Php正则表达式获取用户名和id

如何提取@之后的所有用户名和id并创建链接

Hello @[user name1](id:1) i'm @[user name2](id:20)

将来

找到这个：

@\[(.*?)\]\(id:(.*?)\)

替换为：

<a href="something" id ="\2">\1</a>

您将得到如下结果：

Hello <a href="something" id ="1">user name1</a> i'm <a href="something" id ="20">user name2</a>

你好，我是

此处演示：

使用此

/\[([^)]+)\]\(id:(.*?)\)/.

PHP 检查这个

---

仅获取用户名模式

/\[([^)]+)\]/

PHP

---

用

preg\u replace\u callback

检查此方法：

$data = <<<'EOD'
Hello @[user name1](id:1) i'm @[user name2](id:20)
EOD;

$pattern = '~@\[[^]]+]\K\(id:(\d+)\)~';

$linkList = array();
$path = 'http://the.path.com/index.php?u=';
$count = 0;

$result = preg_replace_callback($pattern, 
              function ($m) use (&$linkList, $path, &$count) {
                  $id = '[' . ++$count . ']';
                  $linkList[] = $id . ': ' . $path . $m[1];
                  return $id;
              }, $data);

$result .= str_repeat(PHP_EOL, 4) . implode(PHP_EOL, $linkList);

echo $result;

$data=
Array ( [0] => user name1 [1] => user name2 ) 
Array ( [0] => 1 [1] => 20 )

/\[([^)]+)\]/

<?php

    $str="Hello @[user name1](id:1) i'm @[user name2](id:20)";
    preg_match_all('/\[([^)]+)\]/', $str, $matches);
    print_r($matches[1]);

?>

Array ( [0] => user name1 [1] => user name2 )

$data = <<<'EOD'
Hello @[user name1](id:1) i'm @[user name2](id:20)
EOD;

$pattern = '~@\[[^]]+]\K\(id:(\d+)\)~';

$linkList = array();
$path = 'http://the.path.com/index.php?u=';
$count = 0;

$result = preg_replace_callback($pattern, 
              function ($m) use (&$linkList, $path, &$count) {
                  $id = '[' . ++$count . ']';
                  $linkList[] = $id . ': ' . $path . $m[1];
                  return $id;
              }, $data);

$result .= str_repeat(PHP_EOL, 4) . implode(PHP_EOL, $linkList);

echo $result;