mysql 通过查询从 order by 中选择平均值

Posted 2023-04-15

【中文标题】mysql 通过查询从 order by 中选择平均值

【英文标题】：mysql select an average from order by with a query

【发布时间】：2015-06-19 10:25:18

【问题描述】：

我有 2 个表 "recipes"、"recipes_items" ，在表 recipes_items 我有一个整数列“重要”。现在我有一个查询，它提取由 recipes_items.important 的总和（重要）排序的食谱，该顺序有一个子查询：

(SELECT sum(...) from recipes_items...) DESC)

但我想要检索作为食谱中的附加列查询表 recipes_items 中“重要”列的平均值 怎么可能？

谢谢 摩西

select  recipes.recipeName,
        recipes.recipeID,
        AVG (recipes_items.importance) AS averageScore
    from  recipes,recipes_items
    WHERE  recipes_items.recipeID=recipes.recipeID
    ORDER BY  
      ( SELECT  SUM(recipes_items.importance) AS sumImportance
            FROM  recipes_items
            WHERE  recipes_items.recipeID = recipes.recipeID
              AND  recipes_items.itemID IN ('1059')
      ) DESC
    LIMIT  20

我想在结果中得到的是从 recipes_items.itemID 中的重要性列排序，这些列与“IN()”部分中的一组 ID 匹配，还希望在其中包含匹配 ID 的数量或总和与recipes_items.itemID 匹配的每一行的ID ...

【问题讨论】：

请提供样本记录和预期答案

您希望将“recipes_items.important”的平均值作为另一列？

是的，完全一样，但通过查询保持顺序相同

【参考方案1】：

您没有显示完整的查询，但您可以在没有子查询的情况下执行此操作，并且仍然包含平均值：

SELECT recipes.id, recipes.name, AVG(recipes_items.important)
FROM recipes
LEFT JOIN recipes_items ON (recipes.id = recipes_items.recipe_id)
GROUP BY recipes.id
ORDER BY SUM(recipes_items.important) DESC

【讨论】：

谢谢@Jim，完整的查询是：select recipes.recipeName, recipes.recipeID,AVG (recipes_items.importance) AS averageScore from recipes,recipes_items WHERE recipes_items.recipeID=recipes.recipeID ORDER BY (SELECT SUM(recipes_items.importance) AS sumImportance FROM recipes_items WHERE recipes_items.recipeID = recipes.recipeID AND recipes_items.itemID IN ('1059')) DESC LIMIT 20

我想在结果中得到的是从 recipes_items.itemID 中的重要性列排序，这些列与“IN()”部分中的一组 ID 匹配，也希望有数字或总和recipes_items.itemID..

的匹配 ID 的每一行中的匹配 ID