PHP和MySQL:从数据库中获取图像

Question

我已经上传了多张图片,并且所有图片的路径都已存储在一起.

使用explode我已将它们分开,现在我希望在旋转木马中回应它们

我使用的代码是:

<?php
    $con=mysqli_connect("localhost","root","","db");
    // Check connection
    if (mysqli_connect_errno())     {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $idd = $_GET['id'];
    echo "<header id='myCarousel' class='carousel slide'>";
        /* Indicators */
        echo"<ol class='carousel-indicators'>";
            echo"<li data-target='#myCarousel' data-slide-to='0' ></li>";
            echo"<li data-target='#myCarousel' data-slide-to='1'></li>";
            echo"<li data-target='#myCarousel' data-slide-to='2'></li>";
        echo"</ol>";

        $sql = "SELECT * FROM register_office WHERE id='".$idd."'";
            $result = mysqli_query($con, $sql);
            if (mysqli_num_rows($result) > 0) 
                {
                    /* Wrapper for slides*/
                    echo"<div class='carousel-inner'>";
                        echo"<div class='item'>";
                            while($row = mysqli_fetch_assoc($result)) 
                                {
                                    $str= $row["offimage"];
                                    $array =  explode('*', $str);
                                    foreach ($array as $item) 
                                        {
                                            echo "<div class='fill'>";
                                            echo "<img src=\"http://example.com/abc/" . $item . "\" height=\"500\" width=\"2000\"/>"; 
                                            echo "</div>";
                                }   
                        echo"</div>";
                    echo"</div>";
                }
                    /*Controls*/
                    echo"<a class='left carousel-control' href='#myCarousel' data-slide='prev'>";
                        echo"<span class='icon-prev'></span>";
                    echo"</a>";
                    echo"<a class='right carousel-control' href='#myCarousel' data-slide='next'>";
                        echo"<span class='icon-next'></span>";
                    echo"</a>";
                echo"</header>";
    ?>

但它只显示一个图像.此外,当我使用下一个控件时,即使我尝试前进或后退,它也显示没有图像.