.", "10/5/2013", "I did something else.\n\nHere is another line.\n\nAnd another.", "5/17/2014", "Lalala.\nMore text on another line."]

日期部分和内容部分是每个单独的条目，交替显示。

我尝试使用[^]代替点，因为JS＆＃39; s与新行不匹配（如Matching multiline Patterns所示），但是这场比赛很贪婪并且也占用了很多，所以结果数组只有1个条目：

var split_pattern = /\b(\d\d?\/\d\d?\/\d\d\d\d)\n([^]+)/gm;
var array_of_mems = contents.match(split_pattern);

// => ["1/12/2011↵I did something else..."]

如果我添加一个问号来获取[^] + ?,根据How to make Regular expression into non-greedy?使得匹配非贪婪，那么我只得到内容部分的第一个字符。

最好的方法是什么？提前谢谢。

Answer 1

(\d{1,2}\/\d{1,2}\/\d{4})\n|((?:(?!\n*\d{1,2}\/\d{1,2}\/\d{4})[\s\S])+)

你可以尝试这个抓住捕获。参见演示。

https://regex101.com/r/sJ9gM7/126

var re = /(\d{1,2}\/\d{1,2}\/\d{4})\n|((?:(?!\n*\d{1,2}\/\d{1,2}\/\d{4})[\s\S])+)/gim;
var str = '1/12/2011\nI did something.\n\n10/5/2013\nI did something else.\n\nHere is another line.\n\nAnd another.\n\n5/17/2014\nLalala.\nMore text on another line.';
var m;

if ((m = re.exec(str)) !== null) {
if (m.index === re.lastIndex) {
re.lastIndex++;
}
// View your result using the m-variable.
// eg m[0] etc.
}

Answer 2

您可以在循环中使用exec()方法来获得所需的结果。

var re  = /^([\d/]+)\s*((?:(?!\s*^[\d/]+)[\S\s])+)/gm, 
matches = [];

while (m = re.exec(str)) {
  matches.push(m[1]);
  matches.push(m[2]);
}

<强>输出

[ '1/12/2011',
  'I did something.',
  '10/5/2013',
  'I did something else.\n\nHere is another line.\n\nAnd another.',
  '5/17/2014',
  'Lalala.\nMore text on another line.' ]

eval.in

JS：匹配条目与捕获组，占新线

2 个答案: