Question

我有两个表，products和image：

SELECT
products.id,
products.name,
MAX(image.angle) AS image_angle,
image.url as image_url,
image.width as image_width,
image.height as image_height
FROM products
JOIN `image` ON (`products`.`id` = `image`.`product_id`)    
WHERE image.width = 100
aND image.height = 200

在图像中存储的图像有不同的尺寸

在product表格中有一列名为image_url，image表格中有url列。 T 他在查询之上给出了正确的角度最大值（来自image表），但是image.url的值不正确。

结果可能是：

    (
        [id] => 308
        [name] => Volvo240
        [image__angle] => 4
        [url] => /image/308/1_100_200.png
        [image__width] => 100
        [image__height] => 200
    )

url从product表中获取值，但我希望url从image表中获取值，并且（在这种情况下）：{ {1}}

我不知道这是否重要，但我使用CodeIgniter和ActiveRecord来创建我的查询。

Answer 1

使用如下的子查询：

SELECT 
    products.id,
    products.name,
    (select 
            MAX(image.angle)
        from
            products
        WHERE
            image.width = 100 aND image.height = 200) AS image_angle,
    image.url as image_url,
    image.width as image_width,
    image.height as image_height
FROM
    products
     JOIN `image` ON (`products`.`id` = `image`.`product_id`) 
WHERE
    image.width = 100 aND image.height = 200

你可以从子查询中删除where条件（根据你的需要）。

Answer 2

如果产品表没有很多行（下面10K），您可以使用以下代码：

SELECT
    products.id,
    products.name,
    (SELECT image.angle
       FROM image
      WHERE `image`.`product_id` = `products`.`id`
        AND `image`.`width` = 100
        AND `image`.`height` = 200
      ORDER BY image.angle DESC
      LIMIT 0, 1) AS image_angle,
    (SELECT image.url
       FROM image
      WHERE `image`.`product_id` = `products`.`id`
        AND `image`.`width` = 100
        AND `image`.`height` = 200
      ORDER BY image.angle DESC
      LIMIT 0, 1) AS image_url,
    100 AS image_width,
    200 AS image_height
FROM 
    products

如果产品表包含大量行，请使用此代码以获得更好的性能：

    SELECT
        products.id,
        products.name,
        image.angle AS image_angle,
        image.url AS image_url,
        image.width AS image_width,
        iamge.height AS image_height
    FROM 
        products
    JOIN 
        (SELECT product_id, max(angle) AS max_angle
           FROM image
          WHERE width = 100 AND height = 200
          GROUP BY product_id) AS t1 ON (t1.product_id = product.id)
    JOIN 
        image ON (image.product_id = t1.product_id AND image.angle = t1.max_angle)
   WHERE 
        image.width = 100 AND 
        image.height = 200

但是上面的代码要求你的图像表的记录在（product_id，angle，width，height）上是唯一的，否则你可能会为一个产品获得多行。

Answer 3

您需要将GROUP BY添加到product.id或image.id. 同时在角度上添加ORDER BY。

SELECT
products.id,
products.name,
products.category_id,
image.angle AS image_angle,
image.url as image_url,
image.width as image_width,
image.height as image_height
FROM products

WHERE image.width = 100
aND image.height = 200
GROUP BY `products.id`
ORDER BY `image.angle` DESC

在两个表中具有相同的列名时获取正确的值 - 基于select in select

3 个答案: