Displaying images from a database.

Anon

hello all,

No all too long ago i read the article about using images and a database. It workes fine. My problem is I am storing the images in the data base by date and want to retrieve all the pics from that date, the example only displays one pic on a plain white screen and I cant seem to figure it out?

any ideas

pete

Anon

You can call the image-view script with paramters, and you can put it inside an IMG tag:
IMG src=get_image.php?Image_ID=3

Anon

Hey,

Thanks for the response but im not sure what you mean?

Maby i can explain what im doing. I have images in a database that are associated with a date. I dynamically crate a page with all the available dates having pictures. When the user clicks the date link, I would like to pull all the pictures out of the database for that date and display them all on one page.

How, I dont know. :-(
Hope you can help

thanks

Pete

Anon

HTML works with URI. The HTML code for your page contains several URI's that point to images.
These images are retreived seperately by the browser once it has finished parsing the html for the IMG tags.

So, these images cannot be supplied by the script that is generating the HTML code. Thus, you'll have to create a new script that can supply the images.

Did you read the column about "storing binary data in a database" on this site?

It uses one script to upload data, and a nother to download.

The script to download the images from the database takes a simple parameter
that is used to look up a record in a database.
In your case, you would supply the image script with a data that it can use to find the image in the database.

get it? :-)

Anon

Did you ever get this to work?
I am trying to let a customer select a from an alphabetic table to pull up movie titles. I would like to be able to pull up an image associated with the movie title as well. I have the image stored as binary in my mysql table.

Any help?

[deleted]

Hi Chris,

Why store the images in the database? Store all your images in a directory with the filenames as the 'auto_increment' field values. So for example if you have a record in the database which is id 256, then you'll have an image called '256.gif' or '256.jpg' in your images directory. Best if you either use one or the other. When your customer clicks on the link for 'Moulin Rouge' it should pass across the id (in this example 256) to another page:

echo "<a href=\"detailspage.php?id=$id\">Moulin Rouge</a>";

You should then have something like this in your details page:

echo "<img src=\"/pathtoimages/$id.gif\">";

Hope this helps. Much better than filling you database full of huge images!

Jason

Anon

Are you hard coding Moulin Rouge in this example?

What if you select the letter 'C' and it brings up all 'C' titles e.g.

Caddyshack
Christmas Vacation
Cool Hand Luke

Since I am retrieving all titles that start with 'C', since retrieving from the database is supposed to be dynamic, how do I know what to put in the echo statement?

echo "<a href=\"detailspage.php?id=$id\">Caddyshack</a>";

I guess what I am trying to ask is how do create a hypertext link from one or more of the data fields that you pull from the mysql database?

Any help is appreciated.
Thanks!

[deleted]

Hi Chris,

I see what you mean. No not hard coded, I just used Moulin Rouge as an example. Try using 3 tables. This way you can put 1 product under many categories.

Table 1. Categories: with a row corresponding to each letter in the alphabet

Table 2. Products: your actual movie products

Table 3. Assignment: which products fall under which category.

When your tables are set up, you need a little bit of code to pull the results back, something like this:

<?php
$query = mysql_query("SELECT p.id,p.title FROM products p JOIN assignment a, categories c WHERE p.id = a.pid AND c.id = a.cid AND c.id = '$cid'"); // $cid is the id called from the alphabetic link
$moviematches = mysql_num_rows($query);
if($moviematches > 0){
while($movie = mysql_fetch_array($query)){
printf("<a href=\"detailspage.php?id=%s\">%s</a>\n", $movie['id'],$movie['title']);
}
} else {
echo "Sorry there were no movies found under this category";
}
?>

Hope you're not confused. This is the way most shops work. An example of my own shop is at "http://www.facilis.co.uk/";

Hope this helps,
Jason