进阶搜索
进阶搜索
提问于 2017-01-09 15:43:47
我们已经做了一个搜索字段，您可以搜索多个成分，并找到食谱。
我们想要根据食谱分类，从搜索框中的大部分成分。
if (isset($_POST['search'])) {
$searchquery = $_POST['search'];
$vals = "'" . str_replace(",", "','", $searchquery) . "'";
$query = mysql_query("SELECT * FROM opskrifter WHERE id IN
(SELECT opskrifterid FROM ingredienser WHERE ing_name IN ('$vals'))") or die("search failed");能把它们分类吗？
编辑:菜谱表
+---------+----------+-------------+------------+------------+--+
| id | name | procedure | category | image_url | |
+---------+----------+-------------+------------+------------+--+
| 1 | Sausage | Fry it | Main dish | www....com | |
| 2 | Pizza | Bake it | Main dish | www....com | |
| 3 | Burger | Eat it | Main dish | www....com | |
+---------+----------+-------------+------------+------------+--+配料表
+---------+----------+-------------+------------+------------+--+
| id | recipeid | ing_num | ing_meas | ing_name | |
+---------+----------+-------------+------------+------------+--+
| 1 | 1 | 1 | stack | sausage | |
| 2 | 2 | 200 | g | wheat | |
| 3 | 2 | 100 | g | beef | |
+---------+----------+-------------+------------+------------+--+更新
我尝试实现了Beginner/Raymond的解决方案：
"SELECT *, COUNT(*) as `total_ingredients`
FROM opskrifter as k
, ingredienser as i
WHERE k.id = i.opskrifterid
AND i.ing_name IN ($vals)
GROUP BY k.id
ORDER BY COUNT(*) DESC"$vals = "'" . str_replace(",", "', '", $searchquery) . "'";和$searchsquery = $_POST['search']; //From the searchfield
不幸的是，搜索只考虑了第一个单词，例如：“盐，意大利面”，它显示了每个含有盐的食谱。但包含这两种成分的配方并不是排序最高的。
我错过了什么？
回答 1
Stack Overflow用户
发布于 2017-01-09 16:08:40
雅各布！
我认为这个查询可以解决您的问题。求你了试试吧。
SELECT k.id
, k.name
, COUNT(*)
FROM opskrifter k
, ingredienser i
WHERE k.id = i.opskrifterid
AND i.ing_name IN ('sausage','beef', 'wheat')
ORDER BY COUNT(*) DESC页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/41551617
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