如何按指定格式显示PHP字符串？附示例原始字符串

没问题！我帮你把这段PHP字符串里的语录内容整理成规范的Markdown格式，还附上PHP代码示例，让你能直接实现这个转换：

• Joy is the serious business of heaven.

  — Lewis, Clive Staples, 1964-01-01

• We were not meant to be somebody--we were meant to know Somebody

  — Piper, John Stephen, 2011-07-17

• He who sings prays twice.

  — Hipponensis, Aurelius Augustinus, 430-02-30

• The task of the modern educator is not to cut down jungles but to irrigate deserts.

  — Lewis, Clive Staples, 1943-09-23

• There is not one blade of grass, there is no color in this world that is not intended to make us...

  — [未完整记录]

<?php
$raw = "'Joy is the serious business of heaven.'.LEWIS,CLIVE STAPLES.1964-01-01. |'We were not meant to be somebody--we were meant to know Somebody'.PIPER,JOHN STEPHEN.2011-07-17.|'He who sings prays twice.'.Hipponensis,Aurelius Augustinus.430-02-30. |'The task of the modern educator is not to cut down jungles but to irrigate deserts.'.LEWIS,CLIVE STAPLES.1943-09-23. |'There is not one blade of grass, there is no color in this world that is not intended to make us...";

// 分割每条语录
$quotes = explode('|', $raw);

echo "# 精选语录集\n\n";
foreach ($quotes as $quote) {
    // 清理首尾空格
    $quote = trim($quote);
    if (empty($quote)) continue;
    
    // 拆分引用内容、作者、日期
    $quote = ltrim($quote, "'");
    $firstDotPos = strpos($quote, "'.");
    if ($firstDotPos === false) {
        // 处理不完整的条目
        echo "- \n  > " . $quote . "\n  > \n  > *— [未完整记录]*\n\n";
        continue;
    }
    
    $quoteText = substr($quote, 0, $firstDotPos);
    $rest = substr($quote, $firstDotPos + 2);
    
    // 分割作者和日期
    $parts = explode('.', $rest);
    $author = trim($parts[0] ?? '');
    $date = trim($parts[1] ?? '');
    
    // 格式化作者名（大写转首字母大写，调整逗号格式）
    $author = ucwords(str_replace(',', ', ', strtolower($author)));
    
    echo "- \n  > " . $quoteText . "\n  > \n  > *— " . $author . ($date ? ", " . $date : "") . "*\n\n";
}
?>

内容的提问来源于stack exchange，提问作者Roy Murwa