blocks|key|2039922|text|替换问号(?)量词("optional")，并在字符类后面添加一个%2B+("one+or+more")：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2039923|@([%5E@+]%2B)|code-block|syntax|javascript|2039924|entityMap^0|5|1|X|1|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Replace the question mark (<code>?</code>) quantifier ("optional") and add in a <code>+</code> ("one or more") after your character class:

<pre><code>@([^@ ]+)
</code></pre>

blocks|key|1254511|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1254512|'/@(\w%2B)/i'|code-block|syntax|javascript|1254513|​|1254514|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try:

<pre><code>'/@(\w+)/i'
</code></pre>

blocks|key|1253373|text|正则表达式|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253374|(%5E%7C\s)(@\w%2B)|code-block|syntax|javascript|1253375|可能就是你要找的东西。|1253376|它基本上是指，行的开始，或者一个空格，然后是一个@符号，后面跟着一个或多个单词字符。|1253377|例如。|1253378|preg_match_all('/(%5E%7C\s)(@\w%2B)/',+'@name1+kdfjd+fkjd+as@name2+@+lkjlkj+@name3',+$result);
var_dump($result[2]);|1253379|为您提供|1253380|Array
++++(
++++++++[0]+=>+@name1
++++++++[1]+=>+@name3
++++)|1253381|entityMap^0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

The regex 

<pre><code>(^|\s)(@\w+)
</code></pre>

Might be what you are after.

It basically means, the start of the line, or a space, then an @ symbol followed by 1 or more word characters.

E.g.

<pre><code>preg_match_all('/(^|\s)(@\w+)/', '@name1 kdfjd fkjd as@name2 @ lkjlkj @name3', $result);
var_dump($result[2]);
</code></pre>

Gives you

<pre><code>Array
 (
 [0] =&gt; @name1
 [1] =&gt; @name3
 )
</code></pre>

blocks|key|401536|text|用%2B替换?是可行的，但并不是您所期望的那样。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|401537|您的表达式与字符串末尾的@name不匹配。|401538|$comment+=+preg_replace('#@(\w%2B)#',+'<a+href="/$1">$0</a>+',+$comment);|code-block|syntax|javascript|401539|这应该可以做你想要的事情。\w%2B代表字母(a-zA-Z0-9)|401540|entityMap^0|1|1|4|1|0|C|5|0|0|D|3|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

Replacing <code>?</code> with <code>+</code> will work but not as you expect.

Your expression does not match <code>@name</code> at the end of string. 

<pre><code>$comment = preg_replace('#@(\w+)#', '&lt;a href="/$1"&gt;$0&lt;/a&gt; ', $comment);
</code></pre>

This should do what you want. <code>\w+</code> stands for letter (a-zA-Z0-9)

blocks|key|1253425|text|我喜欢Petah的回答，但我稍微调整了一下|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1253426|preg_replace('/(%5E%7C\s)@([\w.]%2B)/',+'$1<a+href="/users/$2">@$2</a>',+$text);|code-block|syntax|javascript|1253427|主要区别是：|1253428|1253429|不包括@符号。这只是为了显示，不应该在网址中出现|unordered-list-item|offset|length|style|CODE|1253430|允许.字符(注意：\w包含下划线)|1253431|在替换中，我在开头添加了$1以保留空白|1253432|1253433|entityMap^0|0|0|0|0|3|1|0|2|1|9|2|0|C|2|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|11|8|@$M|12|N|13|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|L|7|14|8|@$M|15|N|16|O|P]|$M|17|N|18|O|P]]|9|@]|A|$]]|$1|S|3|T|5|L|7|19|8|@$M|1A|N|1B|O|P]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]]|W|$]]

I like Petah's answer but I adjusted it slightly
<pre><code>preg_replace('/(^|\s)@([\w.]+)/', '$1&lt;a href=&quot;/users/$2&quot;&gt;@$2&lt;/a&gt;', $text);
</code></pre>
The main differences are:
<ul>
<li>the <code>@</code> symbol is not included. That's for display only, should not be in the URL</li>
<li>allows <code>.</code> character (note: <code>\w</code> includes underscore)</li>
<li>in the replacement, I added <code>$1</code> at the beginning to preserve the whitespace</li>
</ul>

