Question

我有一个多对多的数据库，这三个表，电影，氛围，电影_Ambience：

CREATE TABLE Films (  
id INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),    
Title VARCHAR(255));

CREATE TABLE Ambiences (  
id INT NOT NULL AUTO_INCREMENT,   
PRIMARY KEY(id),  
ambienceName VARCHAR(255));

CREATE TABLE Films_Ambiences (
film_id INT NOT NULL,  
ambience_id INT NOT NULL,  
PRIMARY KEY (film_id, ambience_id),  
FOREIGN KEY (film_id) REFERENCES Films(id) ON UPDATE CASCADE,  
FOREIGN KEY (ambience_id) REFERENCES Ambiences(id) ON UPDATE CASCADE);

我正在使用表格中的信息来搜索特定的电影（例如同时有趣且可怕的电影）。表单只是简单的＆＃39; ticks＆＃39;给定名称旁边。信息由$ _POST发送。

问题在于我不知道会有多少要求。我知道用户可以选择的最大数量，但我不知道他们实际会选择多少或哪些（我可以通过检查isset($_POST['somethin'])来做到这一点但是如果我有，例如20，那将是非常单调的不同的选择。所以我不能做任何事情：

$ambience1 = $_POST["a1"];
$ambience2 = $_POST["a2"];
$ambience3 = $_POST["a2"];
...
...
...

和：

SELECT *,GROUP_CONCAT(ambienceName SEPARATOR ' ') AS ambiences
FROM Films AS f 
INNER JOIN Films_Ambiences as fa ON f.id = fa.film_id           
INNER JOIN Ambiences AS a ON a.id = fa.ambience_id
GROUP BY Title
HAVING (ambiences LIKE '%$ambience1%' AND ambiences LIKE '%$ambience2%' AND ...

我甚至不知道从哪里开始。我可以用SQL或PHP来做吗？

如果你喜欢的话，这里有 SQLFiddle 。

Answer 1

我认为您可以使用PHP来构建SQL，如下所示：

$ambienceWhere = '1=1 ';
for ($i=0;$i<NUMBER_OF_POSSIBLE_AMBIENCES;$i++) {
  if (isset($_POST['a' . $i])) { // or another criteria to avoid processing this one       
    $ambienceWhere .= ' AND '; 
    $ambienceWhere .= ' a.ambiences LIKE \'%' . $_POST['a'. $i] '%\' ';
  }
}
$query = 'SELECT ....... WHERE ('. $ambienceWhere.') .....'

Answer 2

您的表单应该具有元素数组中的氛围，可能是多重选择。然后，这将变成PHP中的数组$_POST['ambience'][]。然后你可以写：

$ambience_query = implode(' AND ', array_map(function($a) use($mysqli) {
    return "'%" . mysqli_real_escape_string($mysqli, $a) . "'";
}, $_POST['ambience']));

$query = "SELECT *, GROUP_CONCAT(ambienceName SEPARATOR ' ') AS ambiences
FROM FROM Films AS f 
INNER JOIN Films_Ambiences as fa ON f.id = fa.film_id           
INNER JOIN Ambiences AS a ON a.id = fa.ambience_id
GROUP BY Title
HAVING $ambience_query";

这是一个非常昂贵的查询。在过滤到你想要的电影之前，它必须计算数据库中每部电影的GROUP_CONCAT(ambienceName)。结构化查询会更好：

SELECT f.*
FROM Films f
INNER JOIN Films_Ambiences fa1 ON f.id = fa1.film_id
INNER JOIN Ambiences a1 ON a1.id = f1.ambience_id
INNER JOIN Films_Ambiences fa2 ON f.id = fa2.film_id
INNER JOIN Ambiences a2 ON a2.id = f2.ambience_id
INNER JOIN Films_Ambiences fa3 ON f.id = fa3.film_id
INNER JOIN Ambiences a3 ON a3.id = f3.ambience_id
...
WHERE a1.ambienceName = '$_POST[ambience][1]'
  AND a2.ambienceName = '$_POST[ambience][2]'
  AND a3.ambienceName = '$_POST[ambience][3]'
  ...

