显示来自db的图像

Question

i have a table

tbl_image 
--------------
imgID(int)
imgName(varchar)
image(blob)

here is code to display image :

<?php
     $query = "SELECT * FROM tbl_image ORDER BY imgID DESC";
     $result = mysqli_query($conn, $query);
     while ($row = mysqli_fetch_array($result)){
          $imgName = $row['imgName'];
          echo '<div class="col-sm-3 gallery-grids-left">
                <div class="gallery-grid">   
                <a class="example-image-link" href="data:image/jpeg;base64,'.base64_encode($row['image'] ).'"
                     data-lightbox="example-set"
                     data-title='.$imgName.'>
                <img src="data:image/jpeg;base64,'.base64_encode($row['image'] ).'"/></a></div></div>';
                    }
                ?>

but I'm really not like to use echo ''; so I changed to

<?php
    $query = "SELECT * FROM tbl_image ORDER BY imgID DESC";
    $result = mysqli_query($conn, $query);
    while ($row = mysqli_fetch_array($result)){
    $imgName = $row['imgName'];
    ?>
        <div class="col-sm-3 gallery-grids-left">
            <div class="gallery-grid">   
                <a class="example-image-link" data-lightbox="example-set"
                    href="<?php echo '<img src="data:image/jpeg;base64,'.base64_encode($row['image'] ).'"; '?>" 
                    data-title="<?php echo imgName;?>">
                    <?php echo '<img src="data:image/jpeg;base64,'.base64_encode($row['image'] ).'"; '?>
                </a>
            </div>
        </div>
<?php }
?>

It just display only image , no title no "image block" like this image demo

Plz help me to show my mistake?and how to fix it. Many thanks,

Answer 1

HTML error in your code: wrong binding of PHP scripting

<?php
    $query = "SELECT * FROM tbl_image ORDER BY imgID DESC";
    $result = mysqli_query($conn, $query);
    while ($row = mysqli_fetch_array($result)){
    $imgName = $row['imgName'];
    ?>
        <div class="col-sm-3 gallery-grids-left">
            <div class="gallery-grid">   
                <a class="example-image-link" data-lightbox="example-set"
                    href="data:image/jpeg;base64,<?php echo base64_encode($row['image'] )?>" 
                    data-title="<?php echo $imgName;?>">
                    <img src="data:image/jpeg;base64,<?php echo base64_encode($row['image']) ?> />
                </a>
            </div>
        </div>
<?php }
?>