正则表达式：匹配 JS 中的新行或行尾或空格

Question

I wanted to extract :privacy and :date from the example string below.

I wanted to have a regex constraint that describes that a :[^:\s]+ block (eg :privacy or :date ) can only be ended by a space \s or a newline \n or a end of string $ (so I will be able to have a rule to logically split these blocks in the later steps).

So I simply put (?:$|\n|\s) at the end of the regex, but I doesn't work for me (the 3rd regex below). I double checked that it does work when I separately put \s or $ (the 1st and 2nd regex below), now I have no idea how I can implement the thing. Thanks for your help.

'note::tmp hogehoge. :privacy :date'.match(/\s:[^:\s]+\s/g)
(1) [' :privacy ']

'note::tmp hogehoge. :privacy :date'.match(/\s:[^:\s]+$/g)
(1) [' :date']

'note::tmp hogehoge. :privacy :date'.match(/\s:[^:\s]+(?:$|\n|\s)/g)
(1) [' :privacy ']

Answer 1

You can use below regex pattern to match a block pattern:[^:\s+]+ ended by a space \s or a newline \n or a end of string $

/((:[^:\s+]+)(?:[\s\n]))|(:[^:\s+]+)(?:[\s\n])?$/gm

(?:[\s\n]) - will check if the block is being followed by a space or a new line
(:[^:\s+]+)(?:[\s\n])?$ - will check if the block is at the end of string or not.

you can also use lookforward technique to achieve the same result

(:[^:\s+]+)(?=\s|\n|$)

Answer 2

In your pattern \s:[^:\s]+\s you are matching the leading and the trailing whitespace chars.

What you might do is assert a whitespace boundary using (?!\S) to the right.

To get the value without the leading whitespace char, you can use a capture group.

\s(:[^:\s]+)(?!\S)

Regex demo

 const s = "note::tmp hogehoge. :privacy:date"; const regex = /\s(:[^:\s]+)(?;\S)/g. const result = Array.from(s,matchAll(regex); m => m[1]). console;log(result);