如何 select 包含 ManyToMany 关系中某些子项的父项?
How to select parent that contains certain children from his ManyToMany relation?
给定这段代码 (Python & TortoiseORM):
class Recipe(Model):
description = fields.CharField(max_length=1024)
ingredients = fields.ManyToManyField(model_name="models.Ingredient", on_delete=fields.SET_NULL)
class Ingredient(Model):
name = fields.CharField(max_length=128)
如何查询同时包含 Ingredient.name="tomato" 和 Ingredient.name="onion" 的所有食谱?
我相信在 Django-ORM 中,可以使用 & 运算符或相交方法对 QuerySet 进行一些交集。
更新 #1
这个查询有效,但在我看来有点混乱,当我想要 f.e 时会出现问题。查询包含超过 2 种成分的所有食谱。
subquery = Subquery(Recipe.filter(ingredients__name="onion").values("id"))
await Recipe.filter(pk__in=subquery , ingredients__name="tomato")
更新 #2
SELECT "recipe"."description",
"recipe"."id"
FROM "recipe"
LEFT OUTER JOIN "recipe_ingredient"
ON "recipe"."id" = "recipe_ingredient"."recipe_id"
LEFT OUTER JOIN "ingredient"
ON "recipe_ingredient"."ingredient_id" = "ingredient"."id"
WHERE "ingredient"."name" = 'tomato'
AND "ingredient"."name" = 'onion'
您可以过滤:
Recipe.objects.filter(
<strong>ingredients__name='tomato'</strong>
).filter(
<strong>ingredients__name='onion'</strong>
)
通过使用两次 .filter(…)
[Django-doc] 调用,我们进行 两次 LEFT OUTER JOIN
s,一次搜索 tomato
,一次搜索 onion
。这显然只适用于 Django ORM,不适用于 Tortoise ORM。
如果我们使用print(qs.query)
(构造查询),我们得到:
SELECT recipe.id, recipe.description
FROM recipe
INNER JOIN recipe_ingredients ON recipe.id = recipe_ingredients.recipe_id
INNER JOIN ingredient ON recipe_ingredients.ingredient_id = ingredient.id
INNER JOIN recipe_ingredients T4 ON recipe.id = T4.recipe_id
INNER JOIN ingredient T5 ON T4.ingredient_id = T5.id
WHERE ingredient.name = tomato
AND T5.name = onion
另一种选择是制作一个单LEFT OUTER JOIN
,并检查项目数是否与项目数匹配,所以:
from django.db.models import Count
items = {'tomato', 'onion'}
Recipe.objects.filter(
<strong>ingredients__name__in=items</strong>
).alias(
<strong>ncount=Count('ingredients')</strong>
).filter(<strong>ncount=len(items)</strong>)
或在 django-3.2 之前:
from django.db.models import Count
items = {'tomato', 'onion'}
Recipe.objects.filter(
ingredients__name__in=items
).<strong>annotate(</strong>
ncount=Count('ingredients')
<strong>)</strong>.filter(ncount=len(items))
因此,这提供了一个如下所示的查询:
SELECT recipe.id, recipe.description
FROM recipe
INNER JOIN recipe_ingredients ON recipe.id = recipe_ingredients.recipe_id
INNER JOIN ingredient ON recipe_ingredients.ingredient_id = ingredient.id
WHERE ingredient.name IN (onion, tomato)
GROUP BY recipe.id
HAVING <strong>COUNT(recipe_ingredients.ingredient_id) = 2</strong>
特别是 HAVING COUNT(recipe_ingredients.ingredient_id)
是这里的关键,因为 WHERE
子句已经将其过滤为仅洋葱和西红柿。
这要求成分的 name
是唯一的(即没有两个 Ingredient
记录具有相同的名称)。您可以使用以下方法使 name
字段唯一:
class Ingredient(Model):
name = fields.CharField(<b>unique=True,</b> max_length=128)
给定这段代码 (Python & TortoiseORM):
class Recipe(Model):
description = fields.CharField(max_length=1024)
ingredients = fields.ManyToManyField(model_name="models.Ingredient", on_delete=fields.SET_NULL)
class Ingredient(Model):
name = fields.CharField(max_length=128)
如何查询同时包含 Ingredient.name="tomato" 和 Ingredient.name="onion" 的所有食谱? 我相信在 Django-ORM 中,可以使用 & 运算符或相交方法对 QuerySet 进行一些交集。
更新 #1
这个查询有效,但在我看来有点混乱,当我想要 f.e 时会出现问题。查询包含超过 2 种成分的所有食谱。
subquery = Subquery(Recipe.filter(ingredients__name="onion").values("id"))
await Recipe.filter(pk__in=subquery , ingredients__name="tomato")
更新 #2
SELECT "recipe"."description",
"recipe"."id"
FROM "recipe"
LEFT OUTER JOIN "recipe_ingredient"
ON "recipe"."id" = "recipe_ingredient"."recipe_id"
LEFT OUTER JOIN "ingredient"
ON "recipe_ingredient"."ingredient_id" = "ingredient"."id"
WHERE "ingredient"."name" = 'tomato'
AND "ingredient"."name" = 'onion'
您可以过滤:
Recipe.objects.filter(
<strong>ingredients__name='tomato'</strong>
).filter(
<strong>ingredients__name='onion'</strong>
)
通过使用两次 .filter(…)
[Django-doc] 调用,我们进行 两次 LEFT OUTER JOIN
s,一次搜索 tomato
,一次搜索 onion
。这显然只适用于 Django ORM,不适用于 Tortoise ORM。
如果我们使用print(qs.query)
(构造查询),我们得到:
SELECT recipe.id, recipe.description
FROM recipe
INNER JOIN recipe_ingredients ON recipe.id = recipe_ingredients.recipe_id
INNER JOIN ingredient ON recipe_ingredients.ingredient_id = ingredient.id
INNER JOIN recipe_ingredients T4 ON recipe.id = T4.recipe_id
INNER JOIN ingredient T5 ON T4.ingredient_id = T5.id
WHERE ingredient.name = tomato
AND T5.name = onion
另一种选择是制作一个单LEFT OUTER JOIN
,并检查项目数是否与项目数匹配,所以:
from django.db.models import Count
items = {'tomato', 'onion'}
Recipe.objects.filter(
<strong>ingredients__name__in=items</strong>
).alias(
<strong>ncount=Count('ingredients')</strong>
).filter(<strong>ncount=len(items)</strong>)
或在 django-3.2 之前:
from django.db.models import Count
items = {'tomato', 'onion'}
Recipe.objects.filter(
ingredients__name__in=items
).<strong>annotate(</strong>
ncount=Count('ingredients')
<strong>)</strong>.filter(ncount=len(items))
因此,这提供了一个如下所示的查询:
SELECT recipe.id, recipe.description
FROM recipe
INNER JOIN recipe_ingredients ON recipe.id = recipe_ingredients.recipe_id
INNER JOIN ingredient ON recipe_ingredients.ingredient_id = ingredient.id
WHERE ingredient.name IN (onion, tomato)
GROUP BY recipe.id
HAVING <strong>COUNT(recipe_ingredients.ingredient_id) = 2</strong>
特别是 HAVING COUNT(recipe_ingredients.ingredient_id)
是这里的关键,因为 WHERE
子句已经将其过滤为仅洋葱和西红柿。
这要求成分的 name
是唯一的(即没有两个 Ingredient
记录具有相同的名称)。您可以使用以下方法使 name
字段唯一:
class Ingredient(Model):
name = fields.CharField(<b>unique=True,</b> max_length=128)