在注释中按名称替换 id 而不替换错误的大小写

Question

我有一条评论包含提到的 userId，如下所示：

$subject = "@1 likes @12 and @123";

现在我想用数组中的真实姓名替换这些 id：

$users = [1 => "Henry", 12 => "Tony", 123 => "Pizza"];

我陷入了 Henry 用户可以像这样替换所有包含 @1 的 id 的情况：

$subjectReplaced = "@Henry likes @Henry2 and @Henry23"

// What I want is:

$subjectReplaced = "@Henry likes @Tony and @Pizza";

请问有人帮帮我吗？

完整代码在这里：

<?php

$subject = "@1 likes @12 and @123";

$users = [1 => "Henry", 12 => "Tony", 123 => "Pizza"];

foreach ($users as $id => $user) {

  $subject = str_replace('@' . $id, '@' . $user, $subject);

}

echo $subject;

Answer 1

您可以subject根据空间分解您的空间并对preg_match您收到的每个令牌进行操作。如果令牌与您的@some_text格式匹配，我们会在subject_data下面的数组中进行替换。最后，我们只需执行 implode() 来获取替换的字符串。<?php$users = ['1' => "Henry", '12' => "Tony", '123' => "Pizza"];$subject = "@1 likes @12 and @123";$subject_data = explode(" ",$subject);foreach($subject_data as $key => $each_data){&nbsp; &nbsp; if(preg_match('/@.+/',$each_data) === 1){&nbsp; &nbsp; &nbsp; &nbsp; $subject_data[$key] = "@" . $users[substr($each_data,1)];&nbsp; &nbsp; }}echo implode(" ",$subject_data);请注意，这也可以解决重叠问题，或者您可以说前缀字符串问题，例如@1和@12。我没有使用str_replace().

Answer 2

另一种方法是使用preg_split来断开字符串，然后遍历断开的字符串的各个部分并用关联的用户 ID 替换 id，最后使用implode来连接字符串部分。$subject = "@1 likes @12 and @123";$users = [1 => "Henry", 12 => "Tony", 123 => "Pizza"];$parts = preg_split("/(\@\d*)/", $subject, 0, PREG_SPLIT_DELIM_CAPTURE);foreach ($parts as &$part) {&nbsp; &nbsp; if (strpos($part, '@') !== 0) {&nbsp; &nbsp; &nbsp; &nbsp; continue;&nbsp; &nbsp; }&nbsp; &nbsp; $part = $users[substr($part, 1, strlen($part) - 1)] ?? '';}$result = implode($parts);echo $result;结果： 亨利喜欢托尼和比萨

Answer 3

你可以用array_walk与str_replace$subject = "@1 likes @12 and @123";$patterns = ['1' => 'Henry','12' => 'Tony' ,'123' => 'Pizza'];$previousKey ='';array_walk($patterns, function($v, $k) use (&$subject, &$previousKey, $patterns){&nbsp; &nbsp; if(!empty($previousKey))&nbsp; &nbsp; &nbsp; $subject = str_replace(str_replace($previousKey, $patterns[$previousKey], $k), $v, $subject);&nbsp;else&nbsp; &nbsp; &nbsp; $subject = str_replace($k, $v, $subject);$previousKey = $k;&nbsp; &nbsp; $previousKey = $k;});echo $subject;