用于从数据库返回图像的PHP代码[重复]

Question

This question already has an answer here:

• When to use single quotes, double quotes, and backticks in MySQL 12 answers

I am fairly new to php and am having some trouble trying to get my images to populate correctly from my database. Any help would be appreciated.

Below is my current code. It is currently returning my default image and not my product image if in the database.

<?php

$query = "SELECT products.giftID, products.gift_name, products.price, products.short_description, product_image.picture FROM products LEFT JOIN product_image ON product_image.giftID='products.giftID' AND products.vendorID='1'";                                 
$result = mysql_query($query) or die ("Database access failed:".mysql_error());

while($row = mysql_fetch_assoc($result)) {
    if($row['picture'] == '') {
        echo "<img src=product_img/image_coming_soon.jpg>";
    } else {
        echo "<img src='".$row['picture'] ."'>";
    }                                       
}
?>  

Formatting the query for readability:

SELECT products.giftID, products.gift_name, products.price,
       products.short_description, 
       product_image.picture
  FROM products
  LEFT JOIN product_image ON product_image.giftID='products.giftID'
                         AND products.vendorID='1'

</div>

Answer 1

Your query joins the product_image table where product_image.giftID='products.giftID'.
I assume you want to match the numeric ID from the products table rather than a literal string.

I suggest removing the quotes from 'products.giftID':

SELECT products.giftID, products.gift_name, products.price, products.short_description, 
       product_image.picture
  FROM products
  LEFT JOIN product_image ON product_image.giftID = products.giftID
                          AND products.vendorID = 1