如何将[link](#)替换为link?

I want to do something like *. actually changing this style []() to this style <a href=""></a>. here is my try:

我想做一些类似*的东西。实际上，将这个样式[]()更改为这个样式。这是我的尝试:

$str = '[link](#)';
$str = str_replace('[','<a href="',$str);     // output: <a href="link](#)
$str = str_replace(']','">',$str);            // output: <a href="link">(#)
$str = str_replace('(','',$str);              // output: <a href="link">#)
$str = str_replace(')','</a>',$str);          // output: <a href="link">#</a>

but now, I need to change link with #, how can I do that ?

但是现在，我需要改变与#的链接，我该怎么做呢?

1 个解决方案

#1

You want to take a look at preg_replace(), with this you can use a regex to replace it, e.g.

您需要查看preg_replace()，您可以使用regex来替换它，例如。

$str = preg_replace("/\[(.*?)\]\((.*?)\)/", "<a href='$2'>$1</a>", $str);

regex explanation:

正则表达式的解释:

\[(.*?)\]\((.*?)\)

\[ matches the character [ literally
\[匹配字符[字面意思]
1st Capturing group (.*?)
- .*? matches any character (except newline)
  - Quantifier: *? Between zero and unlimited times, as few times as possible, expanding as needed [lazy]
  - 量词:* ?在零到无限的时间之间，尽可能少的时间，根据需要扩展(懒惰)
- . * ?匹配任何字符(换行除外)量词:*?在零到无限的时间之间，尽可能少的时间，根据需要扩展(懒惰)
第一个抓捕小组*?匹配任何字符(换行除外)量词:*?在零到无限的时间之间，尽可能少的时间，根据需要扩展(懒惰)
\] matches the character ] literally
[\]匹配字符]字面意思
\( matches the character ( literally
\(匹配字符
2nd Capturing group (.*?)
- .*? matches any character (except newline)
  - Quantifier: *? Between zero and unlimited times, as few times as possible, expanding as needed [lazy]
  - 量词:* ?在零到无限的时间之间，尽可能少的时间，根据需要扩展(懒惰)
- . * ?匹配任何字符(换行除外)量词:*?在零到无限的时间之间，尽可能少的时间，根据需要扩展(懒惰)
第二抓捕小组*?匹配任何字符(换行除外)量词:*?在零到无限的时间之间，尽可能少的时间，根据需要扩展(懒惰)
\) matches the character ) literally
\)匹配字符)字面上

#1