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我發現了一個非常漂亮的小腳本,它幾乎可以完成我正在尋找的任務。它用維基百科的鏈接替換每個單詞列表的發生。問題是我只想要第一次發生聯繫。用鏈接替換列表中每個單詞的第一個出現
這裏是腳本(從this answer):
function replaceInElement(element, find, replace) {
// iterate over child nodes in reverse, as replacement may increase
// length of child node list.
for (var i= element.childNodes.length; i-->0;) {
var child= element.childNodes[i];
if (child.nodeType==1) { // ELEMENT_NODE
var tag= child.nodeName.toLowerCase();
if (tag!='style' && tag!='script') // special case, don't touch CDATA elements
replaceInElement(child, find, replace);
} else if (child.nodeType==3) { // TEXT_NODE
replaceInText(child, find, replace);
}
}
}
function replaceInText(text, find, replace) {
var match;
var matches= [];
while (match= find.exec(text.data))
matches.push(match);
for (var i= matches.length; i-->0;) {
match= matches[i];
text.splitText(match.index);
text.nextSibling.splitText(match[0].length);
text.parentNode.replaceChild(replace(match), text.nextSibling);
}
}
// keywords to match. This *must* be a 'g'lobal regexp or it'll fail bad
var find= /\b(keyword|whatever)\b/gi;
// replace matched strings with wiki links
replaceInElement(document.body, find, function(match) {
var link= document.createElement('a');
link.href= 'http://en.wikipedia.org/wiki/'+match[0];
link.appendChild(document.createTextNode(match[0]));
return link;
});
我一直在試圖修改(沒有成功)使用的indexOf insted的正則表達式(從this answer),我相信這將是比快正則表達式:
var words = ["keyword","whatever"];
var text = "Whatever, keywords are like so, whatever... Unrelated, I now know " +
"what it's like to be a tweenage girl. Go Edward.";
var matches = []; // An empty array to store results in.
//Text converted to lower case to allow case insensitive searchable.
var lowerCaseText = text.toLowerCase();
for (var i=0;i<words.length;i++) { //Loop through the `words` array
//indexOf returns -1 if no match is found
if (lowerCaseText.indexOf(words[i]) != -1)
matches.push(words[i]); //Add to the `matches` array
}
所以我的問題是如何結合這兩個來獲得最高效/最快的結果,而不使用庫?
我看到什麼都「的Lorem」被鏈接,而不是僅僅在第一 – ofko 2012-04-14 21:07:19
每一個元素,如何代碼工作,但它只能更換第一份。它可以修改它,只需要跟蹤哪些表達式被發現。 – GillesC 2012-04-14 21:09:25
http://jsfiddle.net/bW7LW/2/在這裏這個工作,我不得不扭轉元素的循環,否則它會從底部到頂部而不是從頂部到底部。 – GillesC 2012-04-14 21:30:48