有人能纠正这个我错了


Can Some one correct this where i am wrong?

描述->我只是想在我的网页上从数据库中获取图像,我已经给了BLOB类型的数据库,我在哪里存储图像,我如何获取该图像,在这段代码中,我正在获得图像的小图标与损坏的格式。

<?php
 $connection = mysqli_connect("localhost", "****",  "******","****");
 $find_data =mysqli_query($connection,"SELECT * FROM ashish " );
 while($row = mysqli_fetch_assoc($find_data)){
 $current=$row['weblink'];$asdf=$row['webrate'];$imag =$row['weblogo'];
 $ima =$row['weblogo'];
 }
 ?>
 <form action="ashishdb" method="post" enctype="multipart/form-data">
 <div><p>WebSite Link :<input type="text" name="weblink">    &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;     Websites Rating : <input type="text" name="webrate">   &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Website Logo : <input type="file"   name="weblogo"></p>
  <p align="center"><input type="submit" value="Submit" name="btn_file"></p>
  </div>
   <table style="width:100%">
   <tr>
   <th>A</th>
   <th>A</th>
   <th>A</th>
   </tr>
   <tr>
   <td><?php echo $current ?></td>
    <td><?php echo $asdf  ?></td>
     <td><img src=<?php echo $ima ?> ></td> 

        </tr>
       </table>
     </form>
     </body></html>

如果您的"ashish"表"weblogo"列包含图像的路径或url,您可以尝试:

<img src="<?php echo $ima; ?>" >

<img src="<?=$ima?>" >