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display images in php

Posted on 2013-01-01
3
287 Views
Last Modified: 2013-01-16
facing problem in displaying the images that are stored in databases the following code is to display the images

<?php

$link = mysql_connect('localhost', 'root', '');
if (!$link) {
    die('Not connected : ' . mysql_error());
}

$db_selected = mysql_select_db('upload', $link);

if (!$db_selected) {
    die ('Database error : ' . mysql_error());
}

        $sql = "SELECT * FROM images";
 
       
        header("Content-type: image/jpeg");
        echo mysql_result($result, 0,'imageData');
 
     
?>
the following code is to upload the images in the database
<?php

$link = mysql_connect('localhost', 'root', '');
if (!$link) {
    die('Not connected : ' . mysql_error());
}

$db_selected = mysql_select_db('upload', $link);
if (!$db_selected) {
    die ('Database error : ' . mysql_error());
}

$maxFileSize = "1000000";

$image_array            = array("image/jpeg","image/jpg","image/gif","image/bmp","image/pjpeg","image/png");

$fileType = $_FILES['userfile']['type'];

 
$msg = '';


if(@$_POST['Submit'])
{


if (in_array($fileType, $image_array))
{


 if(is_uploaded_file($_FILES['userfile']['tmp_name']))
 {
 
 
        if($_FILES['userfile']['size'] < $maxFileSize)
            {
 

                  $imageData =addslashes (file_get_contents($_FILES['userfile']['tmp_name']));
 
        $sql = "INSERT INTO images ( imageData) VALUES ('$imageData')";
 
             
        mysql_query($sql) or die(mysql_error());
            $msg = " Data successfully uploaded";
         }
     else
       {
     
           $msg = ' Error :  File size exceeds the maximum limit ';
             
           }
   }
}
else
{
$msg = 'Error: Not a valid image ';
}

}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"
  "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
 
<html>
     <body>
       <span style="color:#FF0000"><?php echo $msg; ?></span><br />

        <h3>Select File to upload</h3>
 
        <form enctype="multipart/form-data" action="" method="post">
             <input name="userfile" type="file" />
            <input type="submit" value="Submit"  name="Submit"/>
        </form>
    </body>
</html>
i am able to upload the images here in the database but i am unable to display all the images that are stored in the database.help me out experts
0
Comment
Question by:codeoxygen
  • 2
3 Comments
 
LVL 27

Expert Comment

by:Lukasz Chmielewski
ID: 38735888
Try to display it like this:

$db_selected = mysql_select_db('upload', $link);

if (!$db_selected) {
    die ('Database error : ' . mysql_error());
}

        $sql = "SELECT * FROM images";
        $result = mysql_query($sql);
        $row = mysql_fetch_array($result);
        
        header("Content-type: image/jpeg");
        echo $row['imageData'];
 
     
?>

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0
 
LVL 27

Expert Comment

by:Lukasz Chmielewski
ID: 38735896
BUT I would definetely recommend to store the images as files having a link in the database to the path where they are. The database can rapidly grow to big size. If that is just for learning purposes, the code provided should be ok.
0
 
LVL 9

Accepted Solution

by:
rinfo earned 500 total points
ID: 38735970
You have mentioned
$sql = "INSERT INTO images ( imageData) VALUES ('$imageData')";
besides you have also mentioned
  $sql = "SELECT * FROM images";
which indicates you have just one field imageData in the images table - of course there could be an id field.
Does it also mean you have only one image in the images tables.
Because you have mentioned
   $sql = "SELECT * FROM images";
        header("Content-type: image/jpeg");
        echo mysql_result($result, 0,'imageData');
if there are multiple rows in the table you have to iterate through the result array to
display images one by one.
I think this is where you are failing.
0

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