javascript - 如何用链接替换普通 URL，例如？

Question

我快要成功了。我想知道是否有更好的方法。

Root problem

Fiddle

function replaceURLWithHTMLLinks(text) {
    text = text.replace(/a/g, "--ucsps--");
    text = text.replace(/b/g, "--uspds--");
    var arrRegex = [
        /(\([^)]*\b)((?:https?|ftp|file):\/\/[-A-Za-z0-9+&@#\/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#\/%=~_()|])(\))/ig,
        /(\([^)]*\b)((?:https?|ftp|file):\/\/[-A-Za-z0-9+&@#\/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#\/%=~_()|])(.?\b)/ig,
        /()(\b(?:https?|ftp|file):\/\/[-a-z0-9+&@#\/%?=~_()|!:,.;]*[-a-z0-9+&@#\/%=~_()|])(.?\b)/ig];
    for (i = 0; i < arrRegex.length; i++) {
        text = text.replace(arrRegex[i], "$1a$2b$3");
    }
    text = text.replace(/a([^b]*)b/g, "<a href='$1'>$1</a>");
    text = text.replace(/--ucsps--/g, "a");
    text = text.replace(/--uspds--/g, "b");
    return text;
}
var elm = document.getElementById('trythis');
elm.innerHTML = replaceURLWithHTMLLinks(elm.innerHTML);

有什么想法吗？

Answer 1

位于 CodeReview这个问题得到了很好的回答splendidly .

function replaceURLWithHTMLLinks(text) {
    var re = /(\(.*?)?\b((?:https?|ftp|file):\/\/[-a-z0-9+&@#\/%?=~_()|!:,.;]*[-a-z0-9+&@#\/%=~_()|])/ig;
    return text.replace(re, function(match, lParens, url) {
        var rParens = '';
        lParens = lParens || '';

        // Try to strip the same number of right parens from url
        // as there are left parens.  Here, lParenCounter must be
        // a RegExp object.  You cannot use a literal
        //     while (/\(/g.exec(lParens)) { ... }
        // because an object is needed to store the lastIndex state.
        var lParenCounter = /\(/g;
        while (lParenCounter.exec(lParens)) {
            var m;
            // We want m[1] to be greedy, unless a period precedes the
            // right parenthesis.  These tests cannot be simplified as
            //     /(.*)(\.?\).*)/.exec(url)
            // because if (.*) is greedy then \.? never gets a chance.
            if (m = /(.*)(\.\).*)/.exec(url) ||
                    /(.*)(\).*)/.exec(url)) {
                url = m[1];
                rParens = m[2] + rParens;
            }
        }
        return lParens + "<a href='" + url + "'>" + url + "</a>" + rParens;
    });
}

注意:我在“var re”中的“@”符号有错误 - 我只是将其替换为@@