MySQL選擇查詢，子選擇作為列

Question

我目前正在使用以下PHP代碼：

$pages = $Database->query("SELECT page_id FROM pages WHERE page_parent = '0'");
$pages = $pages->fetchAll(PDO::FETCH_ASSOC);
foreach ($pages as $page_key => $page) {
    $pages[$page_key]['page_children'] = $Database->prepare("SELECT page_id FROM pages WHERE page_parent = :id");
    $pages[$page_key]['page_children']->execute(array(
        ':id' => $page['page_id']
    ));
    $pages[$page_key]['page_children'] = $pages[$page_key]['page_children']->fetchAll(PDO::FETCH_ASSOC);
}
echo '<pre>';
print_r($pages);
die();

但是，它正在做什么應該是很明顯的：

1. 選擇所有沒有父項的“頁面”記錄（page_parent = 0）
2. 選擇所有“頁面”記錄，這些記錄是步驟1中父母的孩子

有沒有一種方法可以將其組合為單個SQL語句？

更新以顯示所需的輸出：

Array
(
    [0] => Array
        (
            [page_id] => 1
            [page_children] => Array
                (
                    [0] => Array
                        (
                            [page_id] => 2
                        )

                    [1] => Array
                        (
                            [page_id] => 3
                        )

                )

        )

)

Answer 1

您可以使用INNER JOIN

  SELECT a.* 
  FROM pages a
  INNER JOIN pages b on a.page_parent = b.page_id 
  where b.parent_page ='0';

如果您還希望沒有孩子的父母可以使用左聯接

  SELECT a.* 
  FROM pages a
  LEFT JOIN pages b on a.page_parent = b.page_id and  b.parent_page ='0';

或者你可以使用工會

  SELECT a.* 
  FROM pages a
  INNER JOIN pages b on a.page_parent = b.page_id 
  where b.parent_page ='0'
  union all
  SELECT *
  FROM pages 
  WHERE page_parent = '0'