asked on

display images in php

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

Lukasz Chmielewski

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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Lukasz Chmielewski

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.

ASKER CERTIFIED SOLUTION

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