2017-04-05 67 views
1

我想獲取所有菜單項存儲在數據庫中作爲一個數組(兩個數組類型在一個遞歸合併)。以下是我想出了這麼遠:從數據庫構建一個多維數組

// function to push associative array elements recursively 
function array_push_assoc($array, $key, $value){ 
    $array[$key] = $value; 
    return $array; 
} 

$menu_array = $dropdown = $single = array(); 

// getting menu items (sections) 
$get_sections = Db::query("SELECT * FROM `sections` ORDER BY `place` ASC"); 

if($get_sections){ 
    foreach($get_sections as $menu_items){ 
     $menu_item_id = $menu_items['key_id']; 
     // getting sub-sections (in any) 
     $get_children = Db::query("SELECT `id_lng`, `title`, `url`, `parent` FROM `sections` WHERE `lng`.`key_id` = `sections`.`id` AND `parent` = '$menu_item_id' AND `key_type` = '1' AND `published`= '1' ORDER BY `place` ASC "); 
     if(mysqli_num_rows($get_children) > 0){ 
      foreach($get_children as $kids){ 
       // here I need to add all "kids" recursively so it will be like: 
       // [Our Services] => Array 
       // (
       //  [Service One] => service-one.php 
       //  [Service Two] => service-two.php 
       // ) 
       $dropdown = array($menu_items['title'] => array($kids['title'] => $kids['url'])); 
      } 
     } else{ 
      // if there are no "kids" to form a dropdown menu - form an array of this type: 
      // [Single Menu Item 1] => singe-menu-item-1.php 

      $single = array_push_assoc($single, $menu_items['title'], $menu_items['url']); 
     } 

我有兩個問題:

1)我已經找到了如何推動關聯數組中循環,但不能肯定該怎麼做多維數組($dropdown

2)我知道如何合併兩個數組,但我需要的是在一個循環中都要加入一個接一個 - 我非常希望得到這個類型的數組:

[Single Menu Item 1] => singe-menu-item-1.php 
[Our Services] => Array 
    (
     [Service One] => service-one.php 
     [Service Two] => service-two.php 
    ) 
[Single Menu Item 2] => singe-menu-item-2.php 
[Single Menu Item 3] => singe-menu-item-3.php 

[Contact us] => Array 
    (
     [email us] => email.php 
     [visit us] => visit.php 
     [call us] => call.php 
    ) 

回答

2

因爲它只有兩個級別,我想你將能夠只建立使用兒童環一推,是這樣的:

<?php 
$menu   = array(); 
$get_sections = Db::query("SELECT * FROM `sections` ORDER BY `place` ASC"); 

if($get_sections){ 
    foreach($get_sections as $menu_items){ 
     $menu_item_id = $menu_items['key_id']; 
     $get_children = Db::query("SELECT `id_lng`, `title`, `url`, `parent` FROM `sections` WHERE `lng`.`key_id` = `sections`.`id` AND `parent` = '$menu_item_id' AND `key_type` = '1' AND `published`= '1' ORDER BY `place` ASC "); 
     if(mysqli_num_rows($get_children) > 0){ 
      # Create a base array 
      $menu[$menu_items['title']] = array(); 
      foreach($get_children as $kids){ 
       # Push current array with new sets of 
       $menu[$menu_items['title']][$kids['title']] = $kids['url']; 
      } 
     } else{ 
      $menu[$menu_items['title']] = $menu_items['url']; 
     } 
    } 
} 

print_r($menu); 

我沒有測試過這個,我只想通過它我的頭......所以記住這一點。

+0

不要擔心不測試它。大多數PHP代碼未經測試。 – Fuser97381

+0

Rasclatt,mate - 該死的......如此怪異的容易......呃......我想的是一個完全錯誤的方向......非常感謝,陛下! – Rossitten

+1

沒問題,有時很容易過度考慮問題。我一直這樣做!如果你想要一個真正的遞歸和更復雜的解決方案,這是一個可以工作的解決方案,但它需要一個不同的方法:http://stackoverflow.com/questions/41031756/create-dynamic-menu-array-using-php/41042166#41042166 – Rasclatt