重建規則的陣列，以多維數組

Question

我有一個從DB從單一的表如下輸出信息的數組：

Array 
(
    [0] => stdClass Object 
     (
      [users_info_id] => 1 
      [user_id] => 374 
      [user_email] => foos@foo.com 
      [address_type] => BT 
      [firstname] => Foo 
      [lastname] => Faa 
      [vat_number] => 
      [country_code] => US 
      [address] => Jajajaja 
      [city] => KOKOKOKOKOKO 
      [state_code] => MD 
      [zipcode] => 20745 
      [phone] => 2401111111 
     ) 
    [1] => stdClass Object 
     (
      [users_info_id] => 1 
      [user_id] => 374 
      [user_email] => foos@foo.com 
      [address_type] => ST 
      [firstname] => Foos 
      [lastname] => Faas 
      [vat_number] => 
      [country_code] => US 
      [address] => JSUSUSUS 
      [city] => LASOSLSL 
      [state_code] => DC 
      [zipcode] => 1234 
      [phone] => 1234567895 
     ) 
     // ... about 500 records... 
) 

什麼我要找的是重新構建陣列的每個塊，輸出將是這樣的：

Array 
(
    [0] => stdClass Object 
     (
      [users_info_id] => 1 
      [user_id] => 374 
      [user_email] => foos@foo.com 
      [phone] => 3213213213 
      [bt] => array (
       [firstname] => Foo 
       [lastname] => Faa 
       [vat_number] => 
       [country_code] => US 
       [address] => Jajajaja 
       [city] => KOKOKOKOKOKO 
       [state_code] => MD 
       [zipcode] => 20745 
       [phone] => 2401111111 
      ) 
      [st] => array (
       [firstname] => Foos 
       [lastname] => Faas 
       [vat_number] => 
       [country_code] => US 
       [address] => JSUSUSUS 
       [city] => LASOSLSL 
       [state_code] => DC 
       [zipcode] => 1234 
       [phone] => 1234567895 
      ) 
     ) 

我甚至不知道如何開始的代碼要做到這一點，另外，如果你注意到，ST和BT鍵從哪個是展會的關鍵address_type來到在第一個陣列中，ST是「送貨地址」，BT是賬單地址，一些用戶有一個船ping和一個帳單，但有一個用戶誰有3個或更多的地址發貨...
任何幫助將不勝感激。

Answer 1

在這種情況下，我會用這樣一個循環：

$outputArray = array(); 

$previousUserId = -1; 

// Loop through all source records. 
foreach ($inputRows as $row) { 

    // If the current row has a different user id thn the row before, 
    // add the output row to the final output array and prepare a new output row. 
    // If the current row has the same user id as the row before, 
    // just add further address information. 
    // This handles also the start situation, $previousUserId = -1. 
    if ($previousUserId != $row->user_id) { 
     if ($previousUserId >= 0) { 
      $outputArray[] = $outputRow; 
     } 
     $outputRow = array(); 

     // Copy main attributes 
     $outputRow['users_info_id'] = $row->users_info_id; 
     $outputRow['user_id'] = $row->user_id; 
     $outputRow['user_email'] = $row->user_email; 
     $outputRow['phone'] = $row->phone; 
    } 

    $previousUserId = $row->user_id; 

    // Create a suitable address subarray and fill it. 
    if ($row->address_type == 'BT') { 
     $outputRow['bt'] = array(); 
     $outputRow['bt']['firstname'] = $row->firstname; 
     $outputRow['bt']['lastname'] = $row->lastname; 
     ... 
    } 
    if ($row->address_type == 'ST') { 
     // dito, but for ['st'] 
     // ... 
    } 
} 

這僅僅是一個結構，你將不得不去完成它。

代碼循環遍歷輸入表的每條記錄，我們稱之爲$inputRows。 user_id更改很重要，因爲這會啓動一個新的輸出行。只要user_id保持不變，代碼只會將更多地址類型添加到當前輸出行。所以，幾個輸入行被分組到一個輸出行。所有的輸出行然後被收集在一個$outputArray。

請注意：

1）你在你的問題顯示轉儲顯示含有數組對象。在我的回答中，我創建了一個包含數組作爲輸出的數組。通常，我更喜歡僅僅使用關聯數組，因爲它們提供了更多的選擇名稱的自由。如果你想使用對象，只需相應地修改代碼即可。 （$outputObject->name = ...而不是$outputObject['name'] = ...）

2）I 假定 user_id條件與將輸入行分組到新的輸出行有關。我希望這是正確的;-D

編輯：如果有幾個記錄地址類型，還可以加上這樣一個附加陣列層：

... 
    ... 
    // Create a suitable address subarray and fill it. 
    if ($row->address_type == 'BT') { 
     // If the array that collects several bts has not been created, create it. 
     if (!isset($outputRow['bt']) { 
      $outputRow['bt'] = array(); 
     } 

     // Create an array with the bt address data 
     $addressData = array(); 
     $addressData['firstname'] = $row->firstname; 
     $addressData['lastname'] = $row->lastname; 
     ... 

     // Add the bt address data to the collection of bts. 
     $outputRow['bt'][] = $addressData; 
    } 
    if ($row->address_type == 'ST') { 
     // dito, but for ['st'] 
     // ... 
    } 

略有更高的要求，我建議外包將地址數據收集到自己的函數的部分，以便整個代碼保持可讀性。