小编Kar*_*ieh的帖子

我正在尝试获取prestashop产品的图像URL,具体取决于它的ID.我正在尝试使用prestashop提供的方法的代码,但我什么都没得到

<?php

//require('config/settings.inc.php');
//require('classes/Image.php');
require('classes/Link.php');
$ItemID=$_GET["catID"];
$db=mysql_connect(constant('_DB_SERVER_'),constant('_DB_USER_'), constant


$image = Image::getCover($ItemID);
$imagePath = Link::getImageLink($product->link_rewrite, $image['id_image'], 'home_default');

这段代码不起作用,我试图添加所需的类,但它们也产生了问题.另外,如果我评论最后两行,则require本身会产生错误.我正在计划手动构建图像URL的其他工作代码正在运行

<?php
require('config/settings.inc.php');
$ItemID=$_GET["catID"];
$db=mysql_connect(constant('_DB_SERVER_'),constant('_DB_USER_'), constant

('_DB_PASSWD_')) or die('Could not connect');
mysql_select_db(constant('_DB_USER_'), $db) or die('');
 if (is_numeric($ItemID)) {
$result = mysql_query("SELECT name,date_add,price , IFNULL(imageURL,'') as imageURL 

FROM product inner join product_lang on product_lang.id_product=product.id_product 

left join app_product_image on product.id_product=app_product_image.id_product where 

id_lang=1 and product.id_product='$ItemID'") or die('Could not query');
$json = array();
    if(mysql_num_rows($result)){
           // $row=mysql_fetch_assoc($result);
        while($row=mysql_fetch_row($result)){
            //  cast results to specific data types

            //$test_data[]=$row;
                $json['name']=$row[0];
                $json['date']=$row[1];
                $json['price']=$row[2]; …