Question

我尝试创建一个脚本，循环显示与1个项目匹配的5个项目，然后显示与5个项目中的每个项目相关联的一个图像。
这是我创建的一个小提琴来说明我的问题。 http://sqlfiddle.com/#!2/17d66/1/0

我想从每个匹配的项目中选择一张图片。这是我提出的代码，但到目前为止它没有用。我使用了两个比较点，一个标签和一个系列。与产品类型和系列相关联的标签是它所属的类别。例如，iPhone 4将具有用于标签的智能手机和用作家庭的iPhone 4。任何帮助将不胜感激，我的代码如下。

<?php
    // main product
$images = "SELECT img_id FROM product_images WHERE img_tag = '$p_type' && img_family = '$p_family' LIMIT 5";
$img_list = mysql_query($images);
while($rows = mysql_fetch_array($img_list))  {
    $img_name = $rows['img_name'];
    $img_tag = $rows['img_tag'];
    $img_family = $rows['img_family'];
    $img_id = $rows['img_id'];
    //echo $img_name;


            <div>
              <img src="view.php?h=400&w=400&imgid=<?php echo $img_id; ?>" />
            </div>


}

         </div>      
  </div>

<div id="left_other">
    <div class="titleHeaders">Other matching items:</div>
    <?php 
    $match_items = "SELECT id,product_name,product_type FROM product_list WHERE name_family = '$p_family' AND id <> '$id' LIMIT 5";
    $match_q = mysql_query($match_items);   
    $count = mysql_num_rows($match_q);

    if($count > 0) {
        //get matching products
        while($items = mysql_fetch_array($match_q)) {
        $id = ($items['id']);
        echo "<p>$id</p>";
        $p_name = ($items['product_name']);
        echo "<p>$p_name</p>";
        $type = ($items['product_type']);
        echo "<p>$type</p>";
        $full = "$p_name $type";
        echo "<p>$full</p>";

        //get photos

            $other_photos = "SELECT * FROM product_images WHERE img_family = '$p_family' AND img_tag <> '$p_type' LIMIT 1"; 
            echo $other_photos;
            $other_q = mysql_query($other_photos);  
            while($pictures = mysql_fetch_array($other_q)) {
                echo "<p>----------------------</p>";
                $theid = ($pictures['img_id']);
                echo "<p>$theid</p>";
                $photo = ($pictures['img_name']);
                echo "<p>$photo</p>";
                $thetype = ($pictures['img_tag']);
                echo "<p>$thetype</p>";
                echo "<p>----------------------</p>";

            if($photo != $newname && $theid != $newid){
                if ($photo == '') {
                    $poly = 'images/pph.jpg';
                    //echo $poly;
                }else{
                    $poly = 'product_images/regular_images/'.$photo;
                    //echo $poly;
                } // end if
                $newname = $photo;
                $newid = $theid;
            }else{
                echo "<p>The photo name is the same!</p>";
            }

Answer 1

您可以使用单个查询来获取结果。这是你如何做到的。

SELECT
  pl.id,
  pl.product_name,
  pl.product_type ,
  pi.product_images
FROM   product_list AS pl
LEFT  JOIN (SELECT product_list_id , product_images FROM product_images LIMIT 1) AS `pi` ON pi.product_list_id = pl.id
WHERE pl.img_tag = '$p_type' AND pl.img_family = '$p_family'
LIMIT 5

但如果有一些表格结构，样本数据和所需的输出，那将是更好的。

修改

尝试此查询，您可以在最后放置

SELECT 
  * 
FROM
  `products` AS p 
  LEFT JOIN (SELECT MAX(img_id), img_name, img_tag, img_family FROM  `images`) AS i ON i.img_family = p.`p_family` AND i.img_tag = p.`p_tag` ;

编辑

SELECT *
FROM `products` AS p
  LEFT JOIN images as i
    ON i.img_family = p.`p_family`
group by p.p_name;

从一个项目的5个图像中选择一个图像

1 个答案: