給定SELECT語句，我想要多個子查詢的結果RECIPE_ID嗎？

Question

我有ingredient 。 ingredientId是709710711這個號我把配料子查詢我和AS match_percentage我是萊特(3*100 / count計數的ingredient 。 ingredientId ，所有來自我的PHP代碼在這里我給這個例子我的解決方案。

現在導致像那些RECIPE_ID我是通過三項ingredient 。 ingredientId是匹配與此表recipe_ingredient 。 ingredientId並保持recipe_ingredient 。 recipeId

我想要RECIPE_ID | 結果為1和3。 只有WHO擁有三個ID 709,710,711，甚至沒有兩個709,710。

這是我的RECIPE_ID查詢：

SELECT 
    `recipe`.`recipeId` AS recipe_id ,
    (select count(`recipe_ingredient`.ingredientId) from `recipe_ingredient` where `recipe`.recipeId = `recipe_ingredient`.recipeId) as ingredientCount ,  
    IF(
        (select (3*100 / count(DISTINCT `recipe_ingredient`.ingredientId)) from `recipe_ingredient` where `recipe_ingredient`.recipeId = `recipe`.recipeId)>99                      
        ,100
        ,round( (select (3*100 / count(DISTINCT `recipe_ingredient`.ingredientId)) from `recipe_ingredient` where `recipe_ingredient`.recipeId = `recipe`.recipeId) ) 
    ) 
    as match_percentage , 
    GROUP_CONCAT(
        DISTINCT `recipe_ingredient`.ingredientId 
        ORDER BY `recipe_ingredient`.ingredientId ASC 
           ) as recipeIngredients 
from `recipe` 
left join `recipe_ingredient` on `recipe_ingredient`.recipeId = `recipe`.recipeId
left join `ingredient` on `ingredient`.ingredientId = `recipe_ingredient`.ingredientId   
where  `recipe`.`recipeId` IN( 
                SELECT 
                `recipe_ingredient`.`recipeId`
                FROM `recipe_ingredient` 
                WHERE `recipe_ingredient`.`ingredientId` 
                IN(
                  SELECT `ingredient`.`ingredientId` AS linkIng 
                  FROM `ingredient` 
                  WHERE `ingredient`.`ingredientId` IN(709,710,711) or `ingredient`.`linkIngredientPerent` IN(709,710,711)
                )
                GROUP BY `recipe_ingredient`.`recipeId` 
                ORDER BY `recipe_ingredient`.`recipeId` ASC
              )  

and (select (3*100 / count(DISTINCT `recipe_ingredient`.ingredientId)) from `recipe_ingredient` where `recipe_ingredient`.recipeId = `recipe`.recipeId) > 24  
group by `recipe`.recipeId 

我的查詢鏈接： http : //sqlfiddle.com/#!2/f4983/2

Answer 1

這是你想要的嗎？ 請在其他情況下進行檢查。

http://sqlfiddle.com/#!2/f4983/8