Question

我的评论中包含提及的userId，如下所示：

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

现在，我想用数组中的真实名称替换这些id：

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

我陷入了亨利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";

有人帮我吗？

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

您可以根据空格爆炸subject，并对收到的每个令牌进行preg_match。如果令牌与您的@some_text格式匹配，我们将在下面的subject_data数组中进行替换。最后，我们只需执行implode（）即可获取替换后的字符串。

<?php

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

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

$subject_data = explode(" ",$subject);

foreach($subject_data as $key => $each_data){
    if(preg_match('/@.+/',$each_data) === 1){
        $subject_data[$key] = "@" . $users[substr($each_data,1)];
    }
}

echo implode(" ",$subject_data);

请注意，这也可以解决重叠问题，也可以说前缀字符串问题，例如@1和@12。我不使用str_replace()的原因之一。

Answer 2

另一种使用preg_split来中断字符串的方法，然后循环遍历中断字符串的各个部分，并用相关的用户ID替换ID，最后使用implode来连接字符串的各个部分。 / p>

$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) {
    if (strpos($part, '@') !== 0) {
        continue;
    }

    $part = $users[substr($part, 1, strlen($part) - 1)] ?? '';
}


$result = implode($parts);

echo $result;

结果： 亨利喜欢托尼和比萨饼

测试here

Answer 3

您可以将array_walk与str_replace一起使用

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

Working example

Answer 4

您可以只使用strtr进行替换。 strtr对这项工作很有用，因为它先替换了较长的字符串，然后不再查看它们，从而避免了可能出现的双重替换问题。

$subject = "@1 likes @12 and @123";
$users = [1 => "Henry", 12 => "Tony", 123 => "Pizza"];
echo strtr($subject, $users);

输出：

@Henry likes @Tony and @Pizza

Demo on 3v4l.org

用注释中的名称替换ID而不替换错误的大小写

4 个答案: