正则表达式 - PHP 字符串中匹配url

先用比较宽泛的正则匹配出所有的url，例如

https?:\/\/\S+

然后对于这堆url依次采用parse_url函数

^(http|https|ftp)\://[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(:[a-zA-Z0-9]*)?/?([a-zA-Z0-9\-\._\?\,\'/\\\+&%\$#\=~])*$


http://regexlib.com/Search.aspx?k=url&c=-1&m=5&ps=20

Java 大概这么写

String str = "接收到的字符串"
String regex = "(http:|https:)//[^[A-Za-z0-9,:\\._\\?%&+\\-=/#]]*";
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(str);
        while (matcher.find()) {
            String url=matcher.group();
            System.out.println(url);
        }

以下字符串通过测试.

String str="http://127.0.0.1:6666/ " +
                "https://www.baidu.com/ " +
                "http://127.0.0.1/metinfo/img/img.php?class1=1&serch_sql=%201=if%28ascii%28substr%28user%28%29,1,1%29%29=114,1,2%29%23\n" +
                "或者\n" +
                "哈哈http://www.baidu.com:85676/metinfo/img/img.php?class1=1&serch_sql=%201=if%28ascii%28substr%28user%28%29,1,1%29%29=114,1,2%29%23 6666都是对的";

输出

http://127.0.0.1:6666/
https://www.baidu.com/
http://127.0.0.1/metinfo/img/img.php?class1=1&serch_sql=%201=if%28ascii%28substr%28user%28%29,1,1%29%29=114,1,2%29%23
http://www.baidu.com:85676/metinfo/img/img.php?class1=1&serch_sql=%201=if%28ascii%28substr%28user%28%29,1,1%29%29=114,1,2%29%23

什么，你问的式PHP？抱歉，我不会PHP。。。
正则一样的，自己动动脑袋吧。