正则表达式 - PHP 字符串中匹配url

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Release: 2016-06-06 20:21:40
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需求:

<code>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)</code>
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URL样例:

<code>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
</code>
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当然简单URL也是要匹配出来的
求解正则

回复内容:

需求:

<code>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)</code>
Copy after login
Copy after login

URL样例:

<code>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
</code>
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Copy after login

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

<code>https?:\/\/\S+</code>
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然后对于这堆url依次采用parse_url函数

<code>^(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</code>
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Java 大概这么写

<code class="java">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);
        }</code>
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以下字符串通过测试.

<code class="java">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";</code>
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输出

<code class="java">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</code>
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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。

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