用特殊字符将句子分隔成包含空格的单词

Question

I want to separate a sentence with special characters into words keeping the spaces . Like so:

"la sílaba tónica es la penúltima".split(...regex...)

to:

["la ", "sílaba ", "tónica ", "es ", "la ", "penúltima"]
    ↑                     ↑      ↑      ↑
  space                 space  space  space

I've tried with a modified version of this answer: https://stackoverflow.com/a/26184632/2083117

With the code from that answer:

"la sílaba tónica es la penúltima".split(/\b(?![\s.])/)

Result:

["la ", "s", "í", "laba ", "t", "ó", "nica ", "es ", "la ", "pen", "ú", "ltima"]
              ↑                  ↑                                  ↑

Those special characters shouldn't split the word.

My version simply adding the special characters I want to keep ( .áéíóúñ,:;? ):

"la sílaba tónica es la penúltima".split(/\b(?![\s.áéíóúñ,:;?])/)

Result:

["la ", "sí", "laba ", "tó", "nica ", "es ", "la ", "penú", "ltima"]
          ↑              ↑                              ↑

Now the characters are included but the word is braking after them.

What would be the right regular expression for this?

Answer 1

Try to match \\S+\\s* instead of split.

 var result = "la sílaba tónica es la penúltima".match(/\\S+\\s*/gi); console.log(result); 

Answer 2

 let splitArray = "la sílaba tónica es la penúltima".split(" ") let splitArrayWithSpaces = splitArray.map((item, index ) => { if(index!== splitArray.length-1) return (item+ " ") else return item }) console.log(splitArrayWithSpaces) 

Answer 3

This az\\xC0-\\xff selects chars and diacritics. I split it by /[^az\\xC0-\\xff]/ . Then I add the space.

Alternatively you can split by /[\\s]/

 let test = "la sílaba tónica es la penúltima".split(/[^az\\xC0-\\xff]/) for(let i=0; i < test.length; i++){test[i]+= " ";} console.log(test)