Recipe Finder with PHP and MySQL based on Ingredients - php

Question

I am developing a cooking recipe-website and i want to create a recipe finder based on the used incredients.
My current finder only works with 3 ingredients right.
The Finder should return the right recipe(s) based on the used incredients (should work with 1-n*)
My Tables:
CREATE TABLE IF NOT EXISTS `INGREDIENTS` (
`ingredients_id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(45) NOT NULL,
PRIMARY KEY (`ingredients_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=17 ;
CREATE TABLE IF NOT EXISTS `INGREDIENTS_POS` (
`ingredients_pos_id` int(11) NOT NULL AUTO_INCREMENT,
`ingredients_id` int(11) NOT NULL,
`ingredients_unit` varchar(20) NOT NULL,
PRIMARY KEY (`ingredients_pos_id`),
KEY `ingredients_detail_fk` (`ingredients_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=17 ;
CREATE TABLE IF NOT EXISTS `RECIPES` (
`recipes_id` int(11) NOT NULL AUTO_INCREMENT,
`title` varchar(50) COLLATE utf8_bin NOT NULL,
`text` varchar(2000) COLLATE utf8_bin NOT NULL,
`count_persons` int(11) NOT NULL,
`duration` int(11) NOT NULL,
`user_id` int(11) NOT NULL,
`date` datetime NOT NULL,
`accepted` int(11) NOT NULL DEFAULT '0',
PRIMARY KEY (`recipes_id`),
KEY `recipes_user_fk` (`user_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=88 ;
CREATE TABLE IF NOT EXISTS `RECIPES_POS` (
`recipes_pos_id` int(11) NOT NULL AUTO_INCREMENT,
`recipes_id` int(11) NOT NULL,
`ingredients_id` int(11) NOT NULL,
`ingredients_value` int(11) NOT NULL,
PRIMARY KEY (`recipes_pos_id`),
KEY `recipe_pos_rec_id` (`recipes_id`),
KEY `recipes_pos_ingredient_fk` (`ingredients_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=58 ;
My buggy Solution (doesn't support count from 1-n):
<?php
include 'db_connect.php';
$q = urldecode(mysql_real_escape_string($_GET['q']));
$parameter = explode ('$',$q);
$var = 0;
//print_r($parameter);
foreach($parameter as $ing)
{
//echo $ing;
$sql = "SELECT ingredients_id FROM INGREDIENTS WHERE name='".$ing."'";
$result = mysql_query($sql,$db) or exit('{"Data":null,"Message":null,"Code":500}');
$row = mysql_fetch_array($result);
$arr_id[$var] = $row['ingredients_id'];
$var++;
}
//print_r($arr_id);
$sql = "SELECT r.recipes_id FROM RECIPES r, RECIPES_POS rp WHERE r.recipes_id = rp.recipes_id ";
foreach($arr_id as $id)
{
$sql .= "AND rp.ingredients_id =".$id . " ";
}
//echo $sql;
$result = mysql_query($sql,$db) or exit('{"Data":null,"Message":null,"Code":500}');
mysql_close($db);
$rec;
while($row = mysql_fetch_array($result))
{
//echo "test";
$_GET['id'] = $row['recipes_id'];
$rec= include('get_recipe_byID.php');
}
//print_r(mysql_fetch_array($result));
if (count($arr_id) == 0)
{
echo '{"Data":null,"Message":null,"Code":404}';
die();
}
?>
I need a better solution for that chase.
Maybe SQL itself will help me to find the right recipes
thx

Answer 1

That query helped me a lot:
select r.recipes_id
from RECIPES r
inner join RECIPES_POS rp on r.recipes_id = rp.recipes_id
where rp.ingredients_id in (4, 6)
group by r.recipes_id
having count(distinct rp.ingredients_id) = 2