正则表达式 - PHP 字符串中匹配url
程序员文章站
2022-06-08 10:53:22
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需求:
客户端传过来一段字符串,需要从字符串中匹配出所有的url,包括域名或IP后面的参数(含端口)
URL样例:
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
或者
http://www.baidu.com/metinfo/img/img.php?class1=1&serch_sql=%201=if%28ascii%28substr%28user%28%29,1,1%29%29=114,1,2%29%23
当然简单URL也是要匹配出来的
求解正则
回复内容:
需求:
客户端传过来一段字符串,需要从字符串中匹配出所有的url,包括域名或IP后面的参数(含端口)
URL样例:
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
或者
http://www.baidu.com/metinfo/img/img.php?class1=1&serch_sql=%201=if%28ascii%28substr%28user%28%29,1,1%29%29=114,1,2%29%23
当然简单URL也是要匹配出来的
求解正则
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。