Question

我有一个图库图像数据库表，按以下方式分类：

＆＃39; corporate＆＃39;，＆＃39; food＆＃39;，＆＃39; park＆＃39;，＆＃39; party＆＃39;，＆＃39; rides＆＃39; ，＆＃39;学校＆＃39;，＆＃39;场地＆＃39;

每张图片都分配了其中一个类别。

我正在建立一个主画廊页面，我想在其中显示数据库中每个类别的最新4个图像。

有人可以协助我如何构建查询吗？

查询开始如下：

Bugz\GalleryImage::

表格结构

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();

        });

我还没有足够的经验与Eloquent一起编写更复杂的查询。

谢谢。

Answer 1

这是单程......

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')
;

中间结果......

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 |
+----------+-----------+-------------------+

因此...

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

此代码将检索每个类别的4张图像。如果你想要检索最新的那个，那么只需在外部查询中通过cluse添加排序。

    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();

如果需要更多帮助评论。希望这能解决你的问题。

Laravel 5 - 检索数据库中每个类别的前4条记录

2 个答案: