如何在php中顯示來自mysql的圖像

Question

因此，我有PHP代碼可以從mysql調用和顯示產品。我在顯示圖片時遇到了問題。當用戶將項目添加到數據庫時（如圖所示的數據庫），當圖片上傳時，其存儲在本地主機和數據庫中的文件夾中..它將自動為圖片存儲創建一個隨機數..所以，我該如何調用要查看的圖片？ 排隊：

img /是localhost中的文件夾。

數據庫經驗：

 <?php // Run a select query to get my letest 6 items // Connect to the MySQL database include "dbconnect.php"; $dynamicList = ""; $sql = mysql_query("SELECT * FROM product ORDER BY proDate DESC LIMIT 6"); $productCount = mysql_num_rows($sql); // count the output amount if ($productCount > 0) { while($row = mysql_fetch_array($sql)){ $proID = $row["proID"]; $proName = $row["proName"]; $proPrice = $row["proPrice"]; $proDate = strftime("%b %d, %Y", strtotime($row["proDate"])); $dynamicList .= ' <div class="single-product"> <div class="product-f-image"> <img src="img/' . $proID . '.jpg" alt=""> <div class="product-hover"> <a href="#" class="add-to-cart-link"><i class="fa fa-shopping-cart"></i> Add to cart</a> <a href="single-product.php?id=' . $proID . '" class="view-details-link"><i class="fa fa-link"></i> See details</a> </div> </div> <h2>' . $proName . '</h2> <div class="product-carousel-price"> <ins>$' . $proPrice . '</ins> <del>$425.00</del> </div> </div> '; } } else { $dynamicList = "no new products"; } mysql_close(); ?> 

 <?php echo $dynamicList; ?> 

Answer 1

    <?php 
    // Run a select query to get my letest 6 items
    // Connect to the MySQL database  
    include "dbconnect.php"; 
    $dynamicList = "";
    $sql = mysql_query("SELECT * FROM product ORDER BY proDate DESC LIMIT 6");
    $productCount = mysql_num_rows($sql); // count the output amount
    if ($productCount > 0) {
        while($row = mysql_fetch_array($sql)){ 
                 $proID = $row["proID"];
                 $proImg = $row["proImg"];
                 $proName = $row["proName"];
                 $proPrice = $row["proPrice"];
                 $proDate = strftime("%b %d, %Y", strtotime($row["proDate"]));
                 $dynamicList .= '

                                <div class="single-product">
                                    <div class="product-f-image">
                                        <img src="img/' . $proImg . '" alt="">
                                        <div class="product-hover">
                                        <a href="#" class="add-to-cart-link"><i class="fa fa-shopping-cart"></i> Add to cart</a>
                                        <a href="single-product.php?id=' . $proID . '" class="view-details-link"><i class="fa fa-link"></i> See details</a>
                                        </div>
                                    </div>

                                <h2>' . $proName . '</h2>

                                    <div class="product-carousel-price">
                                    <ins>$' . $proPrice . '</ins> <del>$425.00</del>
                                    </div>                                 
                                </div>
          ';

        }
    } else {
        $dynamicList = "no new products";
    }
    mysql_close();
    ?>

Answer 2

您確定在img/1.jpg下有圖片嗎？ 您寫道，圖片將有一個隨機數，但是正如我所看到的，您正在使用數據庫主鍵標識符。