Question

在我的图像进入数据库之前，我将它们存储为静态，因此我将它们从数据库中提取出来。所以这是我的问题。虽然静态图像（不在数据库中）可以点击它们并自由地浏览它们，但现在我无法点击它并且我不知道从哪里开始？

图像1-静态图像（HTML）：http://puu.sh/cCywf/70ad438845.jpg

如您所见，我可以点击图片并将其放大

代码：

    <div class="wrapper">
        <div class="main">
            <div class="ser-main">
                <h2 class="style">Gallery of honda</h2>
                <div class="ser-grid-list img_style">
                    <div class="gallery1">
                        <a href="../images/ser_pic1.jpg"><img src="../images/ser_pic1.jpg" alt=""></a>
                    </div>
                </div>
                <div class="ser-grid-list img_style">
                    <div class="gallery1">
                        <a href="../images/ser_pic2.jpg"><img src="../images/ser_pic2.jpg" alt=""></a>
                    </div>
                </div>
                <div class="ser-grid-list img_style">
                    <div class="gallery1">
                        <a href="../images/ser_pic3.jpg"><img src="../images/ser_pic3.jpg" alt=""></a>
                    </div>
                </div>
                <div class="ser-grid-list img_style">
                    <div class="gallery1">
                        <a href="../images/ser_pic4.jpg"><img src="../images/ser_pic4.jpg" alt=""></a>
                    </div>
                </div>
                <div class="clear"></div>
            </div>
        </div>
    </div>

现在我已经让他们来自数据库而且它不允许我点击它们

图片：http://puu.sh/cCyzq/7d53fd37ea.jpg

代码：

            <div class="wrapper">
                <div class="main">
                    <div class="ser-main">
                        <h2 class="style">Gallery of honda</h2>
                        <div class="ser-grid-list img_style">
                            <div class="gallery1">
                                                                                        <?php
 $sql = "SELECT image FROM product where productID=1";
 $result = mysql_query($sql) or die(mysql_error($connection));
 while ($row = mysql_fetch_array($result)) //display the results
 {
 echo "<p><img src='../images/" . ($row['image']) . "'" . " " . " />"; 
 }
?>

                            </div>
                        </div>
                        <div class="ser-grid-list img_style">
                            <div class="gallery1">
                                                                                        <?php
 $sql = "SELECT image FROM product where productID=2";
 $result = mysql_query($sql) or die(mysql_error($connection));
 while ($row = mysql_fetch_array($result)) //display the results
 {
 echo "<p><img src='../images/" . ($row['image']) . "'" . " " . " />"; 
 }
?>

                            </div>
                        </div>
                        <div class="ser-grid-list img_style">
                            <div class="gallery1">
                                                                                        <?php
 $sql = "SELECT image FROM product where productID=3";
 $result = mysql_query($sql) or die(mysql_error($connection));
 while ($row = mysql_fetch_array($result)) //display the results
 {
 echo "<p><img src='../images/" . ($row['image']) . "'" . " " . " />"; 
 }
?>

                            </div>
                        </div>
                        <div class="ser-grid-list img_style">
                            <div class="gallery1">
                                                                                        <?php
 $sql = "SELECT image FROM product where productID=4";
 $result = mysql_query($sql) or die(mysql_error($connection));
 while ($row = mysql_fetch_array($result)) //display the results
 {
 echo "<p><img src='../images/" . ($row['image']) . "'" . " " . " />"; 
 }
?>

                            </div>
                        </div>
                        <div class="clear"></div>
                    </div>
                </div>
            </div>

谢谢！

Answer 1

在评论请求中发布此内容：

而不是：

echo "<p><img src='../images/" . ($row['image']) . "'" . " " . " />";

跟着：

echo "<a href='../images/".$row['image']."'><img src='../images/".$row['image']."' /></a>";

PHP / MYSQL不像HTML一样显示

1 个答案: