我正在使用下面的函数来匹配给定文本中的url,并将它们替换为HTML链接。正则表达式工作得很好,但目前我只替换了第一个匹配。
我怎么能替换所有的URL?我想我应该使用exec命令,但我真的不知道如何做到这一点。
function replaceURLWithHTMLLinks(text) {
var exp = /(\b(https?|ftp|file):\/\/[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|])/i;
return text.replace(exp,"<a href='$1'>$1</a>");
}
我对Travis的代码做了一些小的修改(只是为了避免任何不必要的重新声明-但它对我的需求很好,所以做得很好!):
function linkify(inputText) {
var replacedText, replacePattern1, replacePattern2, replacePattern3;
//URLs starting with http://, https://, or ftp://
replacePattern1 = /(\b(https?|ftp):\/\/[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|])/gim;
replacedText = inputText.replace(replacePattern1, '<a href="$1" target="_blank">$1</a>');
//URLs starting with "www." (without // before it, or it'd re-link the ones done above).
replacePattern2 = /(^|[^\/])(www\.[\S]+(\b|$))/gim;
replacedText = replacedText.replace(replacePattern2, '$1<a href="http://$2" target="_blank">$2</a>');
//Change email addresses to mailto:: links.
replacePattern3 = /(([a-zA-Z0-9\-\_\.])+@[a-zA-Z\_]+?(\.[a-zA-Z]{2,6})+)/gim;
replacedText = replacedText.replace(replacePattern3, '<a href="mailto:$1">$1</a>');
return replacedText;
}
Reg例:
/ (\ b ((https ? | | ftp文件):\ \ / | (www)) [-A-Z0-9 +&@#\/%?=~_|!:,.;]*[- A-Z0-9 +&@#\/%=~_|]*)/ 搞笑
function UriphiMe(text) {
var exp = /(\b((https?|ftp|file):\/\/|(www))[-A-Z0-9+&@#\/%?=~_|!:,.;]*[-A-Z0-9+&@#\/%=~_|]*)/ig;
return text.replace(exp,"<a href='$1'>$1</a>");
}
下面是一些测试字符串:
请登录www.google.com找我
www
请登录www.http://www.com找我
请关注我:http://www.nishantwork.wordpress.com
http://www.nishantwork.wordpress.com
请关注我:http://www.nishantwork.wordpress.com
https://stackoverflow.com/users/430803/nishant
注意:如果你不想将www作为有效的传递,请使用下面的reg ex:
/ (\ b ((https ? | | ftp文件):\ \ / | (www)) [-A-Z0-9 +&@#\/%?=~_|!:,.;]*[- A-Z0-9 +&@#\/%=~_|])/ 搞笑
以下是我的解决方案:
var content = "Visit https://wwww.google.com or watch this video: https://www.youtube.com/watch?v=0T4DQYgsazo and news at http://www.bbc.com";
content = replaceUrlsWithLinks(content, "http://");
content = replaceUrlsWithLinks(content, "https://");
function replaceUrlsWithLinks(content, protocol) {
var startPos = 0;
var s = 0;
while (s < content.length) {
startPos = content.indexOf(protocol, s);
if (startPos < 0)
return content;
let endPos = content.indexOf(" ", startPos + 1);
if (endPos < 0)
endPos = content.length;
let url = content.substr(startPos, endPos - startPos);
if (url.endsWith(".") || url.endsWith("?") || url.endsWith(",")) {
url = url.substr(0, url.length - 1);
endPos--;
}
if (ROOTNS.utils.stringsHelper.validUrl(url)) {
let link = "<a href='" + url + "'>" + url + "</a>";
content = content.substr(0, startPos) + link + content.substr(endPos);
s = startPos + link.length;
} else {
s = endPos + 1;
}
}
return content;
}
function validUrl(url) {
try {
new URL(url);
return true;
} catch (e) {
return false;
}
}
经过几个来源的输入,我现在有一个很好的解决方案。这与编写自己的替换代码有关。
的答案。
小提琴。
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;
});
}