表演在MySQL数据库与输入

Question

的变化量一个简单的搜索我有很多对多分贝这三张表，制片，环境音效，Films_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); 

我使用从表单信息搜索特定的电影（例如，同时有趣和可怕的电影）。表格只是给定名称旁边的“蜱”。信息由$ _POST发送。

问题是我不知道会有多少要求。我知道用户可以选择的最大号码，但我不知道他们会选择多少号码或哪些号码（我可以通过检查isset($_POST['somethin'])来做到这一点，但如果我有20个不同的选项，它会非常单调。我不能做这样事情：

$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能做到这一点与SQL或者更确切地说，PHP

这里有一个SQLFiddle？如果你喜欢。

Answer 1

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

执行relational division查询（即仅匹配所有三个条件匹配的影片）的最佳方法是为每个条件查找单独的行，然后将它们联接在一起。

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) = (?, ?, ?); 

你需要一个额外的JOIN到Films_Ambiences和Ambiences每个搜索词。

您应该有一个ambienceName索引，然后所有三个查找将更有效率。

ALTER TABLE Ambiences ADD KEY (ambienceName); 

我相比，在最近的一次演讲的关系划分不同的解决方案：

• 幻灯片：http://www.slideshare.net/billkarwin/sql-query-patterns-optimized
• 网络研讨会记录：http://www.percona.com/webinars/mysql-query-patterns-optimized

---

回复您的评论：

有没有办法改变这个查询，以便在找到标准后显示其余的氛围？

是的，但你必须加入一个更多的时间来为电影全套的氛围中：

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，从而使结果只有没有更多的氛围需要的电影？

上面的查询应该这样做。

---

查询做什么，我想，是寻找影片，其中包括给定的氛围（所以它也发现，有更多的氛围电影）。

对，查询与电影不匹配，除非它匹配搜索条件中的所有三种气氛。但是这部电影可能有其他的环境，除了搜索标准之外，所有电影的氛围（搜索标准加上其他环境）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，这里的指数的EXPLAIN显示用法：

+----+-------------+--------+--------+----------------------+--------------+---------+-----------------------------+------+-----------------------------------------------------------+ 
| 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              | 
+----+-------------+--------+--------+----------------------+--------------+---------+-----------------------------+------+-----------------------------------------------------------+ 

---

我有一个FILM1这是可怕的，搞笑的，笨。当我搜索一部只有可怕的电影时，我会无论如何都会拍电影。如果我不想要那个怎么办？

噢，好吧，我完全不明白这就是你的意思，这是对这些类型的问题的一个不寻常的要求。

这里有一个解决方案：

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 | 
+----+-------+--------------+ 

有没有必要加入到a_all在这种情况下，因为我们不需要氛围名称的列表，我们只需要氛围的数量，我们可以得到只需加入fa_all即可。

Answer 2

我认为你可以使用PHP来建立你的SQL，这样的事情：

$ambienceWhere = '1=1 '; 
for ($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 3

表单应具备的氛围中元素的数组，或许是多选。然后这将变成一个PHP数组，$_POST['ambience'][]。那么你可以写：

$ambience_query = implode(' AND ', array_map(function($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。它将必须计算数据库中每部影片的GROUP_CONCAT(ambienceName)，然后再筛选出您想要的影片。这将是更好的结构是怎样的查询：

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]' 
    ... 

可以使用类似上面或Josejulio的回答循环来构建这个查询。