php中的SQL强制WHERE in多次查找相同的id

我试图做一个购物车的产品有定制。。。为了避免数据库中的大量查询，我尝试在查询后将结果保存在数组中。查询如下所示：

$sql = 'SELECT * 
            FROM cc_restaurants_menu 
            WHERE menu_id 
            IN ("7","50","50") 
            ORDER BY FIELD (menu_id,"7","50","50")';
$result = $conn->query($sql);

if($result->rowCount() > 0) {
    while($row = $result->fetch()) {
        $names[] = $row['menu_product'] .'<br>';
    }
}

---

提前感谢

我想您可能需要对每个菜单id进行一次查询，然后使用UNION ALL将结果连接在一起

大概是这样的：-

SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "7"
UNION ALL
SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "50"
UNION ALL
SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "50"
ORDER BY FIELD (menu_id,"7","50","50")

$sql = "SELECT * FROM cc_restaurants_menu WHERE menu_id = '".implode("' UNION ALL SELECT * FROM cc_restaurants_menu WHERE menu_id = '", id_array)."' ORDER BY FIELD (menu_id,'".implode("','", id_array)."')";

$sql = "SELECT cc_restaurants_menu.* FROM cc_restaurants_menu INNER JOIN (SELECT ".implode(" AS a_menu_id UNION ALL SELECT ", id_array) AS a_menu_id ) sub0 ON cc_restaurants_menu.menu_id = sub0.a_menu_id ORDER BY FIELD (cc_restaurants_menu.menu_id,".implode(",", id_array).")";

或者在子查询中建立所需的菜单ID，然后将其加入到主表中

SELECT cc_restaurants_menu.* 
FROM cc_restaurants_menu 
INNER JOIN
(
    SELECT 7 AS a_menu_id UNION ALL SELECT 50 UNION ALL SELECT 50
) sub0
ON cc_restaurants_menu.menu_id = sub0.a_menu_id
ORDER BY FIELD (cc_restaurants_menu.menu_id,"7","50","50")

使用内爆生成第一个查询的操作如下：-

SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "7"
UNION ALL
SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "50"
UNION ALL
SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "50"
ORDER BY FIELD (menu_id,"7","50","50")

$sql = "SELECT * FROM cc_restaurants_menu WHERE menu_id = '".implode("' UNION ALL SELECT * FROM cc_restaurants_menu WHERE menu_id = '", id_array)."' ORDER BY FIELD (menu_id,'".implode("','", id_array)."')";

$sql = "SELECT cc_restaurants_menu.* FROM cc_restaurants_menu INNER JOIN (SELECT ".implode(" AS a_menu_id UNION ALL SELECT ", id_array) AS a_menu_id ) sub0 ON cc_restaurants_menu.menu_id = sub0.a_menu_id ORDER BY FIELD (cc_restaurants_menu.menu_id,".implode(",", id_array).")";

使用内爆生成第二个查询的过程如下：-

SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "7"
UNION ALL
SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "50"
UNION ALL
SELECT * 
FROM cc_restaurants_menu 
WHERE menu_id = "50"
ORDER BY FIELD (menu_id,"7","50","50")

$sql = "SELECT * FROM cc_restaurants_menu WHERE menu_id = '".implode("' UNION ALL SELECT * FROM cc_restaurants_menu WHERE menu_id = '", id_array)."' ORDER BY FIELD (menu_id,'".implode("','", id_array)."')";

$sql = "SELECT cc_restaurants_menu.* FROM cc_restaurants_menu INNER JOIN (SELECT ".implode(" AS a_menu_id UNION ALL SELECT ", id_array) AS a_menu_id ) sub0 ON cc_restaurants_menu.menu_id = sub0.a_menu_id ORDER BY FIELD (cc_restaurants_menu.menu_id,".implode(",", id_array).")";

---

选择DISTINCT*

？不，只是尝试了一下，结果是一样的，给出2个结果而不是3如果直接从数据库调用查询，结果是什么？例如，如果我返回产品名称，我会得到“name1”（id 7）和“name2”（id 50）。。。只有2个结果，不是3个这与你的问题无关，但是为什么你要重复50这个值两次呢？这是多余的。我认为，如果你有随机id，这会导致问题，如果id的数量变得非常大，随机数量也是如此。在php中，使用内爆可以相对轻松地创建查询。好的，给我几分钟时间，我需要思考内爆看起来如何。老实说，内爆看起来并不优雅，但正在工作，因此目前它是正确的答案。。。谢谢我做了第一个查询，效果很好