Question

我真的是SQL新手-从1周开始，我的查询仅适用于1条记录

ast = pd.read_csv(directory,encoding = 'UTF-8')
inte = pd.read_csv(directory,encoding = 'UTF-8')
ast.reset_index(inplace= True)
ast.rename(columns = {'index': 'FID'},inplace = True)
final = pd.merge(ast, inte, on = ['FID'], how = 'inner')
final

此查询仅适用于一条记录，但是如果我想获取所有餐点（不带“ and”功能），我仍然只获得一条记录，但是sumpricePrice是添加的所有记录的总和，如何获取此查询一顿饭吗？

我喜欢每顿饭一个总价的记录

我的桌子：

Select m.meal_id, m.name, m.usp, m.description, m.worktime, m.proprietor, m.img, m.url, m.servings, d.name, c.name, s.name, sum(i.price / i.minamount * r.quantity) as mealPrice

from meals m
join recipe r on m.meal_id = r.meal_id
join ingredients i on i.ingredient_id = r.ingredient_id
join difficulty d on d.difficulty_id = m.difficulty_id
join cat c on c.cat_id = m.cat_id
join social s on s.social_id = m.social_id
and m.meal_id = 1

Answer 1

编辑：基于OP的recent edit；您只需做一个GROUP BY。

尝试：

Select m.meal_id, 
       m.name, 
       m.usp, 
       m.description, 
       m.worktime, 
       m.proprietor, 
       m.img, 
       m.url, 
       m.servings, 
       d.name, 
       c.name, 
       s.name, 
       SUM(i.price / i.minamount * r.quantity) as mealPrice
from meals m
join recipe r on m.meal_id = r.meal_id
join ingredients i on i.ingredient_id = r.ingredient_id
join difficulty d on d.difficulty_id = m.difficulty_id
join cat c on c.cat_id = m.cat_id
join social s on s.social_id = m.social_id 
GROUP BY m.meal_id, 
         m.name, 
         m.usp, 
         m.description, 
         m.worktime, 
         m.proprietor, 
         m.img, 
         m.url, 
         m.servings, 
         d.name, 
         c.name, 
         s.name

基于OP问题的Previous Version：

这是一个Window function问题。在较新版本的MySQL（> = 8.0 ）中，您可以轻松地做到这一点。在较旧的版本（您的版本为5.5）中，我们可以使用Session Variables来解决它。

尝试：

Select m.meal_id, 
       m.name, 
       m.usp, 
       m.description, 
       m.worktime, 
       m.proprietor, 
       m.img, 
       m.url, 
       m.servings, 
       d.name, 
       c.name, 
       s.name, 
       (@sum := @sum + (i.price / i.minamount * r.quantity)) as mealPrice
from meals m
join recipe r on m.meal_id = r.meal_id
join ingredients i on i.ingredient_id = r.ingredient_id
join difficulty d on d.difficulty_id = m.difficulty_id
join cat c on c.cat_id = m.cat_id
join social s on s.social_id = m.social_id 
CROSS JOIN (select @sum := 0) AS init

在MySQL版本> = 8.0中，我们可以在完整结果集的分区上将SUM()用作Window函数。在这种情况下，查询将是：

Select m.meal_id, 
       m.name, 
       m.usp, 
       m.description, 
       m.worktime, 
       m.proprietor, 
       m.img, 
       m.url, 
       m.servings, 
       d.name, 
       c.name, 
       s.name, 
       SUM(i.price / i.minamount * r.quantity) OVER() as mealPrice
from meals m
join recipe r on m.meal_id = r.meal_id
join ingredients i on i.ingredient_id = r.ingredient_id
join difficulty d on d.difficulty_id = m.difficulty_id
join cat c on c.cat_id = m.cat_id
join social s on s.social_id = m.social_id

Answer 2

您需要在下面的查询中，交叉联接有助于您获得所需的结果

Select m.meal_id, 
       m.name, 
       m.usp, 
       m.description, 
       m.worktime, 
       m.proprietor, 
       m.img, 
       m.url, 
       m.servings, 
       d.name, 
       c.name, 
       s.name, 
       (@s := @s + (i.price / i.minamount * r.quantity)) as mealPrice

from meals m
join recipe r on m.meal_id = r.meal_id
join ingredients i on i.ingredient_id = r.ingredient_id
join difficulty d on d.difficulty_id = m.difficulty_id
join cat c on c.cat_id = m.cat_id
join social s on s.social_id = m.social_id 
CROSS JOIN ( select @s:= 0) AS t

sql查询仅适用于1条记录

2 个答案: