upload and display an image

whitemoss

Hi All,

i've got some problems in displaying the image from the database.my page already can upload the image/picture into database but couldn't display it..hope anyone can help me because this is urgent matter..i have to submit it in 2 days time
here is my coding:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">
<html>
<head><title>Store binary data into SQL Database</title></head>
<body>

<?php
// code that will be executed if the form has been submitted:

if ($submit) {

// connect to the database
// (you may have to adjust the hostname,username or password)

MYSQL_CONNECT("localhost","kastam","datakastam");
mysql_select_db("kastamdb");

$data = addslashes(fread(fopen($form_data, "r"), filesize($form_data)));

$result=MYSQL_QUERY("INSERT INTO Gambar (description,bin_data,filename,filesize,filetype) ".
    "VALUES ('$form_description','$data','$form_data_name','$form_data_size','$form_data_type')");

$id= mysql_insert_id(); ?>

<img src="getdata.php3?id=$id"> 
<? //print "<p>This file has the following Database ID: <b>$id</b>";

MYSQL_CLOSE();

} else {

// else show the form to submit new data:
?>

<form method="post" action="<?php echo $PHP_SELF; ?>" enctype="multipart/form-data">

<input type="hidden" name="MAX_FILE_SIZE" value="1000000">

<input type="file" name="form_data"  size="40">
<p><input type="submit" name="submit" value="submit">
</form>

<?php
}
?>

<?php
if($id) {

// you may have to modify login information for your database server:


$query = "select bin_data,filetype from binary_data where id=$id";
$result = @MYSQL_QUERY($query);

$data = @MYSQL_RESULT($result,0,"bin_data");
$type = @MYSQL_RESULT($result,0,"filetype");

//Header( "Content-type: $type");
echo $data;

};
?>
</body>
</html>

Oersoep

You display <HTML> stuff, and then start echoing the binary blob.
You should ONLY echo the image mime-types and the binary blob, or the browser doesn't understand the data.

Besides, you might not want to use the old $form_data_name and assume register_globals is on.
Use $FILES and $POST, $GET or $REQUEST instead.

whitemoss

Hi Oersoep,

Actually i didnt get what do u mean..can u please explain it or show, where should i put @ replace what u had said...

thanks

Oersoep

First of all: I'm only guessing. You say it doesn't work, but you didn't give us an error. To get that try typing the URL directly into your browser:

getdata.php3?id=5
(or whatever id you got)

I see .php3
You might want to consider an upgrade. If not, forget what I said about $_FILES and stuff because it won't work on php 3. It's OLD.

Okay, here's an exerpt of your script:

 <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> 
<html> 
<head><title>Store binary data into SQL Database</title></head> 
<body> 

<?php 

if ($submit) { 
   //  save image
} else { 
  // show form
} 
?> 

<?php 
if($id) { 
    // get data
    echo $data; 
}
?> 
</body> 
</html>

I'll tell you where it'll go wrong:

If you echo an <img> tag, the browser tries to fetch the image with a new request.
It expects a header with a mime-type like this:
Content-type: image/jpeg\r\n
(or some other image type)
And then it expects a binary blob.

What it get's when it tries to download the image is this:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> 
<html> 
<head><title>Store binary data into SQL Database</title></head> 
<body> 

*COMPLETE FORM HERE CAUSE THERE'S NO $submit*

*BINARY BLOB*

You understand that is no image. That's HTML with some binary muck attached to it.

Maybe you should make a seperate script that doesn't output anything (not even a newline) exept for the header and the image.

like:
show_image.php

<?php 
if($id) { 

// you may have to modify login information for your database server: 


$query = "select bin_data,filetype from binary_data where id=$id"; 
$result = @MYSQL_QUERY($query); 

$data = @MYSQL_RESULT($result,0,"bin_data"); 
$type = @MYSQL_RESULT($result,0,"filetype"); 

Header( "Content-type: $type"); 
echo $data; 

}
?>