我的代碼正在工作,但我對此不滿意,看起來很髒。 我的問題是,我如何使我的代碼功能不重複。 這裏是我的代碼,如何在PHP中重複查詢數據庫的功能
<?php
// PARENT MENU
$q2 = "SELECT t.*, t2.id as m_id FROM userpage t left join menu t2 on (t2.parent=t.parent and t2.sort=t.sort) WHERE t.parent = (SELECT MIN(parent) from userpage) ORDER BY t.sort ASC";
$r2 = $db->query($q2);
$a2 = $db->fetch_all_array($q2);
foreach($a2 as $k => $v)
{
$indent = "➥";
$tab = " ";
$menu_id = $v['m_id'];
echo $v['pagename'] . "(" . $v['page'] . ")";
// 1ST SUB MENU
if ($menu_id != '')
{
$q3 = "SELECT t.*, t2.id as m_id FROM userpage t left join menu t2 on (t2.parent=t.parent and t2.sort=t.sort) WHERE t.parent = " . $menu_id . " ORDER BY t.sort ASC";
$r3 = $db->query($q3);
$a3 = $db->fetch_all_array($q3);
foreach($a3 as $k3 => $v3)
{
$menu_id2 = $v3['m_id'];
echo $v3['pagename'] . "(" . $v3['page'] . ")";
// 2ND SUB MENU
if ($menu_id2 != '')
{
$q4 = "SELECT t.*, t2.id as m_id FROM userpage t left join menu t2 on (t2.parent=t.parent and t2.sort=t.sort) WHERE t.parent = " . $menu_id2 . " ORDER BY t.sort ASC";
$r4 = $db->query($q4);
$a4 = $db->fetch_all_array($q4);
foreach($a4 as $k4 => $v4)
{
$menu_id3 = $v4['m_id'];
echo $v4['pagename'] . "(" . $v4['page'] . ")";
// 3RD SUB MENU
if ($menu_id3 != '')
{
$q5 = "SELECT t.*, t2.id as m_id FROM userpage t left join menu t2 on (t2.parent=t.parent and t2.sort=t.sort) WHERE t.parent = " . $menu_id3 . " ORDER BY t.sort ASC";
$r5 = $db->query($q5);
$a5 = $db->fetch_all_array($q5);
foreach($a5 as $k5 => $v5)
{
$menu_id4 = $v5['m_id'];
echo $v5['pagename'] . "(" . $v5['page'] . ")";
// 4TH SUB MENU
if ($menu_id4 != '')
{
$q6 = "SELECT t.*, t2.id as m_id FROM userpage t left join menu t2 on (t2.parent=t.parent and t2.sort=t.sort) WHERE t.parent = " . $menu_id4 . " ORDER BY t.sort ASC";
$r6 = $db->query($q6);
$a6 = $db->fetch_all_array($q6);
foreach($a6 as $k6 => $v6)
{
$menu_id5 = $v6['m_id'];
echo $v6['pagename'] . "(" . $v6['page'] . ")";
// 5TH SUB MENU
if ($menu_id5 != '')
{
$q7 = "SELECT t.*, t2.id as m_id FROM userpage t left join menu t2 on (t2.parent=t.parent and t2.sort=t.sort) WHERE t.parent = " . $menu_id5 . " ORDER BY t.sort ASC";
$r7 = $db->query($q7);
$a7 = $db->fetch_all_array($q7);
foreach($a7 as $k7 => $v7)
{
$menu_id6 = $v7['m_id'];
echo $v['pagename'] . "(" . $v['page'] . ")";
} //5th submenu
} //closing if for 5th submenu
} //4th submenu
} //closing if for 4th submenu
} //3rd submenu
} //closing if for 3rd submenu
} //2nd submenu
} //closing if for 2nd submenu
} //1st submenu
} //closing if for 1st submenu
} //parent menu
?>
謝謝。 任何答案是高度讚賞。
這對'遞歸'是個不錯的任務。谷歌它:) –
如果你可以向我們解釋你的代碼是做什麼的,這可能會有所幫助。評論你的代碼是一個普遍的編程勝利。嘗試並做到這一點。 – Martin
謝謝你回答,你們是對的。 我的程序是將所有頁面顯示爲菜單格式。 – Luvz