javascript - split string into array of n words per index - Stack Overflow

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I have a string that I'd like to split in an array that has (for example) 3 words per index. What I'd also like it to do is if it encounters a new line character in that string that it will "skip" the 3 words limit and put that in a new index and start adding words in that new index until it reaches 3 again. example

var text = "this is some text that I'm typing here \n yes I really am"

var array = text.split(magic)

array == ["this is some", "text that I'm", "typing here", "yes I really", "am"]

I've tried looking into regular expressions, but so far I can't really make sense of the syntax that is used in regex.

I have written a way to plicated function that splits my string into lines of 3 by first splitting it into an array of separate words using .split(" "); and then using a loop to add add it per 3 into another array. But with that I can't take the new line character into account.

I have a string that I'd like to split in an array that has (for example) 3 words per index. What I'd also like it to do is if it encounters a new line character in that string that it will "skip" the 3 words limit and put that in a new index and start adding words in that new index until it reaches 3 again. example

var text = "this is some text that I'm typing here \n yes I really am"

var array = text.split(magic)

array == ["this is some", "text that I'm", "typing here", "yes I really", "am"]

I've tried looking into regular expressions, but so far I can't really make sense of the syntax that is used in regex.

I have written a way to plicated function that splits my string into lines of 3 by first splitting it into an array of separate words using .split(" "); and then using a loop to add add it per 3 into another array. But with that I can't take the new line character into account.

• javascript
• arrays
• regex
• string

Share Improve this question Follow edited May 11, 2014 at 19:32 andmcgregor 48533 silver badges1616 bronze badges asked May 11, 2014 at 19:05 Ramin AfsharRamin Afshar 1,01922 gold badges1919 silver badges3434 bronze badges 1

• You can't split with that magic. Loop over the string, count the spaces, push every 3rd space, and \n and there you go. Custom split – Sterling Archer Commented May 11, 2014 at 19:10

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5 Answers 5

Sorted by: Reset to default 4

You can try with this pattern:

var result = text.match(/\b[\w']+(?:[^\w\n]+[\w']+){0,2}\b/g);

since the quantifier {0,2} is greedy by default, it will take a value less than 2 (N-1) only if a newline is found (since newlines are not allowed here: [^\w\n]+) or if you are a the end of the string.

If you're interested in a regexp solution, it goes like this:

   text.match(/(\S+ \S+ \S+)|(\S+ \S+)(?= *\n|$)|\S+/g)
   // result ["this is some", "text that I'm", "typing here", "yes I really", "am"]

Explanation: match either three space separated words, or two words followed by spaces + newline, or just one word (a "word" being simply a sequence of non-spaces).

For any number of words, try this:

text.match(/((\S+ ){N-1}\S+)|(\S+( \S+)*)(?= *\n|$)|\S+/g)

(replace N-1 with a number).

Try something like this:

words = "this is some text that I'm typing here \n yes I really am".split(" ");
result = [];
temp = "";

for (i = 0; i < words.length; i++) {
  if ((i + 1) % 3 == 0) {
    result.push(temp + words[i] + " ");
    temp = "";
  } else if (i == words.length - 1) {
    result.push(temp + words[i]);
  } else {
    temp += words[i] + " ";
  }
}

console.log(result);

Basically what this does is splits the string by words, then loops through each word. Every third word it gets to, it adds that along with what is stored in temp into the array, otherwise it adds the word to temp.

Only if you know there are no words 'left', so the number of words is always a multiple of 3:

"this is some text that I'm typing here \n yes I really am".match(/\S+\s+\S+\s+\S+/g)
=> ["this is some", "text that I'm", "typing here \n yes", "I really am"]

but if you add a word:

"this is some text that I'm typing here \n yes I really am FOO".match(/\S+\s+\S+\s+\S+/g)

the result will be exactly the same, so "FOO" is missing.

here one more way: use this pattern ((?:(?:\S+\s){3})|(?:.+)(?=\n|$))
Demo

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