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我如何能實現搜索功能:我的SQL搜索語法實現
where sub_menu_name like '%".$kws."%'
到我的工作之一:
SELECT DISTINCT p.id, p.sub_menu_id, p.sub_menu_name, m.image_id, i.file_url, m.default_menu_id, p.restaurant_id, p.status, p.sub_menu_price
FROM sub_sub_menu AS p
INNER JOIN menu AS m ON m.default_menu_id = p.sub_menu_id
OR m.id = p.sub_menu_id
INNER JOIN icon AS i ON i.id = m.image_id
WHERE p.restaurant_id = '" . (int) $_SESSION['uid'] . "' "
感謝您的幫助!
在末尾添加'AND'sub_menu_name like'%'。$ kws。「%'' – Mihai