MYSQL PHP API-在另一个表的行中显示和获取多行

Question

我正在为我的移动应用程序创建一个API。 我正在使用PHP MYSQL和Slim框架（与该问题无关）进行开发。

我试图从我的mysql数据库中提取多个“ venues”，并为每个“ venue”检索多个“ venue_images”。 数据库：

venues        venue_images
------        ------------
id PK         image_venue_id FK (to venues)
venue_name    image_path
active

然后，我需要以这种格式输出数据：

{
  "completed_in":0.01068,
  "returned":10,
  "results":[
    {
      "venue_id":"1",
      "venue_name":"NameHere",
      "images": [
         {
           "image_path":"http://www.pathhere.com"
         },
         {
           "image_path":"http://www.pathhere2.com"
         }
      ]
    }
  ]
}

因此，基本上，每个场所对图像进行多次迭代。

我当前的代码是：

$sql = "
    SELECT
         venues.id, venues.venue_name, venues.active,
         venue_images.image_venue_id, venue_images.image_path
    FROM
         venues
    LEFT JOIN 
         venue_images ON venue_images.image_venue_id = venues.id
    WHERE
         venues.active = 1
    LIMIT 0, 10
    ";

    $data = ORM::for_table('venues')->raw_query($sql, array())->find_many();
         if($data) {
        foreach ($data as $post) {
            $results[] = array (
                 'venue_id' => $post->id,
                 'venue_name' => $post->venue_name,
                 'images' => $post->image_path
            );
        }

        //Build full json
        $time = round((microTimer() - START_TIME), 5);
        $result = array(
             'completed_in' => $time,
             'returned' => count($results),
             'results' => $results
        );
        //Print JSON
        echo indent(stripslashes(json_encode($result)));
    } else {
         echo "Nothing found";
    }

我当前的代码有效，但是会产生以下结果：

{
  "completed_in":0.01068,
  "returned":10,
  "results":[
    {
      "venue_id":"1",
      "venue_name":"The Bunker",
      "images":"https://s3.amazonaws.com/barholla/venues/1352383950-qPXNShGR6ikoafj_n.jpg"
    },
    {
      "venue_id":"1",
      "venue_name":"The Bunker",
      "images":"https://s3.amazonaws.com/barholla/venues/1352384236-RUfkGAWsCfAVdPm_n.jpg"
    }
]
}

“地堡”有两个图片。 而不是将图像存储在场所数组中，而是创建带有第二张图像的“ The Bunker”重复行。 就像我之前说过的，我需要在每个场所内迭代多个图像。 任何帮助将非常感激！ 谢谢！

Answer 1

您想使用GROUP_CONCAT

像这样的东西（可能不是100％准确：））

$sql = "
    SELECT
         v.id, v.venue_name, v.active,
         GROUP_CONCAT(i.image_path) as venue_image_string
    FROM
         venues v
    LEFT JOIN 
         venue_images i ON i.image_venue_id = v.id
    WHERE
         v.active = 1
    GROUP BY i.image_venue_id
    LIMIT 0, 10
";

您可能需要摆弄一些东西，但应该把自己放在正确的轨道上（注意：以CSV格式提供venue_image_string）

Answer 2

为什么不能使用多个查询代替..？ 它更快，更简单..！

$sql = "SELECT venues.id, venues.venue_name, venues.active FROM venues WHERE venues.active = 1 LIMIT 0, 10";

    $data = ORM::for_table('venues')->raw_query($sql, array())->find_many();
         if($data) {
        foreach ($data as $post) {
            $results[] = array ();

$sql = "SELECT image_path FROM venue_images WHERE image_venue_id = $post->id"; 
$images = ORM::for_table('venue_images')->raw_query($sql, array())->find_many();

 $results[] = array (
                 'venue_id' => $post->id,
                 'venue_name' => $post->venue_name, 
                 'images' => $images);
        }

        //Build full json
        $time = round((microTimer() - START_TIME), 5);
        $result = array(
             'completed_in' => $time,
             'returned' => count($results),
             'results' => $results
        );
        //Print JSON
        echo indent(stripslashes(json_encode($result)));
    } else {
         echo "Nothing found";
    }