将每个索引的字符串拆分为n个单词的数组 [英] split string into array of n words per index

问题描述

我有一个要分割成一个数组的字符串，该数组每个索引有(例如)3个单词. 我还想做的是，如果在该字符串中遇到换行符，它将跳过" 3个单词的限制，并将其放入新索引中，并开始在该新索引中添加单词，直到再次达到3 .例子

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"]

我已经尝试研究正则表达式，但是到目前为止，我还不能真正理解正则表达式中使用的语法.

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

我已经编写了一种复杂函数的方法，该方法将字符串分成3行，方法是先使用.split(" ");将其分成单独的单词数组，然后使用循环将每3个字符串添加到另一个数组中.但是我不能考虑换行符.

I have written a way to complicated 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.

推荐答案

您可以尝试以下模式:

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

由于量词{0,2}在默认情况下是贪婪的，因此只有找到换行符后，它的值才会小于2 (N-1)(因为此处不允许换行符) :[^\w\n]+)或如果您是字符串的结尾.

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.

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