php mysql查询根据匹配关键字的密度返回结果

Question

I have a form which allows users to search a database table for recipes, they can either type the name of the recipe they are looking for or they can type in some keywords.

I am trying to figure out how i can check the records titles and return them based on the % density of keywords. so say the user searched "sweet and sour chicken" the results would find all the recipes which matched all the words first, then all the recipes with 3 of the words, then 2, then 1. Hope that makes sense :)

here is what i have

$keywords = mysqli_real_escape_string($con,$_POST['recipe-search']);
$search_keywords = str_replace("and", "",$keywords);

//Find The Latest Recipies
$searchTerms = explode(' ', $search_keywords);
$searchTermBits = array();
foreach ($searchTerms as $term) 
{
    $term = trim($term);
    if (!empty($term)) 
    {
        $searchTermBits[] = "title LIKE '%$term%'";
    }
}
$q = "SELECT * FROM tbl_recipes WHERE ".implode(' OR ', $searchTermBits);  

which checks the titles that match each of the keywords but it doesn't find out the density of the keywords in the title. Can anyone help me with this, i'm stumped!

Many Thanks

Answer 1

Try this:

$keywords = mysqli_real_escape_string($con,$_POST['recipe-search']);
$search_keywords = preg_replace('/\s{1,}(and\s*)*/', ' ', $keywords);;

//Find The Latest Recipies
$searchTerms = explode(' ', $search_keywords);
$likeStr = "`title` LIKE '%" . implode("%' OR `title` LIKE '%", $search_terms) . "%'";

$q = "SELECT *,(SELECT COUNT(*) FROM `tbl_recipes` WHERE ". $likeStr ." GROUP BY `title`) as `count` FROM `tbl_recipes` WHERE ". $likeStr .") ORDER BY `count` DESC;";