MySQL - 仅选择许多图像中的一个（随机）图像答案

【问题标题】：MySQL - select only one (random) image of manyMySQL - 仅选择许多图像中的一个（随机）图像
【发布时间】：2013-12-14 13:34:00
【问题描述】：

我被一个mysql查询卡住了，我希望你们能破解它...... 好的，所以我的查询选择了所有具有特色和活动的产品，并将它们显示在首页... 但是当产品有多个图像时会出现问题。该产品被复制然后在首页上。

正如您在结果中看到的那样，最后两个产品是相同的，但由于他分配了两张图片而重复。我想知道，仅检索此产品的一张随机图片的正确方法是什么

Tnx！ ;)

这是我的 MySql 查询：

SELECT prod.id_product, des.name, des.description,
                      price.price_net, price.discount_price, 
                      prod.url, prod.active, prodpic.id_product, prodpic.id_image,
                      img.img_path, thumb.thumb_path
                FROM products AS prod
                JOIN descriptions AS des
                ON prod.id_description = des.id_description
                JOIN prices AS price
                ON prod.id_price = price.id_price
                JOIN products_pictures AS prodpic
                ON prod.id_product = prodpic.id_product
                JOIN product_images AS img
                ON prodpic.id_image = img.id_image
                JOIN product_thumb AS thumb
                ON img.id_product_thumb = thumb.id_product_thumb
               WHERE prod.featured = 1
               AND prod.active = 1

结果是：

      0 => 
    object(stdClass)[21]
      public 'id_product' => string '1285' (length=4)
      public 'name' => string '1 item' (length=23)
      public 'description' => string 'Firs item DESC.' (length=109)
      public 'price_net' => string '32.786890' (length=9)
      public 'discount_price' => null
      public 'url' => string 'tunika-crazy-dots-white' (length=23)
      public 'active' => string '1' (length=1)
      public 'id_image' => string '1285' (length=4)
      public 'img_path' => string 'images/webshop/products/Tunika_Crazy_Dot_527fb1f252ccd.jpg' (length=58)
      public 'thumb_path' => string 'images/webshop/products/resized/Tunika_Crazy_Dot_527fb1f252ccd.jpg' (length=66)
  1 => 
    object(stdClass)[22]
      public 'id_product' => string '1300' (length=4)
      public 'name' => string 'Trol face 1' (length=11)
      public 'description' => string 'Troll face 1... Troll face 1... Troll face 1... Troll face 1... Troll face 1... Troll face 1... Troll face 1... ' (length=112)
      public 'price_net' => string '81.967213' (length=9)
      public 'discount_price' => string '50' (length=2)
      public 'url' => string 'Trol-face-1' (length=11)
      public 'active' => string '1' (length=1)
      public 'id_image' => string '1344' (length=4)
      public 'img_path' => string 'images/webshop/products/Trollface1.jpg' (length=38)
      public 'thumb_path' => string 'images/webshop/products/resized/Trollface1_thumb.jpg' (length=52)
  2 => 
    object(stdClass)[23]
      public 'id_product' => string '1300' (length=4)
      public 'name' => string 'Trol face 1' (length=11)
      public 'description' => string 'Troll face 1... Troll face 1... Troll face 1... Troll face 1... Troll face 1... Troll face 1... Troll face 1... ' (length=112)
      public 'price_net' => string '81.967213' (length=9)
      public 'discount_price' => string '50' (length=2)
      public 'url' => string 'Trol-face-1' (length=11)
      public 'active' => string '1' (length=1)
      public 'id_image' => string '1341' (length=4)
      public 'img_path' => string 'images/webshop/products/Trollface.jpg' (length=37)
      public 'thumb_path' => string 'images/webshop/products/resized/Trollface_thumb.jpg' (length=51)

【问题讨论】：

标签： php mysql distinct var-dump

【解决方案1】：

你可以试试

ORDER BY RAND()

和

GROUP BY id_image

【讨论】：

不知道你是否正确理解了问题...我希望mysql选择beatween $product[1] OR $product[2]并且只输出一个，因为这是同一个产品有两张图片...

【解决方案2】：

正如 Supun Silva 所说，答案很简单 :)

查询是：

SELECT prod.id_product, des.name, des.description,
                      price.price_net, price.discount_price, 
                      prod.url, prod.active, prodpic.id_product, prodpic.id_image,
                      img.img_path, thumb.thumb_path
                FROM products AS prod
                JOIN descriptions AS des
                ON prod.id_description = des.id_description
                JOIN prices AS price
                ON prod.id_price = price.id_price
                JOIN products_pictures AS prodpic
                ON prod.id_product = prodpic.id_product
                JOIN product_images AS img
                ON prodpic.id_image = img.id_image
                JOIN product_thumb AS thumb
                ON img.id_product_thumb = thumb.id_product_thumb
               WHERE prod.featured = 1
               AND prod.active = 1
               GROUP BY prod.id_product

【讨论】：