mysql计算列表中的不同元素答案

【问题标题】：mysql count distinct elements which you have in a listmysql计算列表中的不同元素
【发布时间】：2018-06-21 07:22:10
【问题描述】：

在recipie_ingredient 表中，我有不同的成分。

create table recipe_ingredient (
    id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
    rel_recipe INT(6),
    rel_ingredient INT(6)
  );

INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 32);
INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 99);
INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 123);
INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 123);
INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 227);
INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 395);
INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 403);
INSERT INTO recipe_ingredient (rel_recipe, rel_ingredient) VALUES(1, 403);

根据我的成分，我想获得包含我的成分的食谱。我想在食谱中加入一些独特的成分。在DB Fiddle我创建了一个表并插入演示数据，我还添加了我当前无法正常工作的SQL。

SELECT
  COUNT(distinct(ri.rel_ingredient)) as allIngredient,
  sum((ri.rel_ingredient) in (123,403)) have
FROM recipe_ingredient ri
GROUP BY ri.rel_recipe;

最终结果应该是 allIngredient: 6 和 have: 2. (DB Fiddle link)

【问题讨论】：

标签： mysql group-by count

【解决方案1】：

在您的数据中，您有 2 次成分 123 和 403，如果您只想计算一次，您应该使用该查询

SELECT
  COUNT(distinct(ri.rel_ingredient)) as allIngredient,
  COUNT(distinct(ri.rel_ingredient in (123,403))) have
FROM recipe_ingredient ri
GROUP BY ri.rel_recipe

【讨论】：

不，这也不起作用，检查db-fiddle.com/f/XhZs8yAyvRkubyYCf1GiY/6
@Gasper 此查询工作正常，它返回您在帖子中写的结果 allIngredient: 6 and have: 2
是的，如果您尝试使用 123、403 两种成分，结果还可以，但如果您只尝试一种，您也会得到 2，这是错误的，它应该是 1...

【解决方案2】：

我准备了一个解决方案，但有人可以检查这是否是最快的解决方案。检查 DB Fiddle here。

 SELECT r.id, r.name, 
    count(temp.rel_ingredients) as allIngredients,
    sum(temp.rel_ingredients in (123,403)) as have,
    (sum(temp.rel_ingredients in (123,403)) / count(temp.rel_ingredients)) as missing_ratio
 FROM (
  SELECT
    distinct(ri.rel_ingredient) as rel_ingredients,
    ri.rel_recipe
  FROM recipe_ingredient ri
) AS temp
LEFT JOIN recipe r ON r.id = temp.rel_recipe
GROUP BY temp.rel_recipe
HAVING have != 0
ORDER BY missing_ratio DESC, allIngredients DESC
LIMIT 0, 10;

【讨论】：