select是否另一个select返回行

我试图做一个select，使烹饪食谱select你有的项目。

我有一个名为ingredientsOwn的表，结构如下：

idType (int) amount (int) 

另一个表名为这个结构的食谱 ：

 idRecipe (int) name (varchar) 

另一个表名为recipeIngredients

 idRecipe (int) idType (int) amount (int) 

我想展示你可以用你有的元素做的食谱，我怎么能做到这一点？

我试图只在一个查询中实现它，因为我真的不知道如何去抛出和arrays的节点js。

谢谢

Answer 1

我想解决这个问题的方法是，尝试为每个配方计算你需要的成分数量，然后join成分数量，如果这两个数字相符，就有一个候选配方。

所以，为了获得配方需要的配料数量，你必须做类似的事情（这更像是一个sql服务器的语法，所以请尝试着重于概念，而不是语法）：

 select idRecipe, count(*) as neededIngredientsCount from recipeIngredients group by idRecipe 

为了获得每种食谱的可用配料的数量，您必须join配料成分，以便能够知道每种配方有多less配料。

 select ingredientsOwn.idRecipe, count(*) as matchingIngredientsCount from ingredientsOwn inner join recipeIngredients on ingredientsOwn.idType = recipeIngredients.idType where ingredientsOwn.amount >= recipeIngredients.amount group by ingredientsOwn.idRecipe 

现在你join之前的两个查询来获得你有足够的配料idRecieps，并join他们的食谱表来获得配方名称。

 select r.idRecipe, r.name from ((select idRecipe, count(*) as neededIngredientsCount from recipeIngredients group by idRecipe) as in inner join (select ingredientsOwn.idRecipe, count(*) as matchingIngredientsCount from ingredientsOwn inner join recipeIngredients on ingredientsOwn.idType = recipeIngredients.idType where ingredientsOwn.amount >= recipeIngredients.amount group by ingredientsOwn.idRecipe) as io on in.idRecipe = io.idRecipe and in.neededIngredientsCount = io.matchingIngredientsCount inner join (select * from recipes) as r on r.idRecipe = in.idRecipe) 

希望这有助于，并抱歉无法提供有效的MySQL语法。

Answer 2

 SELECT * FROM recipes INNER JOIN ( select idRecipe from recipeIngredients WHERE recipeIngredients.idType IN ( SELECT ingredientsOwn.idType from ingredientsOwn ) ) as a ON a.idRecipe = recipes.idRecipe