正则表达式以匹配空格或字符串结束

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

我试图在注释文本中找到@username的每个实例，并用一个链接替换它。下面是我到目前为止的PHP:

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

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?

唯一的问题是regex依赖于@username引用后的空格。谁能帮我调整一下，如果它在字符串的末尾，它也会匹配吗?

2 个解决方案

#1

try /@(\w+)/ instead of /@(.+?)\s/.

试试/@(\w+)/代替/@(.+?)\s/。

#2

To detect whitespace or end of string, you would use: /@(.+?)(?=\s|$)/, but unless your usernames contain non-alphanumeric characters, it's simpler to use the \w for word character or \b for word break, i.e.: /@(.+?)\b/ or /@(\w)+/.

要检测空格或字符串的结尾，可以使用:/@(.+?):/ @(. + ?)\ b / / @(\ w)+ /。