upload image in mysql database

faamugol · #1 (**permalink**) 10-16-04, 13:54

Hello gys,

I use the following code to store image in mysql databse..

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head>
<title>Untitled Document</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<?php
if($_POST)
{
if(!is_array($_FILES) || !is_array($_FILES['visitor_photo']))
{
die("no file was uploaded");
}
$tmp = $_FILES['visitor_photo']['tmp_name'];
$fp = fopen($tmp, 'rb');
$image = fread($fp, filesize($tmp));
fclose($fp);
$name = $_POST['visitor_name'];
$db = @mysql_connect("localhost", "myusername", "mypassword") or die (mysql_error());
@mysql_select_db("competence", $db) or die (mysql_error());

// gzip file
//$image = gzdeflate($image, 9);
$image = mysql_real_escape_string($image, $db);
$result = @mysql_query("INSERT INTO `tb_member` (`nom`, `photo`) VALUES ('$name' ,'$image')", $db);

if($result === false)
{
die("could not insert into database");
}
unlink($tmp);
exit("upload successful");
}
?>
<form action="" method="post" enctype="multipart/form-data" name="fjoindre" id="fjoindre">
Name:
<input type="text" name="visitor_name">
Image:
<input name="visitor_photo" type="file" id="visitor_photo">
<input type="submit" value="Send" name="Submit">
</form>
</body>
</html>

but I just need to know what should the type of the field Photo be in the database.

I have defined the table tb_member as:
v_ib : int not null
nom: varchar(40) not null
photo: longtext not null

It seems to work but how can I get this image from my database to show it on my pages.

Thank in advance for all your answers..

savbill · #2 (**permalink**) 10-17-04, 14:00

Most of the programs I've used for image management just store the image name and path in the database and the image is located in a directory on your server. I think it would be slower to extract the image from the database and then display vs. using print or echo to write the path/name into the html output.

/

Dzilly · #3 (**permalink**) 11-03-04, 15:39

This is a method I use to show images stored in a MySQL db using PHP

$type is something like 'image/jpeg'
$name is just the name of the image
$image_data is the binary data of the image

header('Content-type: $type);
header("Content-Disposition: attachment; filename='$name'");
header("Content-Description: PHP Generated Data");
echo $image_data;

This works fine for me, hope you can use it...

nitrog7 · #4 (**permalink**) 11-16-04, 16:27

I believe the type of field is a "blob"

-G

sstoyan · #5 (**permalink**) 11-28-04, 09:44

straight from the mysql docs the types for blobs are

TINYBLOB
TINYTEXT
A BLOB or TEXT column with a maximum length of 255 (2^8 - 1) characters.

BLOB
TEXT
A BLOB or TEXT column with a maximum length of 65535 (2^16 - 1) characters

MEDIUMBLOB
MEDIUMTEXT
A BLOB or TEXT column with a maximum length of 16777215 (2^24 - 1) characters

LONGBLOB
LONGTEXT
A BLOB or TEXT column with a maximum length of 4294967295 or 4G (2^32 - 1) characters

---------------------
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www.viewmoresoft.com
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Nanaki · #6 (**permalink**) 11-29-04, 07:26

I posted the (almost) exact same question some days ago (just below this post), and I'll try the code you posted Dzilly

Code:

$type is something like 'image/jpeg'
$name is just the name of the image
$image_data is the binary data of the image

header('Content-type: $type);
header("Content-Disposition: attachment; filename='$name'");
header("Content-Description: PHP Generated Data");
echo $image_data;

Hope it works.
Do you output is as : image?id=(id from db) ?

And if you'll have to output some text as well, how would you alter the code?
Just add "echo $myTextVariable" ?