blocks|key|1254587|text|我建议在匹配@之前使用后处理，然后使用一个或多个不是空格或@的字符。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1254588|“一个或多个”量词(%2B)+阻止没有提到任何的提及的匹配。|BOLD|1254589|使用lookbehind是一个好主意，因为它不仅可以防止电子邮件地址和其他不需要的子字符串的匹配，还会要求正则表达式引擎主要搜索@，然后检查前面的字符。这应该会提高模式的性能，因为空格的数量应该始终超过注释中提到的数量。|1254590|如果输入文本是多行或可能包含换行符，则添加m模式修饰符将告诉%5E匹配所有行开头。如果换行符和制表符是可能的，那么使用(?<=%5E%7C\s)@([%5E@\s]%2B)会更可靠。|1254591|代码：(Demo)|1254592|$comment+=+"@name+kdfjd+@@+fkjd+as@name+@+lkjlkj+@name";

var_export(
++++preg_replace(
++++++++'/(?<=%5E%7C+)@([%5E@+]%2B)/',
++++++++'<a+href="/$1">@$1</a>',
++++++++$comment
++++)
);|code-block|syntax|javascript|1254593|输出：(单引号来自var_export())|1254594|'<a+href="/name">@name</a>+kdfjd+@@+fkjd+as@name+@+lkjlkj+<a+href="/name">@name</a>'|1254595|entityMap|0|LINK|mutability|MUTABLE|url|https://3v4l.org/Fgvov^0|6|1|T|1|0|A|1|D|8|0|1S|1|0|L|1|U|1|1L|J|0|4|4|0|0|0|9|C|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]|$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|H]]|D|@]|E|$]]|$1|I|3|J|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|K|3|L|5|6|7|1I|8|@$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|1P|8|@]|D|@$9|1Q|A|1R|1|1S]]|E|$]]|$1|O|3|P|5|Q|7|1T|8|@]|D|@]|E|$R|S]]|$1|T|3|U|5|6|7|1U|8|@$9|1V|A|1W|B|C]]|D|@]|E|$]]|$1|V|3|W|5|Q|7|1X|8|@]|D|@]|E|$R|S]]|$1|X|3|-4|5|6|7|1Y|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]]]

I recommend using a lookbehind before matching the <code>@</code> then one or more characters which are not a space or <code>@</code>.
The &quot;one or more&quot; quantifier (<code>+</code>) prevents the matching of mentions that mention no one.
Using a lookbehind is a good idea because it not only prevents the matching of email addresses and other such unwanted substrings, it asks the regex engine to primarily search <code>@</code>s then check the preceding character. This should improve pattern performance since the number of spaces should consistently outnumber the number of mentions in comments.
If the input text is multiline or may contain newlines, then adding an <code>m</code> pattern modifier will tell <code>^</code> to match all line starts. If newlines and tabs are possible, is will be more reliable to use <code>(?&lt;=^|\s)@([^@\s]+)</code>.
Code: (<a href="https://3v4l.org/Fgvov" rel="nofollow noreferrer">Demo</a>)
<pre><code>$comment = &quot;@name kdfjd @@ fkjd as@name @ lkjlkj @name&quot;;

var_export(
 preg_replace(
 '/(?&lt;=^| )@([^@ ]+)/',
 '&lt;a href=&quot;/$1&quot;&gt;@$1&lt;/a&gt;',
 $comment
 )
);
</code></pre>
Output: (single-quotes are from <code>var_export()</code>)
<pre><code>'&lt;a href=&quot;/name&quot;&gt;@name&lt;/a&gt; kdfjd @@ fkjd as@name @ lkjlkj &lt;a href=&quot;/name&quot;&gt;@name&lt;/a&gt;'
</code></pre>

Trying to replace all occurrences of an @mention with an anchor tag, so far I have:

<pre><code>$comment = preg_replace('/@([^@ ])? /', '&lt;a href="/$1"&gt;@$1&lt;/a&gt; ', $comment);
</code></pre>

Take the following sample string:

<pre><code>"@name kdfjd fkjd as@name @ lkjlkj @name"
</code></pre>

Everything matches okay so far, but I want to ignore that single "@" symbol. I've tried using "+" and "{2,}" after the "[^@ ]" which I thought would enforce a minimum amount of matches, but it's not working.

Finding @mentions in string

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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到目前为止，我尝试用锚标记替换所有出现@提及的地方：$comment = preg_replace('/@([^@ ])? /', '<a href="/$1">@$1</a> ', $comment);取以下示例字符串："@name kdfjd fkjd as@name @ lkjlkj @name"到目前为止，一切...

问在字符串中查找@提及
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问在字符串中查找@提及EN