Php 在MySQL中实现SQL INTERSECT时嵌套级别过高

我最后的作品之一是基于Yii的硬件目录。每个项目都可以链接到许多组

CREATE TABLE item_group (
id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
itemId INT(10) UNSIGNED NOT NULL,
groupId INT(10) UNSIGNED NOT NULL,
PRIMARY KEY (id)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci;

只需显示ItemID，其中包含用户选择的所有GroupID。下面是我的问题解决方案：

$groups = isset($_GET['groups']) ? array_merge(array_diff($_GET['groups'], array('0'=>'-')),array()) : array();
$sql = '';
$brackets = '';
$groupMaxKey = count($groups) - 1;
//some code here
for($i=0;$i<=$groupMaxKey;$i++){
    $sql .= "SELECT itemId FROM item_group WHERE groupId='".$groups[$i]."' ";
    if($i != $groupMaxKey){
        $sql .= "AND itemId IN (";
        $brackets .= ")";
    } else {
        $sql .= $brackets;
    }
}

//// 答案确实有效：

SELECT g1.itemId 
FROM ((((((((((((((((((((((((((((((((((((((( item_group g1 
INNER JOIN item_group g2 ON g1.itemId = g2.itemId)
INNER JOIN item_group g3 ON g1.itemId = g3.itemId)
INNER JOIN item_group g4 ON g1.itemId = g4.itemId)
INNER JOIN item_group g5 ON g1.itemId = g5.itemId)
INNER JOIN item_group g6 ON g1.itemId = g6.itemId)
INNER JOIN item_group g7 ON g1.itemId = g7.itemId)
INNER JOIN item_group g8 ON g1.itemId = g8.itemId)
INNER JOIN item_group g9 ON g1.itemId = g9.itemId)
INNER JOIN item_group g10 ON g1.itemId = g10.itemId)
INNER JOIN item_group g11 ON g1.itemId = g11.itemId)
INNER JOIN item_group g12 ON g1.itemId = g12.itemId)
INNER JOIN item_group g13 ON g1.itemId = g13.itemId)
INNER JOIN item_group g14 ON g1.itemId = g14.itemId)
INNER JOIN item_group g15 ON g1.itemId = g15.itemId)
INNER JOIN item_group g16 ON g1.itemId = g16.itemId)
INNER JOIN item_group g17 ON g1.itemId = g17.itemId)
INNER JOIN item_group g18 ON g1.itemId = g18.itemId)
INNER JOIN item_group g19 ON g1.itemId = g19.itemId)
INNER JOIN item_group g20 ON g1.itemId = g20.itemId)
INNER JOIN item_group g21 ON g1.itemId = g21.itemId)
INNER JOIN item_group g22 ON g1.itemId = g22.itemId)
INNER JOIN item_group g23 ON g1.itemId = g23.itemId)
INNER JOIN item_group g24 ON g1.itemId = g24.itemId)
INNER JOIN item_group g25 ON g1.itemId = g25.itemId)
INNER JOIN item_group g26 ON g1.itemId = g26.itemId)
INNER JOIN item_group g27 ON g1.itemId = g27.itemId)
INNER JOIN item_group g28 ON g1.itemId = g28.itemId)
INNER JOIN item_group g29 ON g1.itemId = g29.itemId)
INNER JOIN item_group g30 ON g1.itemId = g30.itemId)
INNER JOIN item_group g31 ON g1.itemId = g31.itemId)
INNER JOIN item_group g32 ON g1.itemId = g32.itemId)
INNER JOIN item_group g33 ON g1.itemId = g33.itemId)
INNER JOIN item_group g34 ON g1.itemId = g34.itemId)
INNER JOIN item_group g35 ON g1.itemId = g35.itemId)
INNER JOIN item_group g36 ON g1.itemId = g36.itemId)
INNER JOIN item_group g37 ON g1.itemId = g37.itemId)
INNER JOIN item_group g38 ON g1.itemId = g38.itemId)
INNER JOIN item_group g39 ON g1.itemId = g39.itemId)
INNER JOIN item_group g40 ON g1.itemId = g40.itemId)
WHERE g1.groupId='1' AND g2.groupId='2' AND g3.groupId='3' AND g4.groupId='4' AND g5.groupId='5' AND g6.groupId='6' AND g7.groupId='7' AND g8.groupId='8' AND g9.groupId='9' AND g10.groupId='10' AND g11.groupId='11' AND g12.groupId='12' AND g13.groupId='13' AND g14.groupId='14' AND g15.groupId='15' AND g16.groupId='16' AND g17.groupId='17' AND g18.groupId='18' AND g19.groupId='19' AND g20.groupId='20' AND g21.groupId='21' AND g22.groupId='22' AND g23.groupId='23' AND g24.groupId='24' AND g25.groupId='25' AND g26.groupId='26' AND g27.groupId='27' AND g28.groupId='28' AND g29.groupId='29' AND g30.groupId='30' AND g31.groupId='31' AND g32.groupId='32' AND g33.groupId='33' AND g34.groupId='34' AND g35.groupId='35' AND g36.groupId='36' AND g37.groupId='37' AND g38.groupId='38' AND g39.groupId='39' AND g40.groupId='40'

---

如果我正确理解这个问题，那么您似乎正在尝试查找不同组中ItemId的交集

我们能做的就是统计我们正在查看的组中的所有ItemID

SELECT itemId, COUNT(groupId) as CNT
FROM item_group
WHERE groupId IN (*GROUP_IDS*)
GROUP BY itemId

SELECT * FROM (
    SELECT itemId, COUNT(groupId) as CNT
    FROM item_group
    WHERE groupId IN (*GROUP_IDS*)
    GROUP BY itemId
) as TMP WHERE CNT = *NUMBER_OF_GROUP_IDS*

现在，只要一个项目不能在同一个组中出现两次，我们就可以从这个查询中拉出CNT等于数字的行 我们正在查看的组的数量

SELECT itemId, COUNT(groupId) as CNT
FROM item_group
WHERE groupId IN (*GROUP_IDS*)
GROUP BY itemId

SELECT * FROM (
    SELECT itemId, COUNT(groupId) as CNT
    FROM item_group
    WHERE groupId IN (*GROUP_IDS*)
    GROUP BY itemId
) as TMP WHERE CNT = *NUMBER_OF_GROUP_IDS*

---

这应该可以做到。

您可以通过使用内部联接来实现这一点。没有理由嵌套这些语句

对于您的示例案例，任何适当查询的示例如下：

选择g1.itemId
来自（（项目组）g1
内部联接项（g1上的g2组。itemId=g2。itemId）
内部连接项目（g1上的g3组。项目ID=g3。项目ID）
其中g1.groupId='31'和g2.groupId='24'以及g3.groupId='35'

我在一个简单的表上测试了这一点，该表有三列

（id、itemId、groupId）

，它可以工作。将这类语句放入循环中非常容易，并且联接的数量没有最大值

为了加快运行速度，您应该为

item_组

表中的

itemId

列编制索引

可以使用以下SQL语句执行此操作：

ALTER TABLE item\u组添加索引（itemId）

您的答案也有效，现在如果在创建关系时防止重复行，效果会更好。我说得对吗？我不确定是我的解决方案还是内部连接的性能更好。这本身就是一个有趣的问题。在我看来，SQL越紧凑，就越容易理解。有时可读性比性能更重要。为什么您使用了Drewch的解决方案，但接受了hackattack的解决方案PGlad它对您有效，如果您添加索引，它将非常快。我还建议为groupId添加一个索引。当我的网站上充斥着产品时，我将尝试回答这两种解决方案的速度有多快。我认为这有助于很多人做出正确的选择