您可以使用类似于上面的循环或Josejulio的答案来构建此查询。

Answer 3

使用关键字搜索谓词LIKE '%pattern%'是sure way to cause poor performance，因为它强制进行表扫描。

执行relational division查询的最佳方法是仅匹配所有三个条件匹配的电影，即为每个条件查找单独的行，然后将它们连接在一起。

SELECT f.*, CONCAT_WS(' ', a1.ambienceName, a2.ambienceName, a3.ambienceName) AS ambiences
FROM Films AS f 
INNER JOIN Films_Ambiences as fa1 ON f.id = fa1.film_id           
INNER JOIN Ambiences AS a1 ON a1.id = fa1.ambience_id
INNER JOIN Films_Ambiences as fa2 ON f.id = fa2.film_id           
INNER JOIN Ambiences AS a2 ON a2.id = fa2.ambience_id
INNER JOIN Films_Ambiences as fa3 ON f.id = fa3.film_id           
INNER JOIN Ambiences AS a3 ON a3.id = fa3.ambience_id
WHERE (a1.ambienceName, a2.ambienceName, a3.ambienceName) = (?, ?, ?);

对于每个搜索字词，您还需要额外的JOIN到Films_Ambiences和Ambiences 。

你应该在ambienceName上有一个索引，然后所有三个查找都会更有效率。

ALTER TABLE Ambiences ADD KEY (ambienceName);

我在最近的一次演讲中比较了关系划分的不同解决方案：

幻灯片：http://www.slideshare.net/billkarwin/sql-query-patterns-optimized

网络研讨会录音：http://www.percona.com/webinars/mysql-query-patterns-optimized

重新评论：


是否有办法更改此查询，以便在找到条件后显示其余的氛围？

是的，但你必须再加入一段时间才能获得电影的全部氛围：

SELECT f.*, GROUP_CONCAT(a_all.ambienceName) AS ambiences FROM Films AS f INNER JOIN Films_Ambiences as fa1 ON f.id = fa1.film_id INNER JOIN Ambiences AS a1 ON a1.id = fa1.ambience_id INNER JOIN Films_Ambiences as fa2 ON f.id = fa2.film_id INNER JOIN Ambiences AS a2 ON a2.id = fa2.ambience_id INNER JOIN Films_Ambiences as fa3 ON f.id = fa3.film_id INNER JOIN Ambiences AS a3 ON a3.id = fa3.ambience_id INNER JOIN Films_Ambiences AS fa_all ON f.id = fa_all.film_id INNER JOIN Ambiences AS a_all ON a_all.id = fa_all.ambience_id WHERE (a1.ambienceName, a2.ambienceName, a3.ambienceName) = (?, ?, ?) GROUP BY f.id;


有没有办法改变这个查询，以便结果只是具有所需氛围但不再需要的影片？

上面的查询应该这样做。


我认为，查询的作用是寻找包含特定氛围的电影（因此它也会找到更具氛围的电影）。

是的，除非匹配搜索条件中的所有三个氛围，否则查询与电影不匹配。但是这部电影可能还有其他超出搜索条件的氛围，所有电影的氛围（搜索条件加上其他内容）都被收集为GROUP_CONCAT(a_all.ambienceName)。

我测试了这个例子：

mysql> INSERT INTO Ambiences (ambienceName) VALUES ('funny'), ('scary'), ('1950s'), ('London'), ('bank'), ('crime'), ('stupid'); mysql> INSERT INTO Films (title) VALUES ('Mary Poppins'), ('Heist'), ('Scary Movie'), ('Godzilla'), ('Signs'); mysql> INSERT INTO Films_Ambiences VALUES (1,1),(1,2),(1,4),(1,5), (2,1),(2,2),(2,5),(2,6), (3,1),(3,2),(3,7), (4,2),(4,3), (5,2),(5,7); mysql> SELECT f.*, GROUP_CONCAT(a_all.ambienceName) AS ambiences FROM Films AS f INNER JOIN Films_Ambiences as fa1 ON f.id = fa1.film_id INNER JOIN Ambiences AS a1 ON a1.id = fa1.ambience_id INNER JOIN Films_Ambiences as fa2 ON f.id = fa2.film_id INNER JOIN Ambiences AS a2 ON a2.id = fa2.ambience_id INNER JOIN Films_Ambiences as fa3 ON f.id = fa3.film_id INNER JOIN Ambiences AS a3 ON a3.id = fa3.ambience_id INNER JOIN Films_Ambiences AS fa_all ON f.id = fa_all.film_id INNER JOIN Ambiences AS a_all ON a_all.id = fa_all.ambience_id WHERE (a1.ambienceName, a2.ambienceName, a3.ambienceName) = ('funny','scary','bank') GROUP BY f.id; +----+--------------+-------------------------+ | id | Title | ambiences | +----+--------------+-------------------------+ | 1 | Mary Poppins | funny,scary,London,bank | | 2 | Heist | funny,scary,bank,crime | +----+--------------+-------------------------+

