Question

我找到一种基于其标记在数据库中选择行的有效方法并且还返回与该行关联的所有其他标记，我遇到了一些麻烦。当我使用不返回行的所有标记的查询时，大约需要0.001秒。我的初始方案更加规范化，并且有另一个表格用于标签的标签，但它最终花费了几秒钟来完成一个查询，所以我结束了删除该表并使其更少规范化，但即使这个解决方案看起来也很慢。 / p>

SELECT c.*
FROM collections c,
     tags t
WHERE t.collection_id=c.id
  AND (t.name IN ("foo",
                  "bar"))
GROUP BY c.id HAVING COUNT(t.id)=2 LIMIT 10

现在我没有想出一个有效的方法来获取该元素的所有其他标签而不会变慢。我目前的解决方案大约慢了10倍，需要0.01秒才能完成，我也觉得它不能很好地扩展（我发现它非常难看）。

SELECT c.*,
       GROUP_CONCAT(t1.name) AS tags
FROM collections c,
     tags t,
     tags t1
WHERE t1.collection_id = c.id
  AND t.collection_id=c.id
  AND (t.name IN ("foo",
                  "bar"))
GROUP BY c.id HAVING COUNT(t.id)=2 LIMIT 10

实际上是否有效率或至少更有效的方法来实现这一目标？非常感谢任何关于这个的建议或提示！

Answer 1

行。请考虑以下内容......

DROP TABLE IF EXISTS ingredients;

CREATE TABLE ingredients 
(ingredient_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,ingredient VARCHAR(30) NOT NULL UNIQUE
);

INSERT INTO ingredients (ingredient_id, ingredient) VALUES
(1, 'Macaroni'),
(2, 'Cheese'),
(3, 'Beans'),
(4, 'Toast'),
(5, 'Jam'),
(6, 'Jacket Potato'),
(7, 'Peanut Butter');


DROP TABLE IF EXISTS recipes;

CREATE TABLE recipes 
(recipe_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY
,recipe VARCHAR(50) NOT NULL UNIQUE
);

INSERT INTO recipes (recipe_id, recipe) VALUES
(1, 'Macaroni & Cheese'),
(2, 'Cheese on Toast'),
(3, 'Beans on Toast'),
(4, 'Cheese & Beans on Toast'),
(5, 'Toast & Jam'),
(6, 'Beans & Macaroni'),
(9, 'Beans on Jacket Potato'),
(10, 'Cheese & Beans on Jacket Potato'),
(12, 'Peanut Butter on Toast');

DROP TABLE IF EXISTS recipe_ingredient;

CREATE TABLE recipe_ingredient 
(recipe_id INT NOT NULL
,ingredient_id INT NOT NULL
,PRIMARY KEY (recipe_id,ingredient_id)
);

INSERT INTO recipe_ingredient (recipe_id, ingredient_id) VALUES
(1, 1),
(1, 2),
(2, 2),
(2, 4),
(3, 3),
(3, 4),
(4, 2),
(4, 3),
(4, 4),
(5, 4),
(5, 5),
(6, 1),
(6, 3),
(9, 3),
(9, 6),
(10, 2),
(10, 3),
(10, 6),
(12, 4),
(12, 7);

SELECT r.*
      , GROUP_CONCAT(CASE WHEN i.ingredient IN ('Cheese','Beans') THEN i.ingredient END) i
      , GROUP_CONCAT(CASE WHEN i.ingredient NOT IN('Cheese','Beans') THEN i.ingredient END) o 
   FROM recipes r 
   LEFT 
   JOIN recipe_ingredient ri 
     ON ri.recipe_id = r.recipe_id 
   LEFT 
   JOIN ingredients i 
     ON i.ingredient_id = ri.ingredient_id 
  GROUP 
     BY recipe_id;

+-----------+---------------------------------+--------------+---------------------+
| recipe_id | recipe                          | i            | o                   |
+-----------+---------------------------------+--------------+---------------------+
|         1 | Macaroni & Cheese               | Cheese       | Macaroni            |
|         2 | Cheese on Toast                 | Cheese       | Toast               |
|         3 | Beans on Toast                  | Beans        | Toast               |
|         4 | Cheese & Beans on Toast         | Cheese,Beans | Toast               |
|         5 | Toast & Jam                     | NULL         | Toast,Jam           |
|         6 | Beans & Macaroni                | Beans        | Macaroni            |
|         9 | Beans on Jacket Potato          | Beans        | Jacket Potato       |
|        10 | Cheese & Beans on Jacket Potato | Cheese,Beans | Jacket Potato       |
|        12 | Peanut Butter on Toast          | NULL         | Toast,Peanut Butter |
+-----------+---------------------------------+--------------+---------------------+

小提琴：http://www.sqlfiddle.com/#!2/45aa0/1

Answer 2

使用显式连接语法（不应该对性能有所不同，因为MySQL应该设法优化它）

SELECT c.*,
       GROUP_CONCAT(t1.name) AS tags
FROM collections c
INNER JOIN tags t ON t.collection_id = c.id
INNER JOIN tags t1 ON t1.collection_id = c.id
WHERE t.name IN ("foo", "bar")
GROUP BY c.id 
HAVING COUNT(t.id) = 2 
LIMIT 10

可能值得为您正在检查的每个标签单独进行INNER JOIN，这样就无需使用HAVING： -

SELECT c.*,
       GROUP_CONCAT(t1.name) AS tags
FROM collections c
INNER JOIN tags t ON t.collection_id = c.id AND t.name = "foo"
INNER JOIN tags t0 ON t.collection_id = c.id AND t0.name = "bar"
INNER JOIN tags t1 ON t1.collection_id = c.id
GROUP BY c.id 
LIMIT 10

然而，您的原始查询看起来并不糟糕，因此可能是索引问题。

高效的MySQL多对多标签查询

2 个答案: