Laravel 5 - 檢索前4個記錄每個類別的數據庫

Question

我有一個由下列分類圖庫圖像的數據庫表：

「企業」，「食品」，「公園」，「方'，'遊樂設施'，'學校'，'場館'

每個圖像都有一個分配給它的類別。

我正在構建一個主要的圖庫頁面，我要在數據庫中顯示來自這些類別中的每個類別的最新4張圖像。

有人可以幫助我如何去建立查詢？

查詢開始如下：

Bugz\GalleryImage:: 

表結構

Schema::create('gallery_images', function (Blueprint $table) { 

      //set the table engine: 
      $table->engine = 'InnoDb'; 

      //define an auto-incrementing primary key: 
      $table->increments('id'); 

      //define the general fields: 
      $table->enum('gallery', array('corporate', 'food', 'park', 'parties', 'rides', 'schools', 'venue'))->default('corporate'); 
      $table->string('title'); 
      $table->string('content')->nullable()->default(null); 

      //define the audit fields: 
      $table->timestamps(); 
      $table->softDeletes(); 

     }); 

我沒有足夠的經驗，但用雄辯的編寫更復雜的查詢。

謝謝。

Answer 1

這裏有一種方法......

DROP TABLE IF EXISTS my_table; 

CREATE TABLE my_table 
(image_id INT NOT NULL ATO_INCREMENT PRIMARY KEY 
, category ENUM('corporate', 'food', 'park', 'parties', 'rides', 'schools', 'venue') NOT NULL 
); 

INSERT INTO my_table (category) VALUES 
('corporate'), 
('food'), 
('park'), 
('parties'), 
('rides'), 
('schools'), 
('venue'), 
('rides'), 
('schools'), 
('venue'), 
('food'), 
('park'), 
('parties'), 
('rides'), 
('corporate'), 
('food'), 
('park'), 
('food'), 
('park'), 
('parties'), 
('rides'), 
('food'), 
('park'), 
('food'), 
('corporate'), 
('rides'), 
('corporate'), 
('parties'), 
('rides'), 
('corporate'), 
('food'), 
('schools'), 
('venue'), 
('venue'), 
('food'), 
('park'), 
('parties') 
; 

中間結論...

SELECT x.* 
    , COUNT(y.image_id) temp_ranks_for_y 
    FROM my_table x 
    JOIN my_table y 
    ON y.category = x.category 
    AND y.image_id >= x.image_id 
GROUP 
    BY x.image_id; 
+----------+-----------+-------------------+ 
| image_id | category | temp_ranks_for_y | 
+----------+-----------+-------------------+ 
|  1 | corporate |     5 | 
|  2 | food  |     8 | 
|  3 | park  |     6 | 
|  4 | parties |     5 | 
|  5 | rides  |     6 | 
|  6 | schools |     3 | 
|  7 | venue  |     4 | 
|  8 | rides  |     5 | 
|  9 | schools |     2 | 
|  10 | venue  |     3 | 
|  11 | food  |     7 | 
|  12 | park  |     5 | 
|  13 | parties |     4 | 
|  14 | rides  |     4 | 
|  15 | corporate |     4 | 
|  16 | food  |     6 | 
|  17 | park  |     4 | 
|  18 | food  |     5 | 
|  19 | park  |     3 | 
|  20 | parties |     3 | 
|  21 | rides  |     3 | 
|  22 | food  |     4 | 
|  23 | park  |     2 | 
|  24 | food  |     3 | 
|  25 | corporate |     3 | 
|  26 | rides  |     2 | 
|  27 | corporate |     2 | 
|  28 | parties |     2 | 
|  29 | rides  |     1 | 
|  30 | corporate |     1 | 
|  31 | food  |     2 | 
|  32 | schools |     1 | 
|  33 | venue  |     2 | 
|  34 | venue  |     1 | 
|  35 | food  |     1 | 
|  36 | park  |     1 | 
|  37 | parties |     1 | 
+----------+-----------+-------------------+ 

所以......

SELECT x.* 
    FROM my_table x 
    JOIN my_table y 
    ON y.category = x.category 
    AND y.image_id >= x.image_id 
GROUP 
    BY x.image_id 
HAVING COUNT(y.image_id) <=4 
ORDER 
    BY category 
    , image_id DESC; 
+----------+-----------+ 
| image_id | category | 
+----------+-----------+ 
|  30 | corporate | 
|  27 | corporate | 
|  25 | corporate | 
|  15 | corporate | 
|  35 | food  | 
|  31 | food  | 
|  24 | food  | 
|  22 | food  | 
|  36 | park  | 
|  23 | park  | 
|  19 | park  | 
|  17 | park  | 
|  37 | parties | 
|  28 | parties | 
|  20 | parties | 
|  13 | parties | 
|  29 | rides  | 
|  26 | rides  | 
|  21 | rides  | 
|  14 | rides  | 
|  32 | schools | 
|  9 | schools | 
|  6 | schools | 
|  34 | venue  | 
|  33 | venue  | 
|  10 | venue  | 
|  7 | venue  | 
+----------+-----------+ 
27 rows in set (0.00 sec) 

Answer 2

此代碼將retrives每個類別的4個圖像。 如果你想檢索最新的一個，那麼只需在外部查詢中加入排序順序。

DB::table('gallery_images AS GI') 
    ->where(function($query) 
     { 
      DB::table('tbl_dept_master AS GI2') 
      ->select(DB::raw(count(1))) 
      ->from('gallery_images As GI2') 
      ->where('GI2.gallery','=','GI.gallery') 
      ->where('GI2.title','>=','GI.title'); 
     }) 
    ->get(); 

如果需要更多幫助評論。 希望這將解決您的問題。