Movie Database Not Returing Large Blob Information

donaldsterling

I have been assigned to the creation of a rather small movie database. If it were a large database I would have taken the extra time to set up links to preview images on the hard drive, but since this database will contain no more than 150 preview images I will be storing them in the Large Blob column in MySql. My employer is fairly forign to the programming and opperation of PHP and MySql so I have set up a form which will allow him to update the database with ease. The problem I am having is retrieving the movie image prievew. I tried a normal query and all I got was the text output of the image, not the image itself, so I was told that I needed a Header to determine the filetype so the image would display properly. I placed the header just before the ECHO of the PICTURE and got this error.

Warning: Cannot modify header information - headers already sent by (output started at C:\Program Files\Abyss Web Server\htdocs\testdb\get.php:7) in C:\Program Files\Abyss Web Server\htdocs\testdb\get.php on line 11

Here is the script.

<?php
$connection = mysql_connect("localhost", "MYALIAS", "MYPASSWORD") or die("Error connecting to database");
mysql_select_db("MOVIES", $connection);
$result = mysql_query("SELECT * FROM movielist ORDER BY movienumber DESC", $connection) or die("error querying database");
$i = 0;
while($result_ar = mysql_fetch_assoc($result)){
echo $result_ar['movienumber'];
echo $result_ar['title'];
echo $result_ar['description'];
echo $result_ar['price'];
Header("Content-type: image/jpeg"); <--This is the error line.
echo $result_ar['picture'];
$i+=1;
}
?>

Any help would be greatly appreciated.

Thank you in advance,
Donald E. Sterling

AstroTeg

In my opinion, it would have bee a ton easier to store the images OUTSIDE of the database. I doubt you would have had this problem.

The problem is you're sending text AND the image content at the same time. This won't work hence the header error.

You have to send the text out. Then you have to send the image data with another PHP script (notice the complexity here?) The 2nd script should be designed to only send the image data. How you do that is up to you. I don't think passing the image data via URL would work the greatest. You might have to pass the ID to the 2nd script and then have it look up the image data and send it (notice the extra query call?).

Databases are for data. Files are for filesystems. Let each one do what they're best at doing.

donaldsterling

Okay. You've convinced me. I wan't to store the images on the hard disk in a folder (preferably called /images/). So here is my plan.

Create a user friendly form.
In form include an upload function for the images.
In form store text for description, title, etc. and image ID in MySql.
Create retrieval script.
In retrieval script query neccesary information for customer.
In retrieval script display results from an array.

Am I leaving anything out?

AstroTeg

The retrieval script would be needed if you were storing the image data in the database.

See, the problem is like this:

Let's say you have:
page_1.php - displays all your records AND you wish to have it display your images too

You'd need something like this as well:
loadimage.php - retrieves 1 image as specified (ideally as a get parameter). This file sends the proper image header and retrieves the desired image data. Then sends the image data.

page_1.php would call loadimage.php for every image you wish to pull from the database (it might look like img src="loadimage.php?image=123" or something like that).

Instead, if you just store the image file name, you can pull the name out of your original query and then insert it directly into an HTML image tag like this: img src="/images/" . $myimage

As for uploading images, there's not much change. Instead of putting the data in a database, you just write it out to disk and then track the file name and any other data you wish to keep on the image. You will need to take care in thinking how you'll handle duplicate image names. In some cases, you might want to overwrite the original. In other cases, you might not. It depends on how you wish your system to work.

donaldsterling

<?
$username = "MYNAME";
$password = "MYPASS";
$database = "MOVIESDB";

$list = $POST["list"];
$movienumber = $POST["movienumber"];
$title = $POST["title"];
$description = $POST["description"];
$price = $POST["price"];
$picdata = $POST["pic"];
$picdata2 = $POST["pic2"];
$picdata3 = $POST["pic3"];
$picdata4 = $_POST["pic4"];

mysql_connect("localhost",$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$picture = basename("$picdata");
$picture2 = basename("$picdata2");
$picture3 = basename("$picdata3");
$picture4 = basename("$picdata4");

if (strlen("$picdata") < 1) {
copy($FILES["pic"]['tmp_name'],
"images/".$FILES["pic"]['name']) ;

echo "Picture 1 uploaded.<br>";

} else {
echo "Picture 1 is empty.<br>";
}
if (strlen("$picdata2") < 1) {
copy($FILES["pic2"]['tmp_name'],
"images/".$FILES["pic2"]['name']) ;

echo "Picture 2 uploaded.<br>";

} else {
echo "Picture 2 is empty.<br>";
}
if (strlen("$picdata3") < 1) {
copy($FILES["pic3"]['tmp_name'],
"images/".$FILES["pic3"]['name']) ;

echo "Picture 3 uploaded.<br>";

} else {
echo "Picture 3 is empty.<br>";
}
if (strlen("$picdata4") < 1) {
copy($FILES["pic4"]['tmp_name'],
"images/".$FILES["pic4"]['name']) ;

echo "Picture 4 uploaded.<br>";

} else {
echo "Picture 4 is empty.<br>";
}

$query = "INSERT INTO $list VALUES
('$movienumber',
'$title',
'$description',
'$price',
'$picture',
'$picture2',
'$picture3',
'$picture4')";
mysql_query($query);

mysql_close();
?>

I run the above script and get the following.

Notice: Undefined index: pic in C:\Program Files\Abyss Web Server\htdocs\testdb\insert.php on line 11

Notice: Undefined index: pic2 in C:\Program Files\Abyss Web Server\htdocs\testdb\insert.php on line 12

Notice: Undefined index: pic3 in C:\Program Files\Abyss Web Server\htdocs\testdb\insert.php on line 13

Notice: Undefined index: pic4 in C:\Program Files\Abyss Web Server\htdocs\testdb\insert.php on line 14
Picture 1 uploaded.
Picture 2 uploaded.
Picture 3 uploaded.
Picture 4 uploaded.

AstroTeg

You'll need to look up a tutorial on handling uploaded files in PHP. Your image data won't be in $POST but in $FILES and requires a little more work. php.net has some good info on this here:

http://us2.php.net/manual/en/features.file-upload.php