正则表达式以匹配空格或字符串结尾 [英] Regex to Match White Space or End of String

问题描述

我正在尝试在注释文本中找到@username的每个实例，并将其替换为链接.到目前为止，这是我的PHP:

I'm trying to find every instance of @username in comment text and replace it with a link. Here's my PHP so far:

$comment = preg_replace('/@(.+?)\s/', '<a href="/users/${1}/">@${1}</a> ', $comment);

唯一的问题是正则表达式取决于@username引用后是否有空格.谁能帮助我进行调整，以便它在字符串末尾也能匹配?

The only problem is the regex is dependent upon there being whitespace after the @username reference. Can anyone help me tweak this so it will also match if it is at the end of the string?

推荐答案

尝试使用/@(\w+)/而不是/@(.+?)\s/.

这篇关于正则表达式以匹配空格或字符串结尾的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！