retrieving images from mysql database using php

Posted on 16th Feb 2014 07:03 pm by admin

So I've been trying to figure out how to store images in a mysql database, and as far as i can tell the images are stored but getting them out seems to be the problem.

when i try to go to the page on my webhost it says that

"Can not select the database: Access denied for user 'testimg'@'localhost' to database 'testImages'"

and when i goto show.php?id=1 on my local mamp install it gives me all kinds of weird symbols

ÿÀ�
â„¢
Ì
�

ÿÄ�È���

�����������
�







����������
�

�!1AQaq"2‘¡±BR#Ábr‚’Ñ¢Â3CS$á²Òsƒ“%ðñc£Ã4â³T56Dt”E&Ód¤´ÄUu•7

�

��
!1AQaqð‘¡"2±ÁÑáñBRÂbr‚¢â#3ÿÚ�
��?�Ü/é -è

i've checked everything in the code a million times and searched google and this forum for anything that can help me but i haven't been able to find something that has helped me understand what exactly is going on and why


heres the upload form

<form enctype="multipart/form-data" action="insert.php" method="post" name="changer">
<input name="MAX_FILE_SIZE" value="1500000" type="hidden">
<input name="image" accept="image/jpeg" type="file">
<input value="Submit" type="submit">
</form>


this is my insert.php that processes the image after its submitted


<?php

include './database.php';


$link = mysql_connect($host, $username, $password);
if (!$link) {
die('Could not connect: ' . mysql_error());
}

mysql_select_db ($database);


if (isset($_FILES['image']) && $_FILES['image']['size'] > 0) {

$tmpName = $_FILES['image']['tmp_name'];

$fp = fopen($tmpName, 'r');
$data = fread($fp, filesize($tmpName));

$data = addslashes($data);
fclose($fp);

$query = "INSERT INTO tbl_images ";
$query .= "(image) VALUES ('$data')";
$results = mysql_query($query, $link);

print "Thank you, your file has been uploaded.";

}else{
print "No image selected/uploaded";
}

mysql_close($link);
?>

and heres the show.php that displays the image using the id

<?php

include './database.php';

@mysql_connect($host, $username, $password) or die("Can not connect to database: ".mysql_error());

@mysql_select_db($database) or die("Can not select the database: ".mysql_error());

$id = $_GET['id'];

if(!isset($id) || empty($id)){
die("Please select your image!");
}else{

$query = mysql_query("SELECT image FROM tbl_images WHERE id='".$id."'");
$row = mysql_fetch_assoc($query);
$content = $row['image'];

header("Content-type: image/jpg");
echo $content;

}

?>

Your Answer:

Login to answer

Other forums

PHP Mysql Staff Induction System
Hi there, I'm pretty new to PHP and Mysql so could really do with being pointed in the right directi

grouping within a foreach?
hello all,

I currently have a list of products spit out from my db via a foreach loop. Right

Syntax error
hi im having a little trobble with this script
-------------------------------------------------

Code error with Index.php
Error: Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/runevid/public_

Give me all your tricks for minimizing jar file size

Hi, I'm coming close to releasing my J2ME game... I am kicking up against the 64k size barrier w

need Array help
This is what I have to do.
$teamname[1] = "Red Sox"
$teamname[2] = "Gian

foreach result into a single variable
Hi,

I have this code...

Code: [Select]foreach ($_POST['Interests'] as $interest =&

bit of help needed
Im about to sort out my registration page for my website by customising a "registration" s

Need help PLEASE
ok i have this warning showing up

Warning: in_array() [function.in-array]: Wrong datatype for

Open link with largest int string first
I have the following links i would like to open either all at once or one-by-one. How would i procee