Cannot display multiple images stored in DB

NICKKKKK

Hello all,

I have a DB table where many images have been stored as longblob datatype. When I try to display them, only the first one shows up.

I am using the following code:

$sql = "SELECT * FROM images";
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result) > 0) {
$row = @mysql_fetch_array ($result);

$image1 = $row["image1"];

}

while ($row = mysql_fetch_array($result))

{
$image1 = $row["image1"];

print $image1;
}

either with while loop or without the result is the same.

What do you think I should do?

thanks
Nick

leatherback

First of all, you could try to not piss people off by stick to one thread on this topic...

Secondly.. How about reading the replies to your earlier question

Third, you are now doing things double:

this should be enough:


$sql = "SELECT * FROM images";
$result = mysql_query ($sql, $conn);
if (mysql_num_rows ($result) > 0) 
  {
  while ($row = mysql_fetch_array($result))
    {
    echo $row["image1"];
    }
}

NICKKKKK

I am sorry for that leatherback. I didnt see it and i thought it was not posted.

I have tried that before and still displays only the first.

leatherback

ok, sorry..

Just got annoyed.. This was the third thread I founf that was double in one assisting session.. Just thought that the whole world was asking questions without reading the answers.

Cheers, will have a look in the other thread, and reply there.

NICKKKKK

no worries leatherback...why can't PHP handle images nicely with DB? ggggrrrr stupid problem and the worse is that it works by storing the images in the folder in the server but my hosting company does not accept that 🙁

leatherback

Hey Nick,

Php and MySQL can easily handle images nicely. If you have no alternative than storing them in the database, bcause of permission issues -I am not 100% sure about that, to be honest, I think it is a paths issue somewhere- you can run it of a database. You will just have to create an additional 'showfile.php' that uses a $_GET variablestring to determine which image to show, retrieve it from the database, and show the image.

That is exactly the system I use for plotting information onto a page which need to be in a graph (e.g., here)

sirlemons

This code should fix your problem(I had a PHP reference book handy):

<?php
	$sql = "SELECT * FROM images"; 
	$result = mysql_query ($sql, $conn); 

$row = mysql_fetch_array( $result );

while ($row = mysql_fetch_array($result)){
	for($num=0;$num<=1000;$num++){
		echo "{@$row[$num]}" 
	}
}
?>

NogDog

Browsers retrieve and display each image via separate HTTP requests. If you output the image data for multiple images in one script, then you are trying to send them all to the browser in one request. What I suspect is happening is that the browser starts receiving the image data, reads the first part of it to determine what kind of image it is (and probably other meta-data like how many bytes it is), then when it gets to the end of the first image's data, it regards any further incoming data as invalid and discards it.

Instead, you will need an 'image server' script to serve up individual images, then your main page will instead output multiple <img> tags that each call that server script, probably with a value in the URL query string to specify the image id, which will then be used in the DB query to retrieve the image data. This means each image's display will require a separate database connection, query, retrieval, and output. This is one of the main reasons it is usually more practical to store the images as files and just use the database to store data about those images, including the file name (and possibly directory) needed to populate your <img> tags.

laserlight

This code should fix your problem

Unfortunately, it is not correct, unless the images table has 1001 columns.

sirlemons

laserlight wrote:
Unfortunately, it is not correct, unless the images table has 1001 columns.

Actually, the code will work as it will echo nothing when it requests to echo something that doesn't exist.

leatherback

sirlemons wrote:
Actually, the code will work as it will echo nothing when it requests to echo something that doesn't exist.

that is the silliest argument I ever heard.

First of all.. If it doesn't give output (Even though your error logs will fill up like nothing) it does not mean the code is correct. the basic reasoning behind your code is flawed.

Second: As before: A page can only contain one type of content. And if you output the content image/jpg then you can only output ONE image for a file. In order to display multiple files, you need to create a 'container file' which display images, and each image is loaded through an image-generation script.

NICKKKKK

Hello guys,

it is must easier to manipulate images using the move_uploaded_file(). I am trying to sort it out with my hosting now.

Thanks
Nick

sirlemons

leatherback wrote:
that is the silliest argument I ever heard.

First of all.. If it doesn't give output (Even though your error logs will fill up like nothing) it does not mean the code is correct. the basic reasoning behind your code is flawed.

Second: As before: A page can only contain one type of content. And if you output the content image/jpg then you can only output ONE image for a file. In order to display multiple files, you need to create a 'container file' which display images, and each image is loaded through an image-generation script.

OK, sorry, I must have gotten mixed up a bit, I thought you wanted to echo some html code that embeds an image on a page using a DB.