PHP が SQL ステートメントのフォーマット方法を実装する方法の詳細な分析

<?php
include(&#39;SqlFormatter.php&#39;);
$sql="SELECT DATE_FORMAT(b.t_create, &#39;%Y-%c-%d&#39;) dateID, b.title memo
  FROM (SELECT id FROM orc_scheme_detail d WHERE d.business=208
  AND d.type IN (29,30,31,321,33,34,3542,361,327,38,39,40,41,42,431,4422,415,4546,47,48,&#39;a&#39;,
  29,30,31,321,33,34,3542,361,327,38,39,40,41,42,431,4422,415,4546,47,48,&#39;a&#39;)
  AND d.title IS NOT NULL AND t_create >=
  DATE_FORMAT((DATE_SUB(NOW(),INTERVAL 1 DAY)),&#39;%Y-%c-%d&#39;) AND t_create
  < DATE_FORMAT(NOW(), &#39;%Y-%c-%d&#39;) ORDER BY d.id LIMIT 2,10) a,
  orc_scheme_detail b WHERE a.id = b.id";
echo SqlFormatter::format($sql);
?>