PHP使用正则表达式在字符串中查找＃

Question

I have a php variable where I need to show #value Values as link pattern. The code looks like this.

$reg_exUrl = "/\#::(.*?)/";

 // The Text you want to filter for urls
$text = "This is a #simple text from which we have to perform #regex    operation";

// Check if there is a url in the text
   if(preg_match($reg_exUrl, $text, $url)) {

   // make the urls hyper links
    echo preg_replace($reg_exUrl, '<a href="'.$url[0].'" rel="nofollow">'.$url[0].'</a>', $text);

   } else {

   // if no urls in the text just return the text
     echo "IN Else #$".$text;

 }

Answer 1

By using \\w, you can match a word contains alphanumeric characters and underscore. Change your expression with this:

$reg_exUrl = "/#(.*?)\w+/"

Answer 2

It's not clear to me exactly what you need match. If you want to replace a # followed by any word chars:

$text = "This is a #simple text from which we have to perform #regex    operation";

$reg_exUrl = "/#(\w+)/";
echo preg_replace($reg_exUrl, '<a href="$0" rel="nofollow">$1</a>', $text);

//Output:
//This is a <a href="#simple" rel="nofollow">simple</a> text from which we have to perform <a href="#regex" rel="nofollow">regex</a>    operation

The replacement uses $0 to refer to the text matched and $1 the first group.

Answer 3

$reg_exUrl = "/\\#::(.*?)/";

This doesn't match because of the following reasons

1. there is no need to escape # , this is because it is not a special character.

2. since you want to match just # followed by some words, there is no need for ::

3. (.*?) tries to match the least possible word because of the quantifier ? . So it won't match the required length of word you need.

If you still want to go by your pattern, you can modify it to

$reg_exUrl = "/#(.*?)\\w+/" See demo

But a more efficient one that still works is

$reg_exUrl = "/#\\w+/" . see demo