PHP While loop seperating unique categories from multiple 'Joined' tables

Posted by Hob on Stack Overflow See other posts from Stack Overflow or by Hob
Published on 2012-11-18T10:49:47Z Indexed on 2012/11/18 10:59 UTC
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I'm pretty new to Joins so hope this all makes sense.

I'm joining 4 tables and want to create a while loop that spits out results nested under different categories.

My Tables

categories
id | category_name

pages
id | page_name | category

*page_content*
id | page_id | image_id

images
id | thumb_path

My current SQL join

<?php $all_photos = mysql_query("
    SELECT * FROM categories JOIN pages ON pages.category = categories.id
    JOIN image_pages ON image_pages.page_id = pages.id
    JOIN images ON images.id = image_pages.image_id
");?>

The result I want from a while loop

I would like to get something like this....

Category 1
page 1
Image 1, image 2, image 3

page 2
Image 2, image 4

Category 2
page 3
image 1

page 4
image 1, image 2, image 3

I hope that makes sense.

Each image can fall under multiple pages and each page can fall under multiple categories.

at the moment I have 2 solutions, one which lists each category several times according to the the amount of pages inside them:

eg. category 1, page 1, image 1 - category 1, page 1, image 2 etc

One that uses a while loop inside another while loop inside another while loop, resulting in 3 sql queries.

    <?php 
        while($all_page = mysql_fetch_array($all_pages)) {
        ?>
            <p><?=$all_page['page_name']?></p>


            <?php $all_images = mysql_query("SELECT * FROM images JOIN image_pages ON image_pages.page_id  = " . $all_page['id'] . " AND image_pages.image_id = images.id");

            ?>

            <div class="admin-images-block clearfix">

            <?php

            while($all_image = mysql_fetch_array($all_images)) {
            ?>


                <img src="<?=$all_image['thumb_path']?>" alt="<?=$all_image['title']?>"/>


            <?php
            }

            ?>
            </div>
            <?php


        }



}

?>

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