Asked  7 Months ago    Answers:  5   Viewed   27 times

Hi I have a image table in my database. These are stored as blob along with details such as image type & name.

I am having a problem showing the image, all I get is a white box with a red cross in it. code:

<?php

include '../connection.php';

$ID = $_GET['id'];

$query = "SELECT * FROM `images` WHERE `image_id` = '$ID'";

$result=mysql_query($query);
$row = mysql_fetch_array($result);

$image = $row['image'];
$image_type= $row['image_type'];   

header("Content-type: $image_type");
print $image; 

exit;

?>

Thanks

 Answers

43

Well here is a short answer.

<?php
include '../connection.php';
$id = (int)$_GET['id'];
$query = "SELECT * FROM `images` WHERE `image_id` = '$id'";

$result=mysql_query($query);
$row = mysql_fetch_array($result);

$image = $row['image'];
$image_type= $row['image_type'];
$size = $row['image_size'];
//alternative
/* list($image, $image_type, $size) = array(
                                       $row['image'],
                                       $row['image_type'],
                                       $row['image_size']
                                      );
*/
$ext = explode('/', $image_type);
$name = $id . '.' . $ext[1]; 

header("Content-type: $image_type");
header("Content-length: $size");
header("Content-Disposition: attachment; filename=$name");

print $image;     
exit;

Check your blobtype to be a least MEDIUMBLOB which is able to store data up to 16M

Wednesday, March 31, 2021
 
rblarsen
answered 7 Months ago
71

you can't print the result from mysqli_query, it is mysqli_resource and for dumping the error you need to change mysql_error() to mysqli_error()

$username = "bob";
$db = mysqli_connect("localhost", "username", "password", "user_data");
$sql1 = "select id from user_information where username='$username'";
$result = mysqli_query($db, $sql1) or die(mysqli_error());
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) { 
    echo $row['id'].'<br>'; 
} 
Saturday, May 29, 2021
 
pamelus
answered 5 Months ago
84

Example taken from W3Schools: PHP Select Data from MySQL

<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM Persons");

echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";

while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?>

It's a good place to learn from!

Friday, June 4, 2021
 
Litty
answered 5 Months ago
39

yo need create the user "pma" in mysql or change this lines(user and password for mysql):

/* User for advanced features */
$cfg['Servers'][$i]['controluser'] = 'pma'; 
$cfg['Servers'][$i]['controlpass'] = '';

Linux: /etc/phpmyadmin/config.inc.php

Tuesday, July 13, 2021
 
ShadowZzz
answered 3 Months ago
32

Even though I posted it in the issue, I completely overlooked the fact that the image was Base64 encoded, meaning it was a string. As is, this value couldn't be used by Jasper, so it had to be decoded. This can be done one of two ways:

Change the expression to:

new java.io.ByteArrayInputStream(javax.xml.DatatypeConverter.parseBase64Binary($F{ImageField}))

or Simply convert the string to binary in the data base. In Sybase, that's the base64_decode function. Once it is converted, you can simply change the type to java.io.InputStream and it'll work.

Saturday, August 7, 2021
 
GilShalit
answered 3 Months ago
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