顺便说一句，这里是EXPLAIN显示索引的用法：

+----+-------------+--------+--------+----------------------+--------------+---------+-----------------------------+------+-----------------------------------------------------------+ | id | select_type | table | type | possible_keys | key | key_len | ref | rows | Extra | +----+-------------+--------+--------+----------------------+--------------+---------+-----------------------------+------+-----------------------------------------------------------+ | 1 | SIMPLE | a1 | ref | PRIMARY,ambienceName | ambienceName | 258 | const | 1 | Using where; Using index; Using temporary; Using filesort | | 1 | SIMPLE | a2 | ref | PRIMARY,ambienceName | ambienceName | 258 | const | 1 | Using where; Using index | | 1 | SIMPLE | a3 | ref | PRIMARY,ambienceName | ambienceName | 258 | const | 1 | Using where; Using index | | 1 | SIMPLE | fa1 | ref | PRIMARY,ambience_id | ambience_id | 4 | test.a1.id | 1 | Using index | | 1 | SIMPLE | f | eq_ref | PRIMARY | PRIMARY | 4 | test.fa1.film_id | 1 | NULL | | 1 | SIMPLE | fa2 | eq_ref | PRIMARY,ambience_id | PRIMARY | 8 | test.fa1.film_id,test.a2.id | 1 | Using index | | 1 | SIMPLE | fa3 | eq_ref | PRIMARY,ambience_id | PRIMARY | 8 | test.fa1.film_id,test.a3.id | 1 | Using index | | 1 | SIMPLE | fa_all | ref | PRIMARY,ambience_id | PRIMARY | 4 | test.fa1.film_id | 1 | Using index | | 1 | SIMPLE | a_all | eq_ref | PRIMARY | PRIMARY | 4 | test.fa_all.ambience_id | 1 | NULL | +----+-------------+--------+--------+----------------------+--------------+---------+-----------------------------+------+-----------------------------------------------------------+


我有一部可怕，有趣，愚蠢的电影1。当我搜索一部只有可怕的电影时，无论如何我都会得到电影1。如果我不想要那个怎么办？

哦，好吧，我完全不明白这就是你的意思，在这些类型的问题上这是一个不寻常的要求。

这是一个解决方案：

mysql> SELECT f.*, GROUP_CONCAT(a_all.ambienceName) AS ambiences FROM Films AS f INNER JOIN Films_Ambiences as fa1 ON f.id = fa1.film_id INNER JOIN Ambiences AS a1 ON a1.id = fa1.ambience_id INNER JOIN Films_Ambiences as fa2 ON f.id = fa2.film_id INNER JOIN Ambiences AS a2 ON a2.id = fa2.ambience_id INNER JOIN Films_Ambiences AS fa_all ON f.id = fa_all.film_id WHERE (a1.ambienceName, a2.ambienceName) = ('scary','stupid') GROUP BY f.id HAVING COUNT(*) = 2 +----+-------+--------------+ | id | Title | ambiences | +----+-------+--------------+ | 5 | Signs | scary,stupid | +----+-------+--------------+

在这种情况下没有必要加入a_all，因为我们不需要环境名称列表，我们只需要环境数量，我们只需加入{{1}即可获得}。

使用可变数量的输入在MySQL数据库中执行简单搜索

3 个答案: