php - 如何通过反复查询数据库在PHP中创建函数

Question

我的代码可以正常工作，但我对此并不满意，它看起来很脏。我的问题是，如何为我的代码创建一个不重复的函数。这是我的代码，

    <?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 = "&#10149;";
    $tab = "&emsp;";
    $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
?>

谢谢。非常感谢任何答案。

Answer 1

您可以更改代码以使用递归。执行此操作的最简单方法仍将多次调用数据库，但您不需要在代码中全部调用它们。

调用下面的函数将写出子菜单及其所有子级/孙级等...

它使用 recursion 工作在这种情况下，这意味着该函数会调用自身来完成其任务。

function subMenu($menu_id) {
    $q = "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";
    $r = $db->query($q);
    $a = $db->fetch_all_array($q);
    foreach($a2 as $k => $v)
    {
        $menu_id2 = $v['m_id'];
        echo $v['pagename'] . "(" . $v['page'] . ")";
        if ($menu_id2 != '') {
            // write out the submenu of this submenu
            subMenu($menu_id2);
        }
    }
}

---

带有缩进计数器。将 0 传递给 indent_count

function subMenu($menu_id, $indent_count) {
    $q = "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";
    $r = $db->query($q);
    $a = $db->fetch_all_array($q);
    foreach($a2 as $k => $v)
    {
        $menu_id2 = $v['m_id'];
        // You need to add in an indent based on $indent_count
        echo $v['pagename'] . "(" . $v['page'] . ")";
        if ($menu_id2 != '') {
            // write out the submenu of this submenu
            subMenu($menu_id2, $indent_count + 1);
        }
    }
}