php regex - 用字符串中的链接替换所有@usernames [英] php regex - Replace all @usernames with link in a string

问题描述

我正在我的网站上实施一个时间轴系统，用户可以在其中使用 @username(例如 twitter)在他们的时间轴中提及其他用户.

I am implimenting a timeline system on my website where users can mention others users in their timeline using @username like twitter.

我想将 @username 转换为链接并将其指向他们的个人资料

I want to convert @username to link and point it to their profile

我的字符串:

$timeline="@fred-ii 's posts on @stackoverflow are intresting."; 

我使用以下代码将@username 替换为 url :

I am using the following code to replace @username with url :

echo preg_replace("/@([^\s]+)/i","<a href='http://example.com/$1'>@$1</a>",$timeline);

它有效，问题是它也匹配空格

it works, the problem is that it matchs spaces also

这个字符串

"@fred-ii 's posts on@stackoverflow";

on 和 @stackoverflow 之间没有空格，我想排除它，

There is no space between on and @stackoverflow ,I want to exclude it,

所以我更新了我的正则表达式

so i updated my regex

/\s+@([^\s]+)/

它起作用了，但它与我的字符串的第一部分(用户名 @fred-ii )不匹配.我认为正则表达式引擎正在寻找字符串开头的空格.

it worked but it didnot match the first part of my string (username @fred-ii ) .I think regex engine is looking for space|s at the start of string.

我需要在我的模式中更改什么以匹配所有 @usernames ?

What do I need to change in my pattern to match all @usernames ?

$timeline="@fred-ii 's posts on @stackoverflow are intresting."; 

推荐答案

你可以使用这个lookbehind assertion:

You can use this lookbehind assertion:

/(?<=\s|^)@\S+/

(?<=\s|^) 确保 @ 之前有空格或行开始.

(?<=\s|^) makes sure there is whitespace or line start before @.

代码:

php> $s = "@fred-ii 's posts on@stackoverflow";

php> echo preg_replace('/(?<=\s|^)@\S+/', "<a href='http://example.com/$0'>$0</a>", $s);
<a href='http://example.com/fred-ii'>@fred-ii</a> 's posts on@stackoverflow

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