如何将文本节点拆分为每个3个单词的跨度？

Let's say I have this div:

假设我有这个div:

<div> Let's say these are the words in my div </div>

I know I can wrap every words in the div into a span this way

我知道我可以通过这种方式将div中的每个单词都包含在内

$("div").contents().wrap('<span class="wrapper"/>');
//output:  <div><span class="wrapper"> Let's say these are the words in my div</span> </div>

However, I would like to achieve this instead:

但是,我想改为:

<div>
    <span class="wrapper">Let's say these</span>
    <span class="wrapper">are the words</span>
    <span class="wrapper">in my div</span>
</div>

Every particular amount of words (this case: every 3 words) are to be divided into a group and each group of words is to be wrapped separately.

每个特定数量的单词(这种情况:每3个单词)将被分成一组,每组单词将被分开包装。

These are what first come to my mind:

这些是我首先想到的:

1) I think it can be achieved by using text() to obtain the string and split(' ') it to form an array with each element contains a word, write a while loop to manipulate the array:

1)我认为可以通过使用text()获取字符串并将其拆分('')来形成一个数组,每个元素包含一个单词,写一个while循环来操作数组:

var a = 0;
var b = array.length;
while (a<b) {
    array[a] = "<span class="wrapper>" +  array[a]";
    a +=2;
    if (a>b) {
        a = b-1;
    }
    array[a] = array[a] + "</span>";
    a++;
}

then just simply .join('') the array to form a string and $("div").html(string);

然后只需简单地.join('')数组形成一个字符串和$(“div”)。html(string);

2) Or I can simply use regular expression after obtaining with text():

2)或者我可以在使用text()获取后简单地使用正则表达式:

do a global search for expressions containing a word + a space + a word + a space + another word

全局搜索包含单词+空格+单词+空格+另一个单词的表达式

/(\w+\s+\w+\s+\w+)/g

replace it with it wrapped in a span

用它包裹在一个跨度中替换它

<span>$1</span>

and html() the output before performing a $("div").contents().eq(2).wrap('<span class="wrapper"/>') for the odd one out if there is any.

和html()输出之前执行$(“div”)。contents()。eq(2).wrap('')的奇数输出(如果有的话)。

These are what I've come up with and I want to know, are there better ways other than these?

这些是我想出来的,我想知道,除了这些之外还有更好的方法吗?

And what's the best way (fastest & require least memory) to achieve it?

什么是实现它的最佳方式(最快和需要最少的内存)?

2 个解决方案

#1

This ought to do what you want:

这应该做你想要的:

Demo here: http://jsfiddle.net/UQk7r/

在这里演示:http://jsfiddle.net/UQk7r/

$("div").each(function() {
    var out = [];
    var words = $(this).text()..trim().split(' ');
    for (var i = 0; i < words.length; i += 3) {
        out.push('<span class="wrapper">' + words.slice(i, i+3).join(' ') + '</span>');
    }
    $(this).html(out.join(' '));
});

Here's the DOM version (no jQuery):

这是DOM版本(没有jQuery):

Demo here: http://jsfiddle.net/cCege/2/

在这里演示:http://jsfiddle.net/cCege/2/

var divs = document.getElementsByTagName('div');
for (var i = 0; i < divs.length; i++) {
    var out = [];
    var words = divs[i].innerText.trim().split(' ');
    for (var j = 0; j < words.length; j += 3) {
        out.push('<span class="wrapper">' + words.slice(j, j+3).join(' ') + '</span>');
    }
    divs[i].innerHTML = out.join(' ');
}

Finally, here's a DOM + RegEx version... this should be your optimal performer:

最后,这是一个DOM + RegEx版本......这应该是你的最佳表现者:

Demo: http://jsfiddle.net/Xgm5q/1/

var divs = document.getElementsByTagName('div');
for (var i = 0; i < divs.length; i++) {
    divs[i].innerHTML = '<span class="wrapper">' 
            + divs[i].innerText.trim().replace(/(([^\s]+\s+){2}[^\s]+)\s+/g, '$1</span> <span class="wrapper">')
            + '</span>';
}

#2

For performance :

表现:

var elems = document.getElementsByTagName('div');

for (var j=elems.length; j--;) {
    var txtArr = elems[j].textContent.replace(/([a-zA-Z]+)\s([a-zA-Z]+)\s([a-zA-Z]+)/g, '$1 $2 $3* ').split('*'),
        txtElm = document.createDocumentFragment(),
        span   = document.createElement('span');

    for (i=0; i<txtArr.length; i++) {
        if (/\S/.test(txtArr[i])) {
            var sp   = span.cloneNode(true),
                text = document.createTextNode(txtArr[i]);

            sp.appendChild(text);
            txtElm.appendChild(sp);
        }
    }

    elems[j].innerHTML = '';
    elems[j].appendChild(txtElm)
}

FIDDLE

#1