javascript - JS : matching entries with capture groups, 占新行

Question

鉴于此文本:

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(或者其他方式？)来得到这个:

["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都是单独的条目，交替出现。

我尝试使用 [^] 而不是点，因为 JS 的 .* 不匹配新行(如 Matching multiline Patterns 所说)，但是匹配是贪婪的并且占用了太多，所以结果数组只有1 条记录:

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..."]

如果我加一个问号得到[^]+?，则根据How to make Regular expression into non-greedy?使匹配非贪婪，然后我只得到内容部分的第一个字符。

最好的方法是什么？提前致谢。

Answer 1

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

您可以尝试这个。抓取捕获的内容。查看演示。

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.
}