我想从phpMyAdmin数据库检索图像并将其显示在HTML表格上

Question

The images are stored as a BLOB type in my database. I want the user to be able to click on a 'View Inventory' button and be taken to a page with an HTML table that displays images for each item.

This is the table in my php file:

echo "<table border='1'>";
        echo "<tr>";
            echo "<th>Categor</th>";
            echo "<th>Description</th>";
            echo "<th>Unit Cost</th>";
            echo "<th>XS Quantity</th>";
            echo "<th>S Quantity</th>";
            echo "<th>M Quantity</th>";
            echo "<th>L Quantity</th>";
            echo "<th>XL Quantity</th>";
            echo "<th>XXL Quantity</th>";
            echo "<th>XXXL Quantity</th>";
            echo "<th>Image</th>";
            echo "<th></th>";
        echo "</tr>";
    while($row=mysqli_fetch_array($result)) {
        echo "<tr>";
            echo "<td>{$row['category']}</td>";
            echo "<td>{$row['description']}</td>";
            echo "<td>$ {$row['unitCost']}</td>";
            echo "<td>{$row['xsQuantity']}</td>";
            echo "<td>{$row['sQuantity']}</td>";
            echo "<td>{$row['mQuantity']}</td>";
            echo "<td>{$row['lQuantity']}</td>";
            echo "<td>{$row['xlQuantity']}</td>";
            echo "<td>{$row['xxlQuantity']}</td>";
            echo "<td>{$row['xxxlQuantity']}</td>";
            echo "<td><img src='Downloads/".$row['apparelImage']."'></td>";
            echo "<td><a href='#'>Add</a></td>";
        echo "</tr>";
    }
    echo "</table>";

Answer 1

If your image is stored as blob, you can't do echo "<td><img src='Downloads/".$row['apparelImage']."'></td>";

You should print a img tag which will handling your image. Pass row ID in src attribute of this tag and get it in script. Then, get the record from database and set headers to show image.

Like:

echo '<img src="script_to_show_image.php?id=' . $row['id'] . '">';

In script_to_show_image.php , after get record from database, you must set headers to print image correctly:

header("Content-Type: image/jpeg");

Then, print the content:

echo $apparelImage;