子字符串拆分 - PHP

Question

所以，我想这是一个非常简单的概念，但我不确定如何实现我的预期结果。我想要的是以“@”符号开头的单词，输出时用 <span> 包围它们。

假设下面是整个字符串：

Mark wants the new app to be released on Friday, but some assets need refining so that they fit the theme @design_team.

我将如何捕获...

@design_team

...子字符串，请记住子字符串中不应包含下划线以外的字符，以帮助保持格式。

请让我知道 PHP 是否可行，如果可行，如何实现。

Answer 1

使用preg_match()

$str = "Mark wants the new app to be released on Friday, but some assets need refining so that they fit the theme @design_team.";
preg_match('/\@[a-zA-Z_]+/', $str, $matches);
print_r($matches);

输出为

Array
(
    [0] => @design_team
)

Answer 2

您可以使用正则表达式来实现这一点。这是一个例子：

$string = 'Hello @php and @regex!';

$matches = [];
preg_match_all('/@(\w+)/', $string, $matches);

var_dump($matches);

输出：

array(2) {
  [0] =>
  array(2) {
    [0] =>
    string(4) "@php"
    [1] =>
    string(6) "@regex"
  }
  [1] =>
  array(2) {
    [0] =>
    string(3) "php"
    [1] =>
    string(5) "regex"
  }
}

延伸阅读：preg_match_all.

Answer 3

使用preg_replace:

$string = preg_replace('/@\w+/', '<span>[=10=]</span>', $string);

\w 匹配单词字符（字母、数字、下划线），+ 匹配它们的序列。并在替换字符串中 [=14=] 获取匹配的子字符串。

Answer 4

我认为使用正则表达式会更容易如果每个字符串有多个@words:

$string = '@Mark wants the new app to be released @Friday, but it needs some @refining';
$didMatch = preg_match_all('/(@[^\W]+)/', $string, $matches);

if($didMatch) { 
    echo "There were " . count($matches[0]) . " matches: <br />";
    print_r($matches[0]);
} else { 
    echo "No @words in string!\n";
}

子字符串拆分 - PHP

Sub-string splitting - PHP

php

substring