查看以blob数据类型的形式从数据库收到的图像到php中的新页面

Question

I have made an code which is used to receive image from database and it's working perfectly. But now as user click's on the image it should be opened on a new page,and i have tried few thing but it's not working...

the below code id used to display the image

  $query=mysqli_query($conn,"select * from images") or die("unable to 
       connect");




$i=1;
                while($row=mysqli_fetch_array($query,MYSQLI_BOTH))
                    {
                    //echo '<img id="my" height="150" width="320" src="data:image;base64,'.$row['image'].'  "> ';
                    echo '
                    <tr>

                    <t>'.@$row["id"].'</td>
                    <a target="_blank" href="09.jpg">
                    <img height="150" width="150" src="data:image/jpeg;base64,'.base64_encode( $row['image'] ).'"/>

                    <td><a href="delete1.php?id='. $row['id'] .'">x</a></td>
                    </td>

        ';

            $i++;


                            }   


                ?>

Now here i want is when user click's on the image , a new page will show an enlarged image...

Answer 1

Don't save image in database, save it as file in server then save the file name into database

You must be save image as two sizes, small and normal, the smal for display in table, when the user click on image, go to normal image

You can save one name in database, example (img1.jpg) And in images directory save (sm-img1.jpg) and (img1.jpg)

The code

<a href="url/mydirectory/$row['image']" target="_blank">
<img width="150" height="150"  src="url/mydirectory/sm-$row['image']" />
</a>

Or

<img width="150" height="150"  src="url/mydirectory/sm-$row['image']" onclick="window.open('url/mydirectory/$row['image']', '_blank